Polymers Class 12 Important Extra Questions Chemistry Chapter 15

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 15 Polymers. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 15 Important Extra Questions Polymers

Polymers Important Extra Questions Very Short Answer Type

Question 1.
Define the term ‘homopolymerisation’ giving an example. (CBSE Delhi 2012)
Answer:
The polymers formed from one type of monomers and having same repeating units are called homopolymers. For example polyvinyl chloride (PVC), polythene.

Question 2.
Give one example of condensation polymer. (CBSE 2013)
Answer:
Nylon-6, 6

Question 3.
Which of the following is a natural polymer? (CBSE 2014)
Buna-S, Proteins, PVC
Answer:
Proteins.

Question 4.
Based on molecular forces what type of polymer is neoprene? (CBSE 2014)
Answer:
Elastomer.

Question 5.
Which of the following is a fibre?
Nylon, Neoprene, PVC (CBSE 2014)
Answer:
Nylon

Question 6.
Draw structure for the polymer used for the manufacture of non-stick utensils. (CBSE2019C)
Answer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 1
Question 7.
What is the primary structural feature necessary for a molecule to make it useful in a condensation polymerisation reaction? (CBSE AI 2010)
Answer:
The monomers must be bifunctional i.e. contain two functional groups.

Question 8.
Arrange the following polymers in the increasing order of tensile strength. Nylon 6, Buna-S, Polythene. (CBSE Sample Paper 2010)
Answer:
Buna – S < Polythene < Nylon 6.

Question 9.
What type of reaction occurs in the formation of Nylon 6,6 polymer? (CBSE Sample Paper 2019)
Answer:
Condensation

Question 10.
Class 12 Chemistry Important Questions Chapter 15 Polymers 2
Is a homopolymer or R copolymer? (CBSE 2019C)
Answer:
Homopolymer

Question 11.
Identify the Monomers of the following polymer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 3
Answer:
Buta-1,3-diene, styrene.
It is obtained by the polymerisation of buta-1, 3-diene and styrene in the ratio of 3: 1 in the presence of sodium.
Class 12 Chemistry Important Questions Chapter 15 Polymers 4

Polymers Important Extra Questions Short Answer Type

Question 1.
Draw the molecular structures of the monomers of
(i) PVC
Answer:
PVC : CH2 = CH – Cl (Vinyl chloride)

(ii) Teflon (CBSE 2010)
Answer:
Teflon : F2C = CF2 (Tetrafluoroethene)

Question 2.
Mention two important uses of each of the following : (CBSE Delhi 2011)
(i) Bakelite
Answer:
Uses of bakelite:

  • Used for making combs, fountain pens, barrels, electrical switches and handles of utensils.
  • Soft bakelites are used for making glue for binding wooden laminated planks in varnishes.

(ii) Nylon 6
Answer:
Uses of Nylon 6:

  • It is used for making tyre cords.
  • It is used for making fabrics and ropes.

Question 3.
Draw the structure of the monomer for each of the following polymers:
(i) Nylon 6
Answer:
The monomer of Nylon-6:

(ii) Polypropene (CBSE Delhi 2012)
Answer:
The monomer of polypropene:
Class 12 Chemistry Important Questions Chapter 15 Polymers 5

Question 4.
Differentiate between thermoplastic and thermosetting polymers. Give one example of each. (CBSE Delhi 2010, 2012) Answer:
Thermoplastics when heated become soft and more or less fluid. These can be moulded into any desired shape. The thermoplastics ‘ can be processed again and again.

The intermolecular forces in thermoplastics are intermediate between those of elastomers and fibres.

On the other hand, thermosetting plastics on heating become hard and insoluble masses. These cannot be moulded into the desired shape and cannot be reprocessed. The intermolecular forces in thermosetting are strong and there are cross-links that hold the molecules in place so that heating does not allow them to move freely.

The common examples are:

  1. Thermoplastics: Polythene, polystyrene and polyvinyl chloride.
  2. Thermosetting: Bakelite and melamine formaldehyde.

Question 5.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Bakelite
Answer:
Phenol and formaldehyde

(ii) Neoprene
Answer:
2-Chloro-1,3-butadiene(chloroprene)

Question 6.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Terylene
Answer:
Ethylene glycol, terephthalic acid

(ii) Nylon-6, 6
Answer:
Hexamethylenediamine and adipic acid.

Question 7.
Write the name of monomers used for getting the following polymers: (CBSE 2014)
(i) Teflon
Answer:
Tetrafluoroethylene

(ii) Buna-N
Answer:
1, 3-butadiene and acrylonitrile

Polymers Important Extra Questions Long Answer Type

Question 1.
Write the structures of monomers used for getting the following polymers:
(i) Nylon-6,6
(ii) Glyptal
(iii) Buna-S
OR
(i) IsClass 12 Chemistry Important Questions Chapter 15 Polymers 6 it a homopolymer or copolymer? Give reason.
(ii) Write the monomers of the following polymer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 7
(iii) What is the role of the Sulphur in the vulcanization of rubber? (CBSE Delhi 2019)
Answer:
(i) Nylon-6,6
Hexamethylenediamine: NH2 – (CH2)6 – NH2 and Adipic acid
Class 12 Chemistry Important Questions Chapter 15 Polymers 8
(ii) Glyptal
Class 12 Chemistry Important Questions Chapter 15 Polymers 9
(iii) Buna – S
CH2 = CH — CH = CH2
1, 3 Butadiene
Class 12 Chemistry Important Questions Chapter 15 Polymers 10
OR
(i) It is a homopolymer because it has only one monomer.
Class 12 Chemistry Important Questions Chapter 15 Polymers 11
(iii) Sulphur makes the rubber hard, tough with greater tensile strength and non-sticky by forming sulphur crosslinks.

Question 2.
(a) Write the names of monomers of the following polymer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 12
Answer:
Ethylene glycol and terephthalic acid or Ethane-1,2- diol and Benzene 1, 4- dicarboxylic acid.

Terylene: It is a polymer of ethylene glycol (ethane-1,2-diol) and terephthalic acid (benzene-1, 4-dicarboxylic acid). It is obtained by heating a mixture of ethylene glycol and terephthalic acid at 420 to 460 K in the presence of zinc acetate-antimony trioxide, [Zn(OOCCH3)2 + Sb2O3] catalyst. It is known as terylene or dacron.
Class 12 Chemistry Important Questions Chapter 15 Polymers 13
(b) What does part 6,6 mean in the polymer Nylon-6,6?
Answer:
It represents 6 carbon atoms present in both the monomer units.

Nylon-6,6: The monomer units of nylon-6,6 are hexamethylenediamine and adipic acid. It is prepared by condensation of hexamethylene diamine with adipic acid under high pressure and high temperature.
Class 12 Chemistry Important Questions Chapter 15 Polymers 14
(c) Give an example of a Biodegradable polymer. (CBSE 2019C)
Answer:
Poly-p-hydroxybutyrate-co-p-hydroxy valerate (PHBV).

It is a copolymer of 3-hydroxybutyric acid and 3-hydroxypentanoic add, in which the monomer units are joined by ester linkages.

Question 3.
Explain the following terms giving a suitable example for each :
(i) Elastomers
Answer:
Elastomers: These are polymers in which the polymer chains are held together by weak intermolecular forces. Because of the presence of weak forces, the polymer can be easily stretched. However, a few cross-linked are also introduced in the chains which impart the property of regaining the original positions after the stretching force is released. A common example is vulcanised rubber.

(ii) Condensation polymers
Answer:
Condensation polymers: A polymer formed by the condensation of two or more than two monomers is called condensation polymer. In this type, each monomer generally contains two functional groups, and condensation takes place by the loss of molecules such as H20, HCl, etc. Common examples are nylon-6, nylon-6,6, bakelite, terylene and alkyl resins.

(iii) Addition polymers (CBSE 2012)
Answer:
Addition polymers: A polymer formed by the direct addition of repeated monomers is called an addition polymer. In this type, the monomers are unsaturated compounds and are generally derivatives of ethene. In addition, polymers have the same empirical formula as their monomers. For example, the addition of polymers polyethene or polypropylene are obtained as:
Class 12 Chemistry Important Questions Chapter 15 Polymers 15

Question 4.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
Buna-S:
1, 3-Butadiene: CH2 = CH – CH == CH2
Styrene: C6H5CH = CH2

(ii) Neoprene
Answer:
Neoprene:
Chloroprene: CH2 = C – CH = CH2

(iii) Nyton-6, 6 (CBSE Delhi 2013)
Answer:
Nylon 6, 6:

  • Adipic acid: HOOC (CH2)4 COOH
  • Hexamethylenediamine: NH2(CH2)6NH2

Question 5.
Write the names and structures of the monomers of the following polymers:
(i) Bakelite
Answer:
Bakelite:

  • Phenol:
    Class 12 Chemistry Important Questions Chapter 15 Polymers 16
  • Formaldehyde: HCHO

(ii) Nylon-6
Answer:
Nylon-6:
Caprolactam
Class 12 Chemistry Important Questions Chapter 15 Polymers 17
(iii) Polythene (CBSE Delhi 2013)
Answer:
Polythene: Ethene: CH2 = CH2.

Question 6.
(i) What is the role of benzoyl peroxide in the polymerisation of ethene?
(ii) Identify the monomers in the following polymer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 18
(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Nylon-6, 6, Polythene, Buna-S
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE 2016)
Answer:
(i) Benzoyl peroxide acts as a free radical generator. In the presence of benzoyl peroxide, phenyl free radical is formed which initiates the chain initiation step.
Class 12 Chemistry Important Questions Chapter 15 Polymers 19
(iii) Buna-S < Polythene < Nylon 6, 6
OR
Mechanism
(i) Chain initiation step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 20
(ii) Chain propagating step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 21
(iii) Chain termination step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 22
Question 7.
Write the names and structures of the monomers of the following polymers:
(a) Neoprene
Answer:
Chloroprene CH2 = C — CH = CH2

(b) Bakelite
Answer:
C6H5OH(Phenol) and HCHO (formaldehyde)

(c) PVC (CBSE 2019C)
Answer:
Polyvinyl chloride CH2 = CH – Cl

Question 8.
Write the structures of monomers used to obtain the following polymers:
(a) Natural rubber
(b) PVC
(c) Nylon-6,6
OR
Write the mechanism of free radical polymerisation of ethene. (CBSE AI 2019)
Answer:
(a) Natural rubber: 2-methyl-1,3-butadiene
Class 12 Chemistry Important Questions Chapter 15 Polymers 23

(b) PVC: vinyl chloride
CH2 = CH – Cl

(c) Nylon-6,6:
HOOC-(CH2)4 – COOH (Adipic acid)

NH2 — (CH2)6 — NH2
(Hexamethylenedlamine)
OR
Chain Initiation Step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 25
Chain propagation:
Class 12 Chemistry Important Questions Chapter 15 Polymers 26
Chain termination step
Class 12 Chemistry Important Questions Chapter 15 Polymers 27
Question 9.
Write the structures of monomers used to obtain the following polymers:
(a) Buna-S
(b) Glyptai
(c) Nylon-6
OR
(a) Arrange the following polymers in increasing order of their intermolecular forces: Polyvinylchloride, Neoprene, Terylene
(b) Write one example of each of
(i) Natural polymer
(ii) Thermosetting polymer
(c) What is the significance of numbers 6,6 in the polymer Nylon-6,6? (CBSE Al 2019)
Answer:
(a) Buna-S
CH2 = CH – CH = CH2 (1,3-Butadiene) and C6H5CH = CH2 (Styrene)

(b) Glyptal
HO – CH2 — CH2 — OH (Ethylene glycol) and
Class 12 Chemistry Important Questions Chapter 15 Polymers 28

(c) Nylon-6
Class 12 Chemistry Important Questions Chapter 15 Polymers 29
OR
(a) Neoprene < Polyvinyl chloride (PVC) < Terylene
(b) (i) Isoprene
(ii) Bakelite
(c) The numbers ‘6, 6’ signifies that both the monomers of nylon-6, 6 have six carbon atoms.

Question 10.
What is a biodegradable polymer? Give an example of a biodegradable aliphatic polyester. (CBSE AI 2013)
Answer:
Polymers that are degraded by micro-organisms within a suitable period so that the polymers and their degraded products do not cause any serious effects on the environment are called biodegradable polymers. For example,

Poly β – hydroxybutyrate – co – β – hydroxy valerate (PHBV)
Class 12 Chemistry Important Questions Chapter 15 Polymers 30
Question 11.
Write the names and structures of the monomers of the following polymers:
(i) Polystyrene
Answer:
Polystyrene: Styrene:
Class 12 Chemistry Important Questions Chapter 15 Polymers 31
(ii) Dacron
Answer:
Dacron:

  • Ethylene glycol: HOCH2CH2OH
  • Terephthalic acid:
    Class 12 Chemistry Important Questions Chapter 15 Polymers 32

(iii) Teflon
Answer:
Teflon:
Tetrafluoroethene: F2C = CF2

Question 12.
Write the names and structures of the monomers of the following polymers:
(a) Teflon
Answer:
Tetrafluoroethylene, CF2 = CF2
Class 12 Chemistry Important Questions Chapter 15 Polymers 33

(b) Terylene
Answer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 34
(c) Buna-N
Answer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 35

Question 13.
Write the structures of monomers used to obtain the following polymers:
(a) Neoprene
(b) PHBV
(c) Bakelite
OR
(a) Arrange the following polymers in decreasing order of their intermolecular forces:
Bakelite, Polythene, Buna-S, Nylon-6,6
(b) Write the monomers of the following polymer:
Class 12 Chemistry Important Questions Chapter 15 Polymers 36
(c) What is the structural difference between high-density polythene (HDP) and low-density polythene (LDP)? (CBSE AI 2019)
Answer:
(a) Neoprene:
Chloroprene
Class 12 Chemistry Important Questions Chapter 15 Polymers 37
(b) PHBV
Class 12 Chemistry Important Questions Chapter 15 Polymers 38
(c) Bakelite
Class 12 Chemistry Important Questions Chapter 15 Polymers 39
OR
(a) Buna-S < potythene < Bakelite < Nylon-6,6

(b) HO — CH2 — CH2 — OH and
Class 12 Chemistry Important Questions Chapter 15 Polymers 40
(c) Low-density polythene (LDP) consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore, it has low density.

High-density polythene (HDP) consists of linear chains and therefore, the molecules can be closely packed. Hence, it has a high density.

Question 14.
What are high density and low-density polythene?
Answer:
High-density polythene is obtained by heating ethene at about 333-343 K under a pressure of 6-7 atm in the presence of a catalyst such as triethylaluminium and titanium tetrachloride (known as Zeigler-Natta catalyst)
Class 12 Chemistry Important Questions Chapter 15 Polymers 41
This polymer consists of linear chains and therefore the molecules can be closely packed in space. It, therefore, has a high density (0.97g/cm3) and a higher melting point (403K). It is harder, tougher and has greater tensile strength than low-density polythene.

Low-density polythene (LDP) is prepared by free radical addition and hydrogen atom abstraction. It consists of highly branched chain molecules. Due to branching, the polythene molecules do not pack well and therefore it has low density (0.92 g/cm3) and low melting point (384 K). Low-density polythene is transparent. It has moderate tensile strength and high toughness. It is chemically inert.

Question 15.
Discuss the mechanism of free radical addition polymerisation.
Answer:
A large number of unsaturated compounds such as alkenes or dienes and their derivatives are polymerised by this process. The polymerisation takes place through the generation of an initiator, which is a molecule that decomposes to form free radicals. The commonly used initiator is t-butyl peroxide.
Class 12 Chemistry Important Questions Chapter 15 Polymers 42
The addition occurs as
Class 12 Chemistry Important Questions Chapter 15 Polymers 43
For example, most of the commercial addition polymers are obtained from alkenes and their derivatives,
Class 12 Chemistry Important Questions Chapter 15 Polymers 44
The general model of polymerisation is:

Chain initiation step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 45
Chain Propagating Step:
Class 12 Chemistry Important Questions Chapter 15 Polymers 46
Chain terminating step
Class 12 Chemistry Important Questions Chapter 15 Polymers 47

Question 16.
List some important differences between natural rubber and vulcanised rubber.
Answer:
Differences between natural rubber and vulcanised rubber:

Natural rubber Vulcanized rubber
1. Natural rubber is soft and sticky. 1. Vulcanized rubber is hard and non-sticky.
2. It has low tensile strength. 2. It has a high tensile strength.
3. It has low elasticity. 3. It has high elasticity.
4. It can be used over a narrow range of temperature (from 10°C to 60°C). 4. It can be used over a wide range of temperature (-40°C to 100°C).
5. It has low wear and tears resistance. 5. It has high wear and tears resistance.
6. It is soluble in solvents like ether, carbon- tetrachloride, petrol etc. 6. It is insoluble in all the common solvents.

Question 17.
(a) What is the difference between two notations: nylon-6 and nylon-6,6? Give their synthesis.
Answer:
Nylon-6 has only one compound having 6-carbon atoms while nylon-6,6 refers to a polymer obtained from 6-carbon atoms of dicarboxylic acid (adipic acid) and 6-carbon atoms of diamine (hexamethylene diamine).

Nylon-6 is synthesised from caprolactam as
Class 12 Chemistry Important Questions Chapter 15 Polymers 48

Nylon-6,6 is synthesised from hexamethylenediamine and adipic acid.
Class 12 Chemistry Important Questions Chapter 15 Polymers 49

(b) How is Buna-S prepared?
Answer:
Buna-S is obtained by copolymerisation of styrene and 1, 3-butadiene.
Class 12 Chemistry Important Questions Chapter 15 Polymers 50

Question 18.
Write the names and structures of the monomers of the following polymers:
(i) Buna-S
Answer:
1, 3-Butadiene: CH2 = CH — CH = CH2
Styrene: C6H5CH = CH2

(ii) Buna-N
Answer:
1, 3-Butadiene: CH2 = CH – CH = CH2
Acrytonitnie: CH2 = CH — CN

(iii) Dacron
Answer:

  • Ethylene glycol:
    Class 12 Chemistry Important Questions Chapter 15 Polymers 51
  • Terephthalic acid:
    Class 12 Chemistry Important Questions Chapter 15 Polymers 52

(iv) Neoprene.
Answer:
Chloroprene:
Class 12 Chemistry Important Questions Chapter 15 Polymers 53

Aldehydes, Ketones and Carboxylic Acids Class 12 Important Extra Questions Chemistry Chapter 12

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 12 Aldehydes, Ketones and Carboxylic Acids.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 12 Important Extra Questions Aldehydes, Ketones and Carboxylic Acids

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Very Short Answer Type

Question 1.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions: ethanal, propanal, propanone, butanone. (CBSE Delhi 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 2.
Which of the following compounds would undergo the Cannizzaro reaction? Benzaldehyde, Cyclohexanone, 2- Methylpentanal. (CBSE Sample Paper 2019)
Answer:
Benzaldehyde

Question 3.
Draw the structure of semicarbazone of cyclopentanone. (CBSE 2019C)
OR
Draw the structure of the product formed when propanal is treated with zinc amalgam and concentrated hydrochloric acid.
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 1
OR
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 2

Question 4.
Write the IUPAC name of the following: (CBSE AI 2012)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 3
Answer:
Pent-2-enal.

Question 5.
Write the structure of 3-methylbutanal. (CBSE Delhi 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 4

Question 6.
Write the structure of the p-Methylbenzal- dehyde molecule. (CBSE Delhi 2013, 2014)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 5

Question 7.
Draw the structure of the compound whose IUPAC name is 4-chloropentan 2-one. (CBSE Delhi 2008, CBSE AI 2011, 2013, 2014)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 6

Question 8.
Draw the structural formula of a 1-phenyl propane-2-one molecule. (CBSE 2010)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 7

Question 9.
Draw the structure of 3-methylbutanal. (CBSE 2011, CBSE Delhi 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 8

Question 10.
Arrange the following compounds in increasing order of their reactivity in nucleophilic addition reactions:
ethanal, propanal, propanone, butanone. (CBSE 2012)
Answer:
butanone < propanone < propanal < ethanal

Question 11.
Write the IUPAC name of (CBSE 2012)
Ph — CH = CH — CHO.
Answer:
3-Phenyl prop-2-en-al.

Question 12.
Rearrange the following compounds in the increasing order of their boiling points:
CH3 – CHO, CH3 – CH2 – OH, CH3 – CH2 – CH3 (CBSE 2013)
Answer:
CH3CH2CH3 < CH3CHO < CH3CH2OH

Question 13.
Ethanol is soluble in water. Why? (CBSE 2013)
Answer:
Ethanol is soluble in water because of hydrogen bonding between the polar carbonyl group of ethanal and water molecules:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 9

Question 14.
Write the IUPAC name of the compound (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 10
Answer:
3-Hydroxybutanoic acid

Question 15.
Write the IUPAC name of the compound (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 11
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 12

Question 16.
Write the IUPAC name of the compound (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 13
Answer:
3-Aminobutanal

Question 17.
Write the IUPAC name of the following compound: (CBSE AI2019)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 14
Answer:
But-3-en-2-one

Question 18.
Write the IUPAC name of the following compound: (CBSE AI 2019)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 15
Answer:
1-Phenylbutan-2-one.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Short Answer Type

Question 1.
Write structures of compounds A and B in each of the following reactions: (CBSE Delhi 2019)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 16
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 17

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 18
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 19

Question 2.
What is Tollen’s reagent? Write one use of this reagent. (CBSE 2010)
Answer:
The ammoniacal solution of silver nitrate is called Tollen’s reagent. It is used as an oxidizing reagent and helps to distinguish between aldehydes and ketones. Aldehydes are oxidized by Tollen’s reagent whereas ketones are not.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 20

Question 3.
Name the reagents used in the following reactions: (CBSE Delhi 2015)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 21
Answer:
Lithium aluminium hydride, LiAlH4

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 22
Answer
Alkaline potassium permanganate (KMnO4), KOH

Question 4.
Distinguish between
C6H5CH = CH – COCH3 and C6H5CH = CH CO CH2 CH3 (CBSE AI 2016)
Answer:
Heat both the compounds with NaOH and l2. C6H5CH = CHCOCH3 gives yellow ppt of iodoform. C6H5CH = CHCOCH2CH3 does not give yellow ppt. of iodoform.

Question 5.
Arrange the following in the increasing order of their reactivity in nucleophilic addition reactions.
CH3CHO, C6H5CHO, HCHO (CBSE AI 2016)
Answer:
C6H5CHO < CH3CHO < HCHO

Question 6.
Write the product in the following reaction: (CBSE AI 2017)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 23
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 24

Question 7.
Write the structure of the products formed: (CBSE Sample Paper 2019)
(a) CH3CH2COOH Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 25
Answer:
CH3 – CH(Cl) – C00H

(b) C6H5COCl Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 26
Answer:
C6H5CHO

Question 8.
Name the reagents in the following reactions: (CBSE AI 2015)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 27
Answer:
Lithium aluminium hydride, LiAlH4

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 28
Answer:
Alkaline potassium permanganate (KMnO4), KOH

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 29
Answer:
CH3 Mg Br, H3O+

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 30
Answer:
Cl2, P (Hell-Volhard-Zelinsky reaction)

Question 9.
Predict the organic products of the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 31
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 32

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 33
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 34

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 35
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 36

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 37
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 38

Question 10.
Describe how the following conversions are carried out:
(i) Toluene to benzoic acid
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 39

(ii) Bromobenzene to benzoic acid (CBSE AI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 40

(iii) Ethylcyanide to ethanoic acid
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 41

(iv) Butan-1-ol to butanoic acid (CBSE Delhi 2010)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 42

Question 11.
Write the structures of A, B, C, and D in the following reaction: (CBSE AI 2016)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 43
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 44

Question 12.
Aromatic carboxylic acids do not undergo Friedel Crafts reaction. Explain. (CBSE Al 2018)
Answer:
Aromatic carboxylic acids do not undergo Friedel Crafts reaction because -COOH group is deactivating and the catalyst aluminum chloride (Lewis acid) gets bonded to the carboxyl group.

Question 13.
pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid. Explain. (CBSE AI 2018)
Answer:
Due to the presence of a strong electron-withdrawing (-NO2) group in 4-nitrobenzoic acid, it stabilizes the carboxylate anion and hence strengthens the acid. Therefore, 4-nitrobenzoic acid is more acidic than benzoic acid and its pKa value is lower.

Aldehydes, Ketones and Carboxylic Acids Important Extra Questions Long Answer Type

Question 1.
Complete the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 45
Or
Write chemical equations for the following reactions:
(i) Propanone is treated with dilute Ba(OH)2.
(ii) Acetophenone is treated with Zn(Hg)/Conc. HCI
(iii) Benzoyl chloride is hydrogenated in presence of Pd/BaSO4. (CBSE Delhi 2019)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 46
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 47
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 48
Or
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 49
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 50
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 51

Question 2.
Predict the products of the following B reactions: (CBSE Delhi 2015)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 52
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 53

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 54
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 55

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 56
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 57

Question 3.
Predict the products of the following reactions: (CBSE 2015)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 58
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 59

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 60
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 61

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 62
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 63

Question 4.
What happens when
(a) Salicylic acid Is treated with (CH3CO)2O/H+?
Write a chemical equation in support of your answer. (CBSE AI 2019)
Answer:
Salicylic acid is treated with acetic anhydride In the presence of H+ to give aspirin.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 64

(b) Phenol is oxidized with Na2Cr2O7/H+?
Answer:
Phenol in the presence of acidified sodium dichromate gives benzoquinone.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 65

(C) Anisole Is treated with CH3Cl/anhydrous AlCl3?
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 66

Question 5.
Draw the structures of the following compounds:
(i) 3-Methylbutanal
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 67

(ii) 4-Chlocopentan-2-one
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 68

(iii) 4-Methylpent-3-en-2-one
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 69

(iv) p-Methyl benzaldehyde (CBSE AI 2014)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 70

Question 6.
How will you bring about the following conversions in not more than two steps?
(i) Propanone to propene (CBSE Delhi 2017)
Answer:
Propanone to propene
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 71
(ii) Propanal to butanone
Answer:
Propanal to butanone
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 72
(iii) Benzaldehyde to benzophenone
Answer:
Benzaldehyde to benzophenone
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 73

(iv) Benzaldehyde to 3-phenyl propan – 1-ol
Answer:
Benzaldehyde to 3- phenyl propan -1 -ol
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 74

(v) Benzaldehyde to α-hydroxyphenyl acetic acid
Answer:
Benzaldehyde to α-hydroxyphenyl acetic acid
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 75

(vi) Ethanol to 3-hydroxybutanal
Answer:
Ethanol to 3-hydroxybutanal
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 76

Question 7.
Convert the following:
(i) Ethanal to propanone. (CBSE AI 2018)
Answer:
Ethanal to propanone
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 77

(ii) Ethanal to lactic acid.
Answer:
Ethanal to lactic acid
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 78
(iii) Ethanal to 2-hydroxy-3-butenoic acid.
Answer:
Ethanal to 2-hydroxy-3-butenoic acid
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 79

(iv) Acetaldehyde to formaldehyde.
Answer:
Acetaldehyde to formaldehyde
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 80
(v) Formaldehyde to acetaldehyde.
Answer:
Formaldehyde to acetaldehyde
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 81

(vi) Acetaldehyde to crotonic acid.
Answer:
Acetaldehyde to crotonic acid
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 82

Question 8.
A compound ‘X’ (C2H4O) on oxidation gives ‘Y’ (C2H4O2). ‘X’ undergoes haloform reaction. On treatment with HCN ‘X’ forms a product ‘V which on hydrolysis gives 2-hydroxy propanoic acid.
(i) Write down structures of ‘X’ and ‘Y’.
(ii) Name the product when ‘X’ reacts with dil NaOH.
(iii) Write down the equations for the reactions involved. (CBSE Sample Paper 2007)
Answer:
Compound ‘X’ (C2H4O) is oxidized to ‘Y’ (C2H4O2). Since it undergoes a haloform reaction, it must be acetaldehyde.
(i) X = CHCHO Y = CH3COOH
On treatment with HCN, X gives cyanohydrin which on hydrolysis gives 2-hydroxypropanoic acid.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 83
(ii) When ‘X’ reacts with dil. NaOH, undergoes an aldol condensation reaction forming aldol which on heating gives but-2-enal.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 84
(iii) Equations for reactions
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 85
Other equations are given above.

Question 9.
Complete the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 86
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 87

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 88
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 89

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 90
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 91

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 92
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 93

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 94
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 95

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 96
(CBSE Sample Paper 2017 – 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 97

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 98
(CBSE Delhi 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 99

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 100
(CBSE AI 2013)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 101

Question 10.
(i) An organic compound (A) has a characteristic odor. On treatment with NaOH, it forms two compounds (B) and (C). Compound (B) has molecular formula C7H80 which on oxidation gives back (A). The compound (C) is a sodium salt of an acid. When (C) is treated with soda lime it yields an aromatic hydrocarbon (D). Deduce the structures of (A), (B), (C) and (D). Write the sequence of reactions involved.
Answer:
The compound A is C6H5CHO, benzaldehyde having a characteristic odor. The reactions are:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 102

(ii) Complete each synthesis by filling the missing starting materials, reagents, or products. (X and Z).
(a) C6H5CHO + CH3CH2CHO NaOH Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 103 X
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 105

(b) CH3(CH2)9 COOC2H5 Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 104 CH3(CH2),CHO
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 106
(iii) How will you bring about the following conversions in not more than two steps?
(a) Toluene to benzaldehyde
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 107
(b) Ethylcyanide to 1-phenylpropanoid. (CBSE Sample Paper 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 108
Question 11.
(i) How do you convert the following?
(a) Ethanal to propanone
(b) Toluene to benzoic acid
OR
(ii) (A), (B), and (C) are three non-cyclic functional isomers of a carbonyl compound with molecular formula C4H8O. isomers (A) and (C) give positive Tollens’ test whereas isomer (B) does not give Tollens’ test but gives positive iodoform test. Isomers (A) and (B) on reduction with Zn(Hg)/conc. HCI gives the same product (D).
(a) Write the structures of (A), (B), (C), and (D).
(b) Out of (A), (B), and (C) isomers, which one is least reactive towards the addition of HCN? (CBSEAI 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 109
OR
(ii) (a) Out of A, B, and C, (A) and (C) give positive Tollens’ test, and therefore, these are aldehydes. (B) does not give Tollens’ test and therefore, it is a ketone, withClass 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 110 a group because it gives positive iodoform test. Thus, the three isomers are:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 111

Question 12.
Predict the products of the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 112
(CBSE Delhi 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 113

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 114
(CBSE Delhi 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 115

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 116
(CBSE AI 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 117

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 118
(CBSE AI 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 119

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 120
(CBSE AI 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 121

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 122
(CBSE AI 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 123

Question 13.
Write the structures of A, B, C, D, and E in the following reactions: (C8SE Delhi 2016)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 124
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 125

Question 14.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanone.
(ii) Pentan-2-one and Pentan-3-one.
(b) Arrange the following compounds in increasing order of their acid strength: Benzoic acid, 4- Nitrobenzoic acid, 3,4 -Dinitrobenzoic acid, 4- Methoxybenzoic acid.
OR
Compare the reactivity of benzaldehyde and ethanal towards nucleophilic addition reactions. Write the cross aldol condensation product between benzaldehyde and ethanal. (CBSE Sample Paper 2019)
Answer:
(a) (i)

Experiments Ethanal Propanone
1. Tollen’s Test: Warm the organic compound with freshly prepared ammoniacal silver nitrate solution (Tollen’s reagent). A bright silver mirror is produced. No silver mirror is formed.
2. Fehling’s Test: Heat the organic compound with Fehling’s reagent. A reddish-brown precipitate is obtained. No precipitate is obtained.
Anyone test

Acetaldehyde (Ethanal) and Acetone (Propanone)
(i) Acetaldehyde gives silver mirror with Tollen’s reagent.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 126
Acetone does not give this test.

(ii) Acetaldehyde gives red ppt with Fehling solution.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 127

(ii) Pentan-2-one and Pentan-3-one

Experiment Pentan-2-one Pentan-3-one
Iodoform Test: The organic compound is heated with iodine in presence of sodium hydroxide solution. A yellow precipitate is obtained. No yellow precipitate is obtained.

Pentan-3-one and Pentan-2-one
(i) Pentan-2-one forms yellow ppt with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
CH3COCH2CH2CH3 + 3I2+ 4NaOH → CH3CH2CH2COONa + CHI3 + 3H20 + 3NaI

(ii) Pentan-2-one gives white ppt with sodium bisulphite while pentan-3-one does not.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 128
Or any other suitable test.

(b) 4- Methoxybenzoic add < Benzoic acid < 4- Nitrobenzoic acid < 3,4-Dinitrobenzoic acid Effect of substituents on acidic strength of acids. The substituents have a marked effect on the acidic strength of carboxylic acids. The nature of substituents affects the stability of the conjugate base (carboxylate ion) and hence affects the acidity of the carboxylic acids.

In general, electron-withdrawing groups (EWG) increase the stability of the carboxylate ion by delocalizing the negative charge and hence increase the acidity of the carboxylic acid. Conversely, electron-donating groups (EDG) decrease the stability of the carboxylate ion by intensifying the negative charge and hence decrease the acidity of the carboxylic acid.
OR
The carbon atom of the carbonyl group of benzaldehyde is less electrophilic than the carbon atom of the carbonyl group present in ethanol. The polarity of the carbonyl group is reduced in benzaldehyde due to resonance hence less reactive than ethanal.

Aromatic Aldehydes and Ketones
In general, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic analogs. For example, benzaldehyde is less reactive than aliphatic aldehydes. This can be easily understood from the resonating structures of benzaldehyde as shown below:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 129
It is clear from the resonating structures that due to the electron releasing (+I effect) of the benzene ring, the magnitude of the positive charge on the carbonyl group decreases, and consequently it becomes less susceptible to the nucleophilic attack. Thus, aromatic aldehydes and ketones are less reactive than the corresponding aliphatic aldehydes and ketones.

However, amongst aromatic aldehydes and ketones, aromatic aldehydes are more reactive than alkyl aryl ketones which in turn are more reactive than diaryl ketones. Thus, the order of reactivity of aromatic aldehydes and ketones is:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 130

Question 15.
(i) Illustrate the following name reaction giving a suitable example:
(a) Clemmensen reduction
(b) Hell-Volhard-Zelinsky reaction

(ii) How are the following conversions carried out?
(a) Ethylcyanide to ethanoic acid
(b) Butan-1-ol to butanoic acid
(c) Benzoic acid to m-bromobenzoic acid .
OR
(i) Illustrate the following reactions giving a suitable example for each.
(a) Cross aldol condensation
(b) Decarboxylation

(ii) Give simple tests to distinguish between the following pairs of compounds:
(a) Pentan-2-one and Pentan-3-one
(b) Benzaldehyde and Acetophenone
(c) Phenol and Benzoic acid (CBSE Delhi 2012)
Answer:
(i) (a) Clemmensen reduction
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 131

(b) Hell-Volhard-Zelinsky reaction
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 132

(ii)Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 133
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 134
OR
(i) (a) Cross Aldol condensation
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 135

(b) Decarboxylation
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 136

(ii) (a) Pentan-2-one and pentan-3-one can be distinguished by iodoform test
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 137
Pentan-3-one does not give this test.

(b) Benzaldehyde and acetophenone: Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent).
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 138
Acetophenone does not react with Tollen’s reagent.

(c) Phenol and Benzoic acid: Benzoic acid reacts with NaHC03 to give effervescence due to the evolution of CO2.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 139
Phenol does not give effervescence.

Question 16.
(i) Write a suitable chemical equation to complete each of the following transformations:
(a) Butan-1-ol to butanoic acid
(b) 4-Methylacetophenone to benzene-1, 4-dicarboxylic acid.

(ii) An organic compound with molecular formula C9H10O forms 2,4-DNP derivative, reduces Toilen’s reagent, and undergoes Cannizzaro’s reaction. On vigorous oxidation, it gives 1, 2-benzene dicarboxylic acid. Identify the compound.
OR
(i) Give chemical tests to distinguish between (a) Propanol and propanone
(b) Benzaldehyde and acetophenone

(ii) Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Methyl tert-butyl ketone (reactivity towards HCN)
(b) Benzoic acid, 3,4-Dinitrobenzoic acid, 4-Methoxybenzoic acid (acid strength)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 140

(ii) The given compound forms a 2,4-DNP derivative. Therefore, it is an aldehyde or ketone. Since it reduces Tollen’s reagent, it must be an aldehyde. The compound undergoes Cannizzaro’s reaction, so it does not contain hydrogen. On vigorous oxidation, it gives 1,2-benzene dicarboxylic acid, which means that it must be containing an alkyl group at 2-position with respect to the CHO group on the benzene ring.
The molecular formula suggests it should be
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 141 (2- Ethytbenza(dehyde)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 142
Answer:
(i) (a) Propanone gives iodoform test but propanol does not. When propanone is heated with aqueous sodium carbonate and iodine solution, a yellow precipitate of iodoform is obtained.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 143
Propanol does not give this test.

(b) Acetophenone forms yellow ppt. of iodoform with an alkaline solution of iodine (iodoform test). Benzaldehyde does not react.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 144

(ii) (a) Methyl tert-butyl ketone < Acetone < Acetaldehyde.
(b) 4-Methoxybenzoic acid < Benzoic add < 3, 4-Dinitrobenzoic acid.
(c) (CH3)2CHCOOH < CH3CH(Br)CH2COOH < CH3CH2CH(Br)COOH

Question 17.
(i) Although phenoxide ion has more resonating structures than carboxylate ion, the carboxylic acid is a stronger acid than phenol. Give two reasons.
(ii) How will you bring about the following conversions?
(a) Propanone to propane
(b) Benzoyl chloride to benzaldehyde
(c) Ethanal to but-2-enal
OR
(i) Complete the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 145
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 146
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 147

(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Ethanal and Propanal
(b) Benzoic acid and Phenol (CBSE AI 2013)
Answer:
(i) The carboxylic acids are stronger acids than phenols. The difference in the relative acidic strengths can be understood if we compare the resonance hybrid structures of carboxylate ion and phenoxide ion.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 148
The resonance hybrids may be represented as:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 149

(a) The electron charge in the carboxylate ion is more dispersed in comparison to the phenoxide ion since there are two electronegative oxygen atoms in the carboxylate ion as compared to only one oxygen atom in the phenoxide ion.

(b) Carboxylate ion is stabilized by two equivalent resonance structures in which negative charge is on more electronegative 0 atoms. But phenoxide ion has non-equivalent resonance structures in which the negative charge is also on a less electronegative carbon atom.

Therefore, the carboxylate ion is relatively more stable as compared to the phenoxide ion. Thus, the release of H+ ions from carboxylic acid is comparatively easier or it behaves as a stronger acid than phenol.

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 150
OR
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 151
(ii) (a) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 152
(b) Benzoic acid reacts with NaHC03 to give effervescence due to the liberation of CO2.
C6H5COOH + NaHCO3 → C6H5COONa + H20 + CO2
Phenol does not give effervescence.
Phenol gives violet color with FeCl3 solution but benzoic acid does not give such color.

Question 18.
(i) How will you convert the following:
(a) Propanone to Propan-2-ol
(b) Ethanal to 2-hydroxypropanoic acid
(c) Toluene to benzoic acid

(ii) Give a simple chemical test to distinguish between:
(a) Pentan-2-one and Pentan-3-one
(b) Ethanal and Propanal
OR
(i) Write the products of the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 153
(ii) Which acid of each pair shown here would you expect to be stronger? (CBSE AI 2013)
(a) F – CH2 – COOH or Cl – CH2 – COOH OH
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 154
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 155

(ii) (a) Pentan-2-one forms yellow ppt. with an alkaline solution of iodine (iodoform test), but pentane-3-one does not give iodoform test.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 156

(b) Ethanal gives a yellow ppt. of iodoform on heating with iodine and NaOH solution whereas propanal does not give.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 157
OR
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 158
(ii) (a) FCH2COOH
(b) CH3COOH

Question 19.
(a) Predict the main product of the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 159
(b) Give a simple chemical test to distinguish between
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 160
(c) Why is alpha (a) hydrogen of carbonyl compounds acidic in nature?
OR
(a) Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH3OH in presence of dry HCI
(ii) Dilute NaOH
(iii) H2N – NH2 followed by heating with KOH in ethylene glycol

(b) Arrange the following compounds in increasing order of their property as indicated:
(i) F – CH2COOH, O2N – CH2COOH, CH3COOH, HCOOH – acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards the addition of HCN (CBSE AI 2019)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 161
(b) On adding NaOH/l2 and heat, acetophenone gives yellow ppt of iodoform but benzophenone does not.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 162
(c) α – hydrogen of carbonyl compounds is acidic because of the electron-withdrawing nature of the carbonyl group.
OR
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 163
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 164

(b) (i) CH3COOH < HCOOH < FCH2COOH < NO2-CH3 COOH
(ii) Acetophenone < Benzaldehyde < Acetone < Acetaldehyde

Question 20.
(a) An organic compound with the molecular formula C7H60 forms a 2,4-DNP derivative, reduces Tollen’s reagent, and undergoes the Cannizzaro reaction. On oxidation, it gives benzoic acid. Identify the compound and state the reactions involved.
(b) Give chemical tests to distinguish between the following pair of compounds:
(i) Phenol and propanol
(ii) Benzoic acid and benzene
OR
(a) Predict the products of the following:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 165
(b) Arrange the following in increasing order of acidic character: (CBSE 2019C)
HCOOH, CF3COOH, ClCH2COOH, CCl3COOH
Answer:
(a) Compound = Benzaldehyde or C6H5CHO
Reaction
Reaction with 2,4-DNP
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 166

With Tollens reagent
RCHO + 2[Ag(NH3)2 ]+ + 30H → RCOO + 2 Ag + 2H2O + 4NH3 (where R = – C6H5)
Cannizzaro
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 167
(b) (i) FeCl3 test: Add 2 drops of neutral FeCl3 to both the compounds separately. Phenol gives violet color.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 168
(ii) Sodium bicarbonate test: Add NaHCO3 to both the compounds: Benzoic acid reacts with NaHCO3 to give effervescence due to liberation of CO2.
C6H5COOH + NaHCO3 → C6H5COONa + H2O + CO2
OR
(a) A = CH3COOH
B = CH3COCl
C = CH3CONH2
D = CH3NH2

(b) HCOOH < ClCH2COOH < CCl3COOH < CF3COOH

  • the electron-withdrawing substituents disperse the negative charge on carboxylate ion and stabilize it and thus, increase acidity.
  • the electron-releasing substituents intensify the negative charge of the carboxylate ion, destabilize it and thus, decrease the acidity.

Question 21.
(i) Give chemical tests to distinguish between:
(a) Propanal and propanone
(b) Benzaldehyde and acetophenone

(ii) How would you obtain:
(a) But-2-enal from the ethanal
(b) Butanoic acid from butanol
(c) Benzoic acid from ethylbenzene? (CBSE 2011)
Answer:
(i) (a) Propanal gives a silver mirror with Tollen’s reagent.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 169
Propanone does not give this test.

(b) Benzaldehyde forms a silver mirror with ammoniacal silver nitrate solution (Tollen’s reagent). Acetophenone does not react.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 170
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 171

Question 22.
(i) Describe the following giving linked chemical equations:
(a) Cannizzaro reaction
Answer:
Aldehydes that do not contain any a-hydrogen atom (e.g. benzaldehyde, formaldehyde) undergo self- oxidation and reduction reaction on treatment with cone, solution of caustic alkali. In this reaction, one molecule is oxidized to acid while another molecule is reduced to an alcohol.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 172

(b) Decarboxylation
Answer:
The process of removal of a molecule of C02 from a carboxylic acid is called decarboxylation. It is usually carried out by heating a mixture of carboxylic acid or its sodium salt with soda lime (NaOH + CaO).
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 173

(ii) Complete the following chemical equations: (CBSE Delhi 2011)
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 174
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 175

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 176
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 177

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 178
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 179

Question 23.
(i) Give chemical tests to distinguish between the following:
Benzoic acid and ethyl benzoate.
Answer:
(a) When treated with NaHCO3 solution, benzoic acid gives brisk effervescence while ethyl benzoate does not
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 180

(b) Ethyl benzoate on boiling with an excess of NaOH gives ethyl alcohol which on heating with iodine gives yellow ppt. of iodoform.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 181

(ii) Complete each synthesis by giving missing reagents or products in the following:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 182
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 183

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 184
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 185

Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 186
(CBSE 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 187

Question 24.
(i) Write the products formed when CH3CHO reacts with the following reagents:
(a) HCN
(b) H2N – OH
(C) CH3CHO in the presence of dilute NaOH
(ii) Give simple chemical tests to distinguish between the following pairs of compounds:
(a) Benzoic acid and Phenol
(b) Propanal and Propanone
OR
(i) Account for the following:
(a) Cl – CH2COOH is a stronger acid than CH3COOH.
(b) Carboxylic acids do not give reactions of the carbonyl group.
(ii) Write the chemical equations to illustrate the following name reactions:
Rosenmund reduction
(iii) Out of CH3CH2—CO—CH2 – CH3 and CH3CH2 — CH2 — CO — CH3, which gives iodoform test? (CBSE 2014)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 188
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 189

(ii) (a) Add neutral FeCl3 in both the solutions, phenol gives violet color with FeCl3 solution, but benzoic acid does not give such color.
(b) Add an ammoniacal solution of silver nitrate (Tollen’s reagent) in both the solutions, propanal gives silver mirror whereas propanone does not.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 190
OR
(0) Chlorine is an electron-withdrawing atom (-I effect). It withdraws the electrons from carbon to which it is attached and therefore, this effect is transmitted throughout the chain. As a result, electrons are withdrawn more strongly towards oxygen of 0—H bond and promotes the release of the proton.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 191
Consequently, ClCH2COOH is a stronger acid than CH3COOH.

(b) Carboxylic acids do not give the characteristic reactions of the carbonyl group (>C = 0) as given by aldehydes and ketones. In carboxylic acids, the carbonyl group is involved in resonance, as follows:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 192
Therefore, it is not a free group. But no resonance is possible in aldehydes and ketones. They give the characteristic reactions of the group.

(ii) Rosenmund reduction. Acid chlorides are converted to corresponding aldehydes by catalytic reduction. The reaction is carried out by passing through a hot solution of the acid chloride in the presence of palladium deposited over barium sulfate (partially poisoned with sulfur or quinoline).
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 193
The poisoning of palladium catalyst decreases its activity and it does not allow the further reduction of an aldehyde into alcohol.

(iii) CH3CH2CH2COCH3 gives iodoform test
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 194

Question 25.
(i) Write the structures of A, B, C, and D in the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 195

(ii) Distinguish between:
(a) C6H5 – CH = CH – COCH3 and C6H5 – CH = CH – CO CH2CH3
(b) CH3CH2COOH and HCOOH

(iii) Arrange the following in the increasing order of their boiling points:
CH3CH2OH, CH3COCH3, CH3COOH
OR
(i) Write the chemical reaction involved in the Etard reaction.
(ii) Arrange the following in the increasing order of their reactivity towards nucleophilic addition reaction:
CH3 – CHO, C6H5COCH3, HCHO
(iii) Why pKa of Cl-CH2-COOH is lower than the pKa of CH3COOH?
(iv) Write the product in the following reaction.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 196
(v) A and B are two functional isomers of compound C3H6O. On heating with NaOH and l2, isomer A forms a yellow precipitate of iodoform whereas isomer B does not form any precipitate. Write the formulae of A and B. (CBSE 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 197
(ii) (a) Heat both the compounds with NaOH and l2, C6H5 CH = CHCOCH3 gives yellow ppt. of iodoform. C6H5CH = CHCOCH2CH3 does not give yellow ppt. of iodoform.
(b) Add ammoniacal AgNO3 solution (Tollen’s reagent), HCOOH gives silver mirror while CH3CH2COOH does not.

(iii) CH3COCH3 < CH3CH2OH < CH3COOH
OR
(i) Etard’s reaction
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 198

(ii) C6H5COCH3 < CH3CHO < HCHO

(iii) Chlorine is an electron-withdrawing group (-I inductive effect).
It withdraws the electrons from the C to which it is attached and this effect is transmitted throughout the chain. As a result, the electrons are withdrawn more strongly towards 0 of the O-H bond and promote the release of the proton. Consequently, Cl — CH2COOH is more acidic than acetic acid.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 199
Therefore, pKa of ClCH2COOH is less than that of CH3COOH.

(iv) CH3 CH2 CH = CH CH2 CN Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 200 CH3 CH2 CH = CH CH2 CHO

(v) Two isomers are CH3COCH3 and CH3CH2CHO
Since A forms yellow ppt. of iodoform on heating with NaOH and l2, it is CH3COCH3.
CH3COCH3 + 3I2 + 4NaOH → CHI3 + CH3COONa + 3Nal + 3H2O
B does not give iodoform, it is CH3CH2CHO.
A: CH3COCH3 B: CH3CH2CHO

Question 26.
Write the structures of A, B, C, D, and E in the following reactions:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 201
OR
(i) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(ii) Draw the structure of the semicarbazone of ethanal.
(iii) Why pKa of F-CH2-COOH is lower than that of Cl – CH2 – COOH?
(iv) Write the product in the following reaction:
CH3 – CH = CH – CH2CN Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 202
(v) How can you distinguish between propanal and propanone? (CBSE Delhi 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 203
OR
(i) Cannizzaro’s reaction: Aldehydes that do not contain a-hydrogen undergo self-oxidation and reduction on treatment with cone. KOH.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 204
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 205
(iii) Fluorine is more electronegative than Cl and therefore, will have a greater electron-withdrawing effect (-I effect). As a result, the carboxylate ion will be stabilized to a larger extent and therefore fluoro acetic acid will be a stronger acid than chloroacetic acid. Hence, the pKa of FCH2COOH will be lower than that of ClCH2COOH.

(iv) CH3 — CH = CH — CH2 CN Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 202 CH3 CH = CH CH2 CHO

(v) Propanal gives red ppt with Fehling solution but propanone does not.
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 206

Question 27.
(i) Give reasons :
(a) HCHO is more reactive than CH3-CHO towards the addition of HCN. (CBSE 2018C)
(b) pKa of O2N—CH2—COOH is lower than that of CH3—COOH.
(c) Alpha hydrogen of aldehydes and ketones is acidic in nature.

(ii) Give simple chemical tests to distinguish between the following pairs of compounds :
(a) Ethanal and Propanal
(b) Pentan-2-one and Pentan-3-one
OR
(i) Write the structure of the product(s) formed :
(a) CH3 — CH2 — COOH Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 207
(b) C6H5COCl Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 208
(c) 2HCHO Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 209

(ii) How will you bring the following conversions in not more than two steps :
(a) Propanone to propene
(b) Benzyl chloride to phenyl ethanoic acid
Answer:
(i) (a) Due to + I effect of the methyl group is CH3CHO.
(b) Due to – I effect of the nitro group in nitroacetic acid.
(c) Due to the strong electron-withdrawing effect of the carbonyl group and resonance stabilization of the conjugate base.

(ii) (a) Add NaOH and l2 to both the compounds and heat, ethanal gives yellow ppt of iodoform.
(b) Add NaOH and l2 to both the compounds and heat, pentan-2-one gives yellow ppt of iodoform.
OR
Class 12 Chemistry Important Questions Chapter 12 Aldehydes, Ketones and Carboxylic Acids 210

The p-Block Elements Class 12 Important Extra Questions Chemistry Chapter 7

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 7 The p-Block Elements.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 7 Important Extra Questions The p-Block Elements

The p-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
Which one of PCl4+ and PCl4 is not likely to exist and why? (CBSE Delhi 2012)
Answer:
PCl4 is not likely to exist because lone pair of PCl3 can be donated to Cl+ and not to Cl.

Question 2.
Of PH3 and H2S which is more acidic and why? (CBSE Delhi 2012)
Answer:
H2S is more acidic because of weak S-H bond. PH3 behaves as a Lewis base because of the presence of lone pair of electrons on P.

Question 3.
Which is a stronger reducing agent, SbH3 or BiH3 and why? (CBSE Al 2012)
Answer:
BiH3 is a stronger reducing agent than SbH3. This is because BiH3 is less stable than SbH3 because of larger size of Bi than Sb.

Question 4.
What is the covalency of nitrogen in N2O5? (CBSE Delhi 2013)
Answer:
Four.

Question 5.
Name two poisonous gases which can be prepared from chlorine gas. (CBSE AI 2013)
Answer:

  1. Phosgene (COCl2)
  2. Mustard gas (ClCH2CH2SCH2CH2Cl)

Question 6.
What is the basicity of H3PO3 and why? (CBSE AI 2013)
Answer:
H3PO3 contains two P-OH bonds and therefore can give 2H+ ions. Its basicity is two.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 1

Question 7.
Why does ammonia act as a Lewis base? (CBSE 2014)
Answer:
Nitrogen atom in NH3 has one lone pair of electrons which is available for donation. Therefore,it acts as a Lewis base.

Question 8.
What is the oxidation number of phosphorus in H3PO2 molecule? (CBSE Delhi 2010)
Answer:
+1

Question 9.
Out of white phosphorus and red phosphorus, which one is more reactive and why? (CBSE 2015)
Answer:
White phosphorus is more reactive because of angular strain in the P4 molecules where the angles are only 60°.

Question 10.
What is the basicity of H3PO4? (CBSE Delhi 2015)
Answer:
Basicity of H3PO4 is three because it has three ionisable hydrogen atoms.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 2

Question 11.
On heating Pb(NO3)2 a brown gas is evolved which undergoes dimerisation on cooling. Identify the gas. (CBSE 2016)
Answer:
NO2.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 3
2NO2 → N2O4

Question 12.
Solid PCl5 is ionic in nature. Give reason. (CBSE AI 2016)
Answer:
PCl5 is ionic in the solid state because it exists as [PCl4]+ [PCl6] in which the cation is tetrahedral and anion is octahedral.

Question 13.
Write the structural difference between white P and red P. (CBSE Delhi 2014)
Answer:
White P consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ∠PPP = 60°. Red P also consists of P4 tetrahedra units but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bonds.

Question 14.
What is the basicity of H3PO2 acid and why? (CBSE AI 2012)
Answer:
H3PO2 has the structure:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 4
It has only one ionisable hydrogen and therefore, its basicity is one.

Question 15.
Name the promoter used In Haber’s process. (CBSE Sample Paper 2017-18)
Answer:
Molybdenum.

Question 16.
NF3 is an exothermic compound whereas NCl3 is not. Explain. (CBSE AI 2011, 2012)
Answer:
NF3 is an exothermic compound while NCl3 is not because:

  • The bond dissociation enthalpy of F2 is lower than that of Cl2.
  • Size of F is small as compared to Cl and therefore F forms stronger bonds with nitrogen releasing large amount of energy.

Therefore, overall energy is released during the formation of NF3 and energy is absorbed during the formation of NCl3.

Question 17.
N-N single bond is weaker than P-P single bond. (CBSE Delhi 2014)
Answer:
N-N single bond is weaker than P-P single bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

Question 18.
What happens when orthophosphorous acid is heated? (CBSE Sample Paper 2017-18)
Answer:
Phosphoric acid and phosphine are formed.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 5

Question 19.
Write the order of thermal stability of the hydrides of group 16 elements. (CBSE Delhi 2017)
Answer:
The thermal stability of the hydrides of group 16 elements decreases from H2O to H2Te as:
H2O > H2S > H2Se > H2Te

Question 20.
The two O – O bond lengths in ozone molecule are equal. Why? (CBSE AI 2013, CBSE Delhi 2014)
Answer:
Ozone is a resonance hybrid of two structures and therefore, the two O-O bond lengths are equal.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 6

Question 21.
SF6 is not easily hydrolysed. (CBSE Delhi 2005)
OR
SF6 is kinetically inert substance. Explain. (CBSE Al 2011, CBSE Delhi 2011)
Answer:
SF6 is chemically inert and therefore, does not get hydrolysed. Its inert nature is due to the presence of stearically protected sulphur atom which does not allow thermodynamically favourable hydrolysis reaction.

Question 22.
Which of the following compounds has a lone pair of electrons at the central atom? (CBSE Sample Paper 2011)
H2S2O8, H2S2O7, H2SO3, H2SO4.
Answer:
H2SO3

Question 23.
Ozone is thermodynamically unstable. Explain. (CBSE Sample Paper 2011)
Answer:
Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (∆H is -ve) and increase in entropy (∆S is +ve). These two factors reinforce each other resulting negative ∆G (∆G = ∆H – T∆S) for its conversion to oxygen.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 7

Question 24.
Which neutral molecule would be isoelectronic with ClO? Is that molecule a Lewis base? (CBSE Delhi 2008)
Answer:
ClO is isoelectronic with ClF. Yes, it is a Lewis base.

Question 25.
Which compound of xenon has distorted octahedral shape? (CBSE Sample Paper 2012)
Answer:
XeF6 has distorted octahedral shape.

Question 26.
Iodide ions can be oxidised by oxygen in acidic medium. Give chemical equation to support this. (CBSE Sample Paper 2011)
Answer:
Iodide can be oxidised by oxygen in acidic medium:
4I(aq) + O2(g) + 4H+(oq) > 2l2(s) + 2H2O(l).

Question 27.
What happens when XeF6 undergoes complete hydrolysis? (CBSE Sample Paper 2017-18)
Answer:
XeO3 is formed.
XeF6 + 3H2O → XeO3 + 6HF

Question 28.
Which xenon compound is isostructural with lCl4? (CBSE Sample Paper 2011)
Answer:
XeF4

Question 29.
Why conductivity of silicon increases on doping with phosphorus? (CBSE Delhi 2019)
Answer:
When silicon is doped with phosphorus, an extra electron is introduced after forming four covalent bonds. This extra electron gets delocalised and serves to conduct electricity. Hence, conductivity increases.

The p-Block Elements Important Extra Questions Short Answer Type

Question 1.
What inspired N. Bartlett for carrying out reaction between Xe and PtF6? (CBSE Delhi 2013)
Answer:
In 1962, N. Bartlett noticed that platinum hexafluoride PtF6, is a powerful oxidising agent which combines with molecular oxygen to form ionic compound, dioxygenyl hexafluoroplatinate (V), O2+ [PtF6].
O2(S) + PtF6(g) → O2+ [PtF6]
This indicates that PtF6 has oxidised O2 to O2+. Now, oxygen and xenon have some similarities:

  • The first ionisation energy of xenon gas (1170 kJ mol-1) is fairly close to that of oxygen (1177 kJ mol-1).
  • The molecular diameter of oxygen and atomic radius of xenon are similar (4Å). These prompted Bartlett to carry out the reaction between Xe and PtF6.

Question 2.
Calculate the number of lone pairs on central atom in the following molecule and predict the geometry. (CBSE Sample Paper 2019)
Answer:
XeF4 has square planar structure. It involves sp3d2 hybridisation of Xenon in which two positions are occupied by lone pairs.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 8
Structure of XeF4 (Square planar)

Question 3.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE AI 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X, e.g. ClF, ICl3, BrF5, IF7, etc.

Question 4.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalides. Give reason. (CBSE 2013, CBSE Delhi 2015)
Answer:
Nitrogen belongs to second period (n = 2) and has only s and p-orbitals. It does not have d-orbitals in its valence shell and therefore, it cannot extend its octet. That is why nitrogen does not form pentahalides.

Question 5.
Why does NO2 dimerise? Explain. (CBSE 2014, CBSE Delhi 2014)
Answer:
NO2 contains odd number of valence electrons. It behaves as a typical molecule. In the liquid and solid state, it dimerises to form stable N2O4 molecule, with even number of electrons. Therefore, NO2 is paramagnetic, while N2O4 is diamagnetic in which two unpaired electrons get paired.

Question 6.
Draw the structure of O3 molecule. (CBSE Delhi 2010)
Answer:
Ozone has angular structure as:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 9
It is resonance hybrid of the following structures:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 10

Question 7.
Complete the following equations: (CBSE 2014)
(i) Ag + PCl5
(ii) CaF2 + H2SO4
Answer:
(i) 2Ag + PCl5 → 2AgCl + PCl3
(ii) CaF2 + H2SO4 → CaSO4 + 2HF

Question 8.
Write balanced chemical equations for the following processes:
(i) XeF2 undergoes hydrolysis.
(ii) MnO2 is heated with cone. HCl.
OR
Arrange the following in order of property indicated for each set:
(i) H2O, H2S, H2Se, H2Te – increasing acidic character
(ii) HF, HCl, HBr, HI – decreasing bond enthalpy (CBSE Delhi 2019)
Answer:
(i) 2XeF2 + 2H2O → 2Xe + 4HF + O2
(ii) MnO2 + 4HCl → MnCl2 + 2H2O + Cl2
OR
(i) H2O < H2S < H2Se < H2Te
(ii) HF > HCI > HBr > HI

Question 9.
Explain the following giving an appropriate reason in each case.
(i) O2 and F2 both stabilise higher oxidation states of metals but O2 exceeds F2 in doing so.
(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach. (CBSE 2012)
Answer:
(i) This is because fluorine can show only -1 oxidation state, whereas oxygen can show -2 oxdiation state as well as positive oxidation states such as +2, +4 and +6.

(ii) Structures of xenon fluorides cannot be explained by Valence Bond approach because xenon is a noble gas and its octet is complete. Therefore, according to Valence Bond theory, xenon should not combine with fluorine to form fluorides due to its stable outermost shell configuration.

Question 10.
Draw the structures of the following molecules:
(i) XeOF4
(ii) H3PO3 (CBSE 2013)
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 11
Structure of XeOF4

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 12
Structure of H3PO3

Question 11.
How are interhalogen compounds formed? What general compositions can be assigned to them? (CBSE 2013)
Answer:
The compounds containing two or more halogen atoms are called interhalogen compounds. They may be assigned general composition as XX’n, where X is halogen of larger size and X’ is of smaller size and X’ is more electronegative than X,
e.g. ClF, ICl3, BrF5, IF7, etc.

Question 12.
Draw the structures of the following molecules: (CBSE 2013)
(i) N2O5
(ii) XeF2
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 13

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 14
Structure of XeF2 (Linear)

Question 13.
What happens when
(i) PCl5 is heated?
(ii) H3PO3 is heated?
Write the reactions involved. (CBSE Delhi 2013)
Answer:
(i) On heating, PCl5 sublimes and decomposes on strong heating:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 15

(ii) On heating, H3PO3 decomposes into phosphoric acid and phosphine.
4H3PO3 → 3H3PO4 + PH3

Question 14.
Complete the following chemical equations:
(i) Ca3P2 + H2O →
(ii) Cu + H2SO4(conc.) →
OR
Arrange the following in the order of property indicated against each set:
(i) HF, HCl, HBr, HI – increasing bond dissociation enthalpy.
(ii) H2O, H2S, H2Se, H2Te – increasing acidic character. (CBSE Delhi 2014)
Answer:
(i) Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3
(ii) Cu + 2H2SO24;(conc.) → CuSO4
+ SO2 + 2H2O
OR
(i) HI < HBr < HCl < HF
(ii) H2O < H2S < H2Se < H2Te

Question 15.
Complete the following equations:
(i) P4 + H2O →
(ii) XeF4 + O2F2 → (CBSE 2014)
Answer:
(i) P4 + H2O → no reaction
(ii) XeF4 + O2F2 → XeF6 + O6

Question 16.
Draw the structures of the following:
(i) XeF2
(ii) BrF3
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 16
Structure of XeF2 (Linear)

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 17
Structure of BrF3 (T Shape)

Question 17.
(a) Draw the structure of XeF4.
(b) What happens when CaF2 reacts with cone. H2SO4? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 18

(b) HF is formed.
CaF2 + H2SO4 → 2HF + CaSO4

Question 18.
(a) Draw the structure of H2S2O7.
(b) What happens when carbon reacts with cone. H2SO4? Write balanced chemical equation. (CBSE 2019C)
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 19

(b) Carbon dioxide is formed.
C + 2H2SO4(conc.) → CO2 + 2 SO2 + 2H2O

Question 19.
Draw the structures of the following molecules: (CBSE 2014)
(i) XeO2
(ii) H2SO4
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 20

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 21

Question 20.
Write the structure of the following: (HPO3)3 (CBSE 2016)
Answer:
(HPO3)3
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 22

Question 21.
Why is N2 less reactive at room temperature? (CBSE Al 2015, H.P.S.B. 2015, 2016, Meghalaya S.B. 2017)
Answer:
In molecular nitrogen, there is a triple bond between two nitrogen atoms (NsN) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 22.
Phosphine has lower boiling point than ammonia. Give reason. (CBSE Al 2008,
CBSE Delhi 2013)
Answer:
Ammonia exists as associated molecule due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH3, phosphine (PH3) molecules are not associated through hydrogen bonding in liquid state.
This is because of low electronegativity of P than N. As a result, the boiling point of PH3 is lower than that of NH3.

Question 23.
Draw structures of the following:
(a) XeF4
(b) S2O82- (CBSE AI 2019)
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 23

(b)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 24

Question 24.
(i) Draw the structure of phosphinic acid (H3PO2).
(ii) Write a chemical reaction for its use as reducing agent. (CBSE Sample Paper 2011)
Answer:
(i) Phosphinic acid, H3PO2.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 25

(ii) H3PO2 reduces Ag+ ion to Ag which shows its reducing nature.
H3PO2 + 4AgNO3 + 2H2O → 4Ag + 4HNO3 + H3PO4

Question 25.
Among the hydrides of Group 15 elements, which have the
(a) lowest boiling point?
(b) maximum basic character?
(c) highest bond angle?
(d) maximum reducing character? (CBSE AI 2018)
Answer:
(a) PH3
(b) NH3
(c) NH3
(d) BiH3

Question 26.
The bond angles (O-N-O) are not of the same value in NO2 and NO2+. Give reason. (CBSE Delhi 2012)
Answer:
In NO2 there is one electron on N. Therefore, in NO2+, there is no electron on N atom and there is a lone pair of electrons on N atom in NO2.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 26

Question 27.
H2PO2 is a stronger reducing agent than H3PO3. Give reason. (CBSEAI 2014, 2016)
Answer:
The reducing character of the acid is due to H atoms bonded directly to P atom. In H3PO2, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO2 is a stronger reducing agent than H2PO3.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 27

Question 28.
Bi(V) is a stronger oxidising agent than Sb(V). Give reason. (CBSE Delhi 2014)
Answer:
On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

Question 29.
Dioxygen is a gas while sulphur is a solid at room temperature. Why?
(CBSE AI 2013, 2018)
Answer:
Oxygen exists as a stable diatomic molecule and is, therefore, a gas. On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has strong tendency to form multiple bonds with itself and forms strong O=O bonds rather than O-O bonds.

On the other hand, sulphur-sulphur double bonds (S=S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O=O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule.
Therefore, oxygen exists as a diatomic gas and sulphur exists as S8 solid.

Question 30.
Draw structures of the following:
(a) H2S2O7
(b) HClO3 (CBSE Al 2019)
Answer:
(a) H2S2O7
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 28

(b) HClO3
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 29

Question 31.
Complete and balance the following equations:
(a) C + H2SO4 (cone.) →
(b) XeF2 + PF5
OR
Write balanced chemical equations involved in the following reactions:
(a) Fluorine gas reacts with water.
(b) Phosphine gas is absorbed in copper sulphate solution. (CBSE AI 2019)
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 30

(b)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 31

OR

(a) 2F2 + 2H2O → 4HF + O2

(b)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 32

Question 32.
Sulphuric acid has low volatility. Give chemical reactions in support of this. (CBSE Sample Paper 2011)
Answer:
Sulphuric acid has low volatility and decomposes the salts of volatile acids forming its own salts:
2MX + H2SO4 → 2HX + M2SO4
(M = metal, X = F, Cl, NO3)
e.g.,NaCl + H2SO4 → NaHSO4 + HCl
KNO3 + H2SO4 → KHSO4 + HNO3.

Question 33.
Electron gain enthalpies of halogens are largely negative. Why? (CBSE AI 2017)
Answer:
The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns2np6).

Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have largely negative electron gain enthalpies.

Question 34.
Why does R3P = 0 exist but R3N = 0 does not (R = alkyl group). (CBSE AI 2014)
Answer:
R3N = 0 does not exist because nitrogen cannot have covalency more than four. Moreover, R3P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bond whereas nitrogen cannot form dπ-pπ bond.

Question 35.
Write any two oxoacids of sulphur and draw their structures. (CBSE Al 2019)
Answer:
(i) Sulphuric acid
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 33

(ii) Peroxodisulphuric acid
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 34

Question 36.
Complete and balance the following equations:
(a) NH3 (excess) + Cl2
(b) XeF6 + 2H2O →
OR
Write balanced chemical equations involved in the following reactions:
(a) XeF4 reacts with SbF5.
(b) Ag is heated with PCl5. (CBSE AI 2019)
Answer:
(a)
8NH3 + 3Cl2 → 6NH4Cl + N2
excess
(b) XeF6 + 2H2O → XeO2F2 + 4HF

OR

(a) XeF4 + SbF5 → [XeF3]+[SbF6]
(b) 2Ag + PCl5 → 2AgCl + PCl3

Question 37.
Complete and balance the following equations:
(a) S + H2SO4 (cone.) →
(b) PCl3 + H2O →
Write balanced chemical equations involved in the following reactions:
(a) Chlorine gas reacts with cold and dilute NaOH
(b) Calcium phosphide Is dissolved in water. (CBSE AI 2019)
Answer:
(a) S + 2H2SO4(conc.) → 3SO2 + 2H2O
(b) PCl3 +3H2O → H3PO3 + 3HCl
OR
(a) Cl2 + ZNaOH → NaCl + NaOCl + H2O
(b) Ca3P2 + 6H2O → 3Ca(OH)2 + 2PH3

Question 38.
Bond enthalpy of fluorine Is lower than that of chlorine. Why? (CBSE AI 2018)
Answer:
Fluorine atoms are smalL and internuclear distance between two F atoms In F2 molecule is also small (1 .43Å). As a result, electron-electron repulsions among the lone pairs on two fluorine atoms in F2 molecule are large. These Llarge electron-electron repulsions weaken the F-F bond. However, the electron-electron repulsions among lone pairs in Cl2 molecule are less because of Larger Cl atoms and larger Cl-Cl distance (1.99Å). Therefore, the bond enthalpy of fluorine is lower than that of chlorine.

The p-Block Elements Important Extra Questions Long Answer Type

Question 1.
Give reasons for the following:
(a) Dioxygen is a gas but sulphur a solid.
(b) NO (g) released by jet aeroplanes is slowly depleting the ozone layer.
(c) Interhalogens are more reactive than pure halogens. (CBSE Al 2019)
Answer:
(a) Due to small size and high electronegativity, oxygen atom forms pπ-pπ double bond, O = O. The intermolecular forces in oxygen are weak van der Waal’s forces and therefore, oxygen exists as a gas. On the other hand, sulphur does not form stable pπ-pπ bonds and does not exist as S2. It is linked by single bonds and forms polyatomic complex molecules having eight atoms per moledule (S8) and has puckered ring structure. Therefore, S atoms are strongly held together and it exists as a solid.

(b) Nitrogen oxide (NO) gas emitted from the exhaust of jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen.
NO + O3 → NO2 + O2
Since jet aeroplanes fly in the stratosphere, near ozone layer, these are responsible for the depletion of ozone layer.

(c) This is because covalent bond between dissimilar atoms in interhalogens (X- Y) is polar and weaker than between similar atoms (X-X and Y-Y). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping between similar atoms.

Question 2.
Give reasons for the following:
(i) Where R is an alkyl group, R3P = 0 exists but R3N = 0 does not.
(ii) PbCl4 is more covalent than PbCl2.
(iii) At room temperature, N2 is much less reactive. (CBSE AI 2013)
Answer:
(i) R3N = 0 does not exist because nitrogen cannot have covalency more than four. But R3P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ-pπ bonds whereas nitrogen cannot form dπ-pπ bond.

(ii) Pb4+ has smaller size than Pb2+. As a result, it can polarise the Cl ion to greater extent than Pb2+. Consequently, PbCl4 is more covalent than PbCl2.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

Question 3.
Give reasons for the following:
(i) Oxygen is gas but sulphur is a solid,
(ii) O3 acts as a powerful oxidising agent.
(iii) BiH3 is the strongest reducing agent amongst all the hydrides of Group 15 elements. (CBSE Al 2013)
Answer:
(i) Oxygen exists as a stable diatomic molecule and is, therefore, a gas.
On the other hand, sulphur exists in solid state as S8 molecules and have puckered ring structure. The main reason for this different behaviour is that oxygen atom has good tendency to form multiple bonds with itself and forms strong o = o bonds than o – o bonds. On the other hand, sulphur- sulphur double bonds (S = S) are not very strong. As a result, catenated -O-O-O- chains are less stable as compared to O = O molecule while catenated -S-S-S- chains are more stable as compared to S = S molecule. Therefore, oxygen exists as a diatomic gas and sulphur exists as S8 solid.

(ii) Ozone acts as a powerful oxidising agent because it has higher energy content and decomposes readily to give atomic oxygen as:
O3 → O2 + O
Therefore, ozone can oxidise a number of non-metals and other compounds. For example,
PbS (s) + 4O3 (8) → PbSO4 (s) + 4O2 (g)
2I (aq) + H2O(l) + O3(g) → 2OH (aq) + I32(s) + O2(g)

(iii) The stability of group 15 hydrides decreases from NH3 to BiH3. Therefore, BiH3 is most unstable hydride among group 15 hydrides because of lowest M-H bond strength. As a result, BiH3 is the strongest reducing agent.

Question 4.
Give reasons for the following:
(i) Though nitrogen exhibits +5 oxidation state, it does not form pentahalide.
(ii) Electron gain enthalphy with negative sign of fluorine is less than that of chlorine.
(iii) The two oxygen-oxygen bonds lengths in ozone molecule are identical. (CBSE Al 2013)
Answer:
(i) Nitrogen does not have vacant d-orbitals in its valence shell. Therefore, it cannot extend its valency beyond 3. Thus, it cannot form pentahalides.

(ii) The less negative electron gain enthalpy of fluorine as compared to chlorine is due to very small size (72 pm) of fluorine atom. As result, there are strong inter- electronic repulsions in the relatively small 2p sub-shell of fluorine and therefore, the incoming electron does not feel much attraction. Thus, its electron gain enthalpy is small.

(iii) Ozone is resonance hybrid of two structures and therefore, the two oxygen-oxygen bonds are identical.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 35

Question 5.
How would you account for the following:
(i) H2S is more acidic than H2O.
(ii) The N-O bond in NO2 is shorter than the N-O in NO3.
(iii) Both O2 and F2 stabilise higher oxidation states but the ability of oxygen to stabilise the higher oxidation states exceeds that of fluorine. (CBSE 2011)
Answer:
(i) Since size of S is more than that of O, the distance between central atom and hydrogen atoms is more in H2S than in H2O. Therefore, bond dissociation enthalpy of H2S is less and bond cleavage is more easy than in H2O. Therefore, H2S is more acidic than H2O.

(ii) In the resonance structures of NO2, two bonds are sharing a double bond, while in NO3, three bonds are sharing a double bond.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 36
As a result, the bond in NO2 will be shorter than in NO3.

(iii) The larger tendency of oxygen to stabilise the higher oxidation state is because of the tendency of oxygen to form multiple bonds with metal.

Question 6.
How would you account for the following:
(i) NF3 is an exothermic compound but NCl3 is not.
(ii) The acidic strength of compounds increases in the order:
PH3 < H2S < HCl
(iii) SF6 is kinetically inert. (CBSE 2011)
Answer:
(i) NF3 is an exothermic compound while NCl3 is an endothermic compound because
(a) The bond dissociation enthalpy of F2 is lower as compared to Cl2.
(b) Size of F is small as compared to Cl and therefore, F forms stronger bond with N releasing large amount of energy.
Therefore, overall energy is released during the formation of NF3 and energy is absorbed during the formation of NCl3.

(ii) The bond dissociation enthalpy decreases in the order:
P-H > S-H > Cl-H
Therefore, acidic character follows the order:
PH3 < H2S < HCl

(iii) SF6 is kinetically inert because sulphur is sterically protected by six F atoms and it is a coordinatively saturated compound. Therefore, it does not undergo thermodynamically favourable hydrolysis reaction.

Question 7.
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF4
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 37

(ii) H2SO4
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 38

(b) White phosphorus consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Answer:
Red phosphorus also consists of P4 tetrahedra units, but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bond.

Question 8.
Account for the following:
(i) Bi(V) is a stronger oxidising agent than Sb(V).
(ii) N-N single bond is weaker than P-P single bond.
(iii) Noble gases have very low boiling points. (CBSE Delhi 2014)
Answer:
(i) On moving down the group, the stability of +5 oxidation state decreases while the stability of +3 oxidation state increases due to inert pair effect. Therefore, +5 oxidation state of Bi is less stable than +5 oxidation state of Sb. Thus, Bi(V) is a stronger oxidising agent than Sb(V).

(ii) N-N single bond is weaker than single P-P bond because of high interelectronic repulsions of non-bonding electrons due to small bond length.

(iii) Noble gases are monoatomic gases and are held together by weak vander Waals forces. Therefore, these are liquified at very low temperatures. Hence they have low boiling points.

Question 9.
Give reasons for the following: (CBSE 2014)
(i) (CH3)3 P = 0 exists but (CH3)3 N = 0 does not.
(ii) Oxygen has less electron gain enthalpy with negative sign than sulphur.
(iii) H3PO2 is a stronger reducing agent than H3PO3.
Answer:
(i) (CH3)3 N = 0 does not exist because
nitrogen cannot have covalency more than four. Moreover, (CH3)3 P = 0 exists because phosphorus can extend its covalency more than 4 as well as it can form dπ – pπ bonds whereas nitrogen cannot form dπ – pπ bonds.

(ii) This is due to small size of oxygen atom so that its electron cloud is distributed over a small region of space and therefore, it repels the incoming electron. Hence the electron gain enthalpy of oxygen is less negative than sulphur.

(iii) In H3PO3, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO3 is stronger reducing agent than H3PO3.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 39

Question 10.
Assign reason for the following:
(i) H3PO2 is a stronger reducing agent than H3PO4.
(ii) Sulphur shows more tendency for catenation than oxygen.
(iii) Reducing character increases from HF to HI. (CBSE 2016)
Answer:
(i) The reducing character of the acid is due to H atoms bonded directly to P atom. In H3PO2, there are two P-H bonds whereas in H3PO3, there is one P-H bond. Therefore, H3PO2 is a stronger reducing agent than H3PO3.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 40

(ii) Sulphur shows more tendency for catenation than oxygen because of stronger S-S bonds than O-O bonds.

(iii) Due to decrease in bond enthalpy down the group, the thermal stability decreases from HF to HI.
As a result of decrease in stability from HF to HI, the reducing character increases down the group as:
HF < HCl < HBr < HI

Question 11.
What happens when
(i) (NH4)2Cr2O7 is heated?
(ii) PCl5 is heated?
(iii) H3PO3 is heated? Write the equations involved. (CBSE AI 2015, CBSE Delhi 2008, 2013, 2017)
Answer:
(i) On heating (NH4)2Cr2O7, nitrogen gas is evolved.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 41

(ii) On heating, PCl5 first sublimes and then decomposes on strong heating:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 42

(iii) On heating, H3PO3 disproportionates to give orthophosphoric acid and phosphine.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 43

Question 12.
(a) Suggest a quantitative method for estimation of the gas which protects us from U.V. rays of the sun.
(b) Nitrogen oxides emitted from the exhaust system of supersonic jet aeroplanes slowly deplete the concentration of ozone layer in upper atmosphere. Comment. (CBSE Sample Paper 2011)
Answer:
(a) The gas which protects us from U.V. rays of the sun is ozone. It reacts with I ions to give iodine as:
O3 + 2I + H2O → O2 + l2 + 2OH
l2 liberated is titrated against sodium thiosulphate solution and amount of O3 can be estimated.

(b) The release of nitrogen oxides (NOx) into stratosphere by the exhaust system of supersonic jet aeroplanes deplete the concentration of O2 because NO reacts with O3 to give O2.
NO (g) + O2 (g) → NO2 (S) + O2 (s)
Therefore, NO is slowly depleting the concentration of ozone.

Question 13.
Give reasons for the following:
(a) Acidic character decreases from N2O3 to Bi2O3.
(b) All the P-Cl bonds in PCl5 are not equivalent.
(c) HF is a weaker acid than HCl in an aqueous solution. (CBSE Al 2019)
Answer:
(a) On moving down the group, metallic character increases. Therefore acidic character of oxides decreases from nitrogen to bismuth.

(b) PCl5 has trigonalbipyramidal geometry.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 44
In this geometry, axial bonds are larger than equatorial bonds because of large repulsions at axial positions. Therefore, all P-Cl bonds in PCl5 are not equivalent.

(c) The H-F bond has high bond dissociation enthalpy than H-Cl bond. Therefore, H-F bond is stronger and can be dissociated into H+ and F ions with difficulty. Hence, it is weaker acid than HCl.

Question 14.
Give reasons for the following:
(a) O-O single bond is weaker than S-S single bond.
(b) Tendency to show -3 oxidation state decreases from Nitrogen (N) to Bismuth (Bi).
(c) Cl2 acts as a bleaching agent. (CBSE AI 2019)
Answer:
(a) O-O single bond is weaker than S-S single bond because oxygen has smaller size and has high inter-electronic repulsion.

(b) Tendency to show -3 oxidation state decreases because metallic character and size increase down the group.

(c) Cl2 acts as a bleaching agent in the presence of moisture
Cl2 + H2O → 2HCl + [O]
Coloured matter + [O] → Colourless matter
The nascent oxygen is responsible for the bleaching action of Cl2.

Question 15.
(a) Draw the structures of the following molecules:
(i) XeOF4
(ii) H2SO4
(b) Write the structural difference between white phosphorus and red phosphorus. (CBSE Delhi 2014)
Answer:
(a) (i) XeOF4
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 45

(ii) H2SO4
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 46

(b) White phosphorus consists of P4 units in which four P atoms lie at the corners of a regular tetrahedron with ΔPPP = 60°.
Red phosphorus also consists of P4 tetrahedra units, but it has polymeric structure consisting of P4 tetrahedra linked together by covalent bond.

Question 16.
Account for the following:
(i) PCl5 is more covalent than PCl3.
(ii) Iron on reaction with HCl forms FeCl2 and not FeCl3.
(iii) The two O-O bond lengths in the ozone molecule are equal. (CBSE Delhi 2014)
Answer:
(i) In PCl5, the oxidation state of P is +5 while in PCl3, it is +3. As a result of higher oxidation state of the central atom, PCl5 has larger polarising power and can polarise the chloride ion (Cl) to a greater extent than in the corresponding PCl3. Since larger the polarisation, larger is the covalent character; PCl5 is more covalent than PCl3.

(ii) Iron reacts with HCl to form ferrous chloride with the evolution of hydrogen gas.
Fe + 2HCl → FeCl2 + H2

(iii) The two O-O bond lengths in ozone molecule are equal and are intermediate between single and double bonds. This is because of resonance between two structures (a) and (b) and actual molecule is resonance hybrid of these two structures:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 47

Question 17.
Account for the following:
(a) Moist SO2 decolourises KMnO4 solution.
(b) In general, interhalogen compounds are more reactive than halogens (except fluorine).
(c) Ozone acts as a powerful oxidising agent. (CBSE Sample Paper 2019)
Answer:
Moist sulphur dioxide behaves as a reducing agent. It reduces MnO4 to Mn2+.
5 SO2 + 2MnO4 + 2H2O → 5 SO42- + 4H+ + 2Mn2+

(b) X-X’ bond in inter halogens is weaker than X-X bond in halogens except F-F bond. This is due to the fact that the overlapping of orbitals of two dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(c) Due to the ease with which it liberates atoms of nascent oxygen, Ozone acts as a powerful oxidising agent.
O3 → O2 + O

Question 18.
(a) What happens when
(i) chlorine gas is passed through a hot concentrated solution of NaOH?
(ii) sulphur dioxide gas is passed through an aqueous solution of Fe(lll) salt?
Answer:
(a) (i) Sodium chlorate and sodium chloride are formed.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 48

(ii) Fe (III) salt is reduced to Fe (II) salt.
2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+

(b) Answer the following:
(i) What is the basicity of H3PO3 and why?
(ii) Why does fluorine not play the role of a central atom in interhalogen compounds?
(iii) Why do noble gases have very low boiling points? (CBSE Delhi 2011)
Answer:
(i) H3PO3 is dibasic and has basicity of
two because it has two P-OH bonds which are ionisable. The third H atom is linked to P and is non-ionisable.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 49
H3PO3 ⇌ HPO32- + 2H+

(ii) Fluorine does not play the role of a central atom in interhalogen compounds because it is highly electronegative. Moreover, it has only one electron less than the octet and does not have vacant d-orbitals in its valence shell. Therefore, it can form only one bond with other halogen atoms and cannot act as central atom in interhalogen compounds.

(iii) Noble gases have very low boiling points because only weak van der Waals’ forces are present between the atoms of the noble gases in the liquid state.

Question 19.
(a) Give reasons for the following:
(i) Bond enthalpy of F2 is lower than that of Cl2.
(ii) PH3 has lower boiling point than NH3.
Answer:
(a) (i) Due to small size of F atom, there are strong repulsions between the non-bonding electrons of F atoms in the small sized F2 molecule. Therefore, bond enthalpy of F2 is lower than relatively larger Cl2 molecule in which repulsions between non-bonding electrons are less.

(ii) Ammonia exists as associated molecules due to its tendency to form hydrogen bonding. Therefore, it has high boiling point. Unlike NH3, phosphine (PH3) molecules are not associated through hydrogen bonding in liquid state. This is because of low electronegativity of P than N. As a result, the boiling point of PH3 is lower than that of NH3.

(b) Draw the structures of the following molecules:
(i) BrF3
(ii) (HPO3)3
(iii) XeF4
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 50
BrF3 (T shape)

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 51
(HPO3)3

(iii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 52
XeF4

OR

(a) Account for the following:
(i) Helium is used in diving apparatus.
(ii) Fluorine does not exhibit positive oxidation state.
(iii) Oxygen shows catenation behaviour less than sulphur.
Answer:
(a) (i) Helium alongwith oxygen (helium- oxygen mixture) is used by deep sea divers in preference to nitrogen-oxygen mixture because of its low solubility in blood.

(ii) Fluorine is the most electronegative element and therefore, it shows an oxidation state of -1 only. It does not show any positive oxidation state.

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

(b) Draw the structures of the following molecules.
(i) XeF2
(ii) H2S2O8 (CBSE Delhi 2013)
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 53
XeF2 (Linear)

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 54

Question 20.
(a) Give reasons for the following:
(i) Sulphur in vapour state shows paramagnetic behaviour.
(ii) N-N bond is weaker than P-P bond.
(iii) Ozone is thermodynamically less stable than oxygen.
Answer:
(a) (i) In the vapour state, sulphur exists as S2 molecules. S2 molecule has two unpaired electrons in anti-bonding molecular orbitals (πx and πy) and hence shows paramagnetism.

(ii) The N-N bond is weaker than P-P bond because of high interelectronic repulsions of non-bonding electrons due to small N-N bond length (109 pm).

(iii) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is negative) and increase in entropy (ΔS positive). These two factors make large negative ΔG for its conversion to oxygen.

(b) Write the name of gas released when Cu is added to
(i) dilute HNO3 and
(ii) conc. HNO3
Answer:
(i) With dilute HNO3; NO (nitric oxide)
3Cu + 8HNO3(dil) → 3Cu(NO3)2 + 2NO + 4H2O

(ii) With cone. HNO3: NO2 (nitrogen dioxide)
Cu + 4HNO3(conc) → Cu(NO3)2 + 2NO2 + 2H2O

OR

(a) (i) Write the disproportionation reaction of H3PO3.
(ii) Draw the structure of XeF4.
Answer:
(i) 4H3PO3 → 3H3PO4 + PH3

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 55

(b) Account for the following:
(i) Although fluorine has less negative electron gain enthalpy, yet F2 is strong oxidising agent.
(ii) Acidic character decreases from N2O3 to Bi2O3 in group 15.
Answer:
(i) Fluorine is the strongest oxidising agent, although it has less negative electron gain enthalpy. This is because of its smaller size, large bond dissociation enthalpy of F2 and high exothermic hydration enthalpy of small F ion.

(ii) On moving down the group, the metallic character increases. Therefore, the acidic character of oxides decreases down the group from N2O3 to Bi2O3

(c) Write a chemical reaction to test sulphur dioxide gas. Write chemical equation involved. (CBSE Delhi 2019)
Answer:
SO2 decolourises pink violet colour of acidified potassium permanganate solution.
2KMnO4 + 5SO2 + 2H2O → K2SO4 + 2MnSO4 + 2H2SO4

Question 21.
(a) Complete the following chemical reaction equations:
(i) P4 + SO2Cl4
(ii) XeF6 + H4O →
Answer:
(i) P4 + 10SO2Cl2 → 4PCl5 + 10SO2

(ii) XeF6 + H2O → XeOF4 + 2HF

(b) Predict the shape and the asked angle (90° or more or less) in each of the following cases:
(i) SO32- and the angle O – S – O
(ii) ClF3 and the angle F – Cl – F
(iii) XeF2 and the angle F – Xe – F
Answer:
(i) SO32-
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 56
The O-S-O angle is more than 90°.

(ii) ClF3
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 57
The F-Cl-F bond angle is equal to 90°.

(iii)
XeF2
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 58
F-Xe-F bond angle is equal to 180° (greater than 90°).

OR

(a) Complete the following chemical equations:
(i) NaOH + Cl2
(hot and cone.)
(ii) XeF4 + O2F2
Answer:
(i) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
(Hot and conc.)

(ii) XeF4 + O2F2 → XeF6 + O2

(b) Draw the structures of the following molecules:
(i) H3PO2
(ii) H2S2O7 (CBSE 2012)
Answer:
(i) H3PO2
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 59

(ii) H2S2O7
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 60

Question 22.
(a) Draw the molecular structures of following compounds:
(i) XeF6
(ii) H2S2O8
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 61

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 62

(b) Explain the following observations:
(i) The molecules NH3 and NF3 have dipole moments which are of opposite directions.
(ii) All the bonds in PCl5 molecule are not equivalent.
(iii) Sulphur in vapour state exhibits paramagnetism.
Answer:
(i) Both NH3 and NF3 have pyramidal shape with one lone pair on N atom.
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 63
The lone pair on N is in opposite direction to the N-F bond moments and therefore, it has very low dipole moment (about 0.234 D). But ammonia has high dipole moment because its lone pair is in the same direction as the N-H bond moments.

(ii) PCl5 has trigonal bipyramidal structure in which there are three P-Cl equatorial bonds and two P-Cl axial bonds. The two axial bonds are being repelled by three bond pairs at 90° while the three equatorial bonds are being repelled by two bond pairs at 90°. Therefore, axial bonds are repelled more by bond pairs than equatorial bonds and hence are larger (219 pm).

(iii) In vapour state, sulphur partly exists as S2 molecule and S2 molecule like O2 has two unpaired electrons in anti-bonding π* molecular orbitals. Therefore, it is paramagnetic.

OR

(a) Complete the following chemical equations:
(i) XeF4 + SbF5
(ii) Cl2 + F2 (excess ) →
Answer:
(i) XeF4 + SbF5 → XeF4.SbF5 → [XeF3]+ [SbF6]

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 64

(b) Explain each of the following:
(i) Nitrogen is much less reactive than phosphorus.
(ii) The stability of +5 oxidation state decreases down group 15.
(iii) The bond angles (O – N – O) are not of the same value in NO2 and NO2+. (CBSE Delhi 2012)
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is non-polar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) The stablity of +5 oxidation state decreases down the group because of inert pair effect. Therefore, the +5 oxidation state of Bi is less stable than that of Sb.

(iii) In NO2, there is one electron on N while in NO2+ there is no electron and in NO2+, there is a lone pair of electrons on N as shown below:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 65
NO2+ molecule is linear and has bond angle of 180°. The repulsion by single electron is less as compared to repulsion by a lone pair of electrons. Therefore, bond pairs in NO2 are forces more closer than in NO2.

Question 23.
(a) Draw the structure of the following molecule: H3PO2
Answer:
(a)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 66

(b) Explain the following observations:
(i) Nitrogen is much less reactive than phosphorus.
(ii) Despite having greater polarity, hydrogen fluoride boils at a lower temperature than water.
(iii) Sulphur has greater tendency for catenation than oxygen in the same group.
Answer:
(i) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N = N) and it is nonpolar in character.
Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(ii) Though electronegativity of F is more than O, yet water forms more extensive hydrogen bonding than HF. In H2O, each oxygen is tetrahedrally surrounded by two covalent bonds and two hydrogen bonds, (-O….H). In HF there is only one hydrogen bond (-F….H).

(iii) Sulphur has strong tendency for catenation, i.e. forming bonds with itself in comparison to oxygen. This is because of stronger S-S single bond than O-O single bond.

OR

(a) Draw the structures of the following molecules:
(i) N2O5
(ii) HClO4
Answer:

(a)
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 68

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 69

(b) Explain the following observations:
(i) H2S is more acidic than H2O.
(ii) Fluorine does not exhibit any positive oxidation state.
(iii) Helium forms no real chemical compound. (CBSE 2012)
Answer:
(i) The size of S is more than that of O. Therefore, the distance between S and H, i.e. S-H bond length, is more than O-H bond length. As a result, the bond dissociation enthalpy of S-H will be less and it will be easier to break the bond in H2S than O-H bond in water. Therefore, H2S will be more acidic than H2O.

(ii) Fluorine is the most electronegative element and therefore, it shows oxidation state of – 1 only. It does not show any positive oxidation state.

(iii) Helium does not form compounds because it has very high ionisation enthalpy and smallest size. Its electron gain enthalpy is also almost zero.

Question 24.
Write balanced equations for the following reactions:
(a) P4 + NaOH + H2O → CBSE Delhi 2009)
(b) As4 + Cl2(excess) →
(c) P4O10 + H2O →
(d) Ca3P2 + H2O → (CBSE Delhi 2008, 2014)
(e) POCl3 + H2O →
(f) HgCl2 + PH3 → (CBSE AI 2010)
(g) Ag + PCl5 → (CBSE AI 2014)
Answer:
(a) P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2
(b) As4 + 10Cl4(excess) → 4AsCl5
(c) P4O10 + 6H4O → 4H3PO4
(d) Ca3P2 + 6H2O → 2PH3 + 3 Ca(OH)2
(e) POCl3 + 3H2O → H3PO4 + 3HCl
(f) 3HgCl2 + 2PH3 → Hg3P2 + 6HCl
(g) 2Ag + PCl5 → 2 AgCl + PCl3

Question 25.
Complete the following reactions
(i) NaOH + Cl2
hot and conc.
(ii) NH3 + Cl2 (excess) → (CBSE Delhi 2017)
(iii) NaNO2 + HCl →
(iv) F2(g) + H2O(l) → (CBSE Delhi 2008)
(v) K2CO3 + HCl →
(vi) Cl2 + H2O → (CBSE Delhi 2017)
(vii) F2 + 2Cl → (CBSE Delhi 2017)
Answer:
(i) 6NaOH + 3Cl2 → 5NaCl + NaClO3 + 3H2O
hot and conc.
(ii) NH3 + 3Cl2 (excess) → NCl3 + 3HCl
(iii) 2NaNO2 + 2HCl → 2NaCl + H2O + NO2 + NO
(iv) 2F2(g) + 2H2O(l) → 4H+(aq) + 4F (aq) + O2(g)
(v) K2CO3 + 2HCl → 2KCl + CO2 + H2O
(vi) 2Cl2 + 2H2O → 4HCl + O2
(vii) F2 + 2Cl → 2F + Cl2

Question 26.
Complete the following reactions:
(i) XeF4 + SbF5 → (CBSE Delhi 2012)
(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 70
(CBSE Delhi 2012, CBSE Al 2012, 2014)
(iii) XeF2 + H2O →
(iv) XeF4 + H2O →
(v) XeF6 + H2O → (CBSE Delhi 2012)
(vi) XeF6 + 2H2O → (CfiSE Delhi 2017)
(vii) XeF6 + 3H2O → (CBSE Delhi 2017)
Answer:
(i) XeF4 + SbF5 → [XeF3]+ [SbF6]
(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 71
(iii) XeF2 + H2O → 2Xe + 4HF + O2
(iv) XeF4 + H2O → 4Xe + 2XeO3 + 24HF + 3O2
(v) XeF6 + H2O → XeOF4 + 2HF
(vi) XeF6 + 2H2O → XeO2F2 + 4HF
(vii) XeF6 + 3H2O → XeO3 + 6HF

Question 27.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl5 is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
Answer:
(a) (i) Ozone is thermodynamically unstable with respect to oxygen because it results in liberation of heat (ΔH is -ve) and increase in entropy ΔS is +ve). These two factors reinforce each other resulting in negative ΔG (ΔG = ΔH – TΔS) for its conversion to oxygen.

(ii) PCl5 has trigonal bipyramidal structure and is not very stable. It splits up into more stable tetrahedral and octahedral structures which are stable as
PCl5 ⇌ [PCl4]+ [PCl6]
Therefore, it exists as ionic.

(iii) Due to small size and high electronegativity, fluorine cannot act as central atom in higher oxoacids.

(b) Draw the structure of
(i) BrF5
(ii) XeF4
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 72
Square Pyramidal

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 73

OR

(i) Compare the oxidising action of F2 and Cl2 by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
Answer:
F2 is stronger oxidising agent than Cl2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 74
The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F2 than for Cl2. Hence, F2 is stronger oxidising agent than Cl2.

(ii) Write the conditions to maximise the yield of H2SO4 by contact process.
Answer:
Conditions for maximum yield of H2SO4 by Contact Process:
(a) Low temperature (optimum temperature 720 K)
(b) High pressure (optimum pressure 2 bar)
(c) Presence of catalyst (V2O5 catalyst).

(iii) Arrange the following in the increasing order of property mentioned:
(a) H3PO3, H3PO4, H3PO2 (Reducing character)
(b) NH3, PH3, ASH3, SbH3, BiH3 (Base strength) (CBSE 2016)
Answer:
(a) H3PO2 > H3PO3 > H3PO4
(b) NH3 > PH3 > ASH3 > SbH3 > BiH3

Question 28.
(a) Account for the following:
(i) Acidic character increases from HF to HI.
(ii) There is large difference between the melting and boiling points of oxygen and sulphur.
(iii) Nitrogen does not form pentahalide.
Answer:
(a) (i) In gaseous state, hydrogen halides are covalent. But in aqueous solution, they ionise and behave as acids. The acidic strength of these acids decreases in the order:
HI > HBr > HCl > HF
Thus, HF is the weakest acid and HI is the strongest acid among these hydrogen halides.

The above order of acidic strength is reverse of that expected on the basis of electronegativity. Fluorine is the most electronegative halogen, therefore, the electronegativity difference will be maximum in HF and should decrease gradually as we move towards iodine through chlorine and bromine.

Thus, HF should be most ionic in nature and consequently it should be strongest acid. Although many factors contribute towards the relative acidic strengths, the major factor is the bond dissociation energy. The bond dissociation energy decreases from HF to HI so that HF has maximum bond dissociation energy and HI has the lowest value.

Since H-I bond is weakest, it can be dissociated into H+ and I ions readily while HF can be dissociated with maximum difficulty. Thus, HI is the strongest acid while HF is the weakest acid among the hydrogen halides.

(ii) Oxygen molecule is held by weak van der Waals forces because of the small size and high electronegativity of oxygen. On the other hand, sulphur molecules do not exist as S2 but form polyatomic molecules having eight atoms per molecule (S8) linked by single bonds. Therefore, S atoms are strongly held together by intermolecular forces and its melting point is higher than that of oxygen. Hence, there is large difference in melting and boiling points of oxygen and sulphur.

(iii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation enthalpy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(b) Draw the structures of the following:
(i) CIF3
(ii) XeF4
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 75
T – Shaped molecule

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 76
Square planar molecule

OR

(i) Which allotrope of phosphorus is more reactive and why?
Answer:
White phosphorus is most reactive of all the allotropes because it is unstable due to angular strain on P4 molecule with bond angle of 60°.

(ii) How are the supersonic jet aeroplanes responsible for the depletion of ozone layers?
Answer:
Nitrogen oxide emitted from exhausts of supersonic jet aeroplanes readily combines with ozone to form nitrogen dioxide and diatomic oxygen. Since supersonic jets fly in the stratosphere near ozone layer, they are responsible for the depletion of ozone layer.

(iii) F2 has lower bond dissociation enthalpy than Cl2. Why?
Answer:
Because of small size of fluorineatoms, there are strong electron-electron repulsions between the lone pairs of electrons on F atoms. Hence, bond dissociation enthalpy of F2 is lower than that of Cl2.

(iv) Which noble gas is used in filling balloons for meteorological observations?
Answer:
Helium is used for filling balloons for meteorological observations because it is non-inflammable.

(v) Complete the equation: (CBSE 2015)
XeF2 + PF5
Answer:
XeF2 + PF5 → [XeF]+ [PF6]

Question 29.
(a) Account for the following:
(i) Interhalogens are more reactive than pure halogens.
(ii) N2 is less reactive at room temperature.
(iii) Reducing character increases from NH3 to BiH3.
Answer:
(i) Interhalogen compounds are more reactive than component halogens.
This is because covalent bond between dissimilar atoms in interhalogen compounds is polar and weaker than between similar atoms in halogens (except F-F). This is due to the fact that the overlapping of orbitals of dissimilar atoms is less effective than the overlapping of orbitals of similar atoms.

(ii) In molecular nitrogen, there is a triple bond between two nitrogen atoms (N ≡ N) and it is non-polar in character. Due to the presence of a triple bond, it has very high bond dissociation energy (941.4 kJ mol-1) and therefore, it does not react with other elements under normal conditions and is very unreactive. However, it may react at higher temperatures.

(iii) The reducing character of hydrides of group 15 depends upon the stability of the hydride. On going down the group, the size of the central atom increases and therefore, its tendency to form stable covalent bond with small hydrogen atom decreases.

The greater unstability of a hydride, the greater is its reducing character. Since the stability of the group 15 hydrides decreases from NH3 to BiH3, hence reducing character increases.

(b) Draw the structures of the following:
(i) H4P2O7 (Pyrophosphoric acid)
(ii) XeF4
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 77

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 78

OR

(a) Which poisonous gas is evolved when white phosphorus is heated with conc. NaOH solution? Write the chemical equation involved.
Answer:
Phosphine, PH3
P4 + 3NaOH + 3H2O → 3NaH2PO3 + PH3

(b) Which noble gas has the lowest boiling point?
Answer:
Helium

(c) Fluorine is a stronger oxidising agent than chlorine. Why?
Answer:
Fluorine has lower bond dissociation enthalpy of F-F bond than Cl-Cl bond of Cl2 and high enthalpy of hydration because of smaller size of F ion. As a result it has greater tendency to accept electron in solution and is stronger oxidising agent than chlorine.

(d) What happens when H3PO3 is heated?
Answer:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 79

(e) Complete the equation:
PbS + O3 → (CBSE 2015)
Answer:
PbS + 4O3 → PbSO4 + 4O2

Question 30.
(a) Account for the following observations:
(i) SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
(ii) Chlorine water is a powerful bleaching agent.
(iii) Bi(V) is a stronger oxidising agent than Sb(V).
Answer:
(a) (f) S atom in SF4 is not sterically protected as it is surrounded by only four F atoms, so attack of water molecules can take place easily. In contrast, S atom in SF6 is protected by six F atoms. Thus attack by water molecules cannot take place easily.

(ii) Chlorine water produces nascent oxygen (causes oxidation) which is responsible for bleaching action.
Cl2 + H2O → 2HCl + O

(iii) Due to inert pair effect Bi(V) can accept a pair of electrons to form more stable Bi (III) (+3 oxidation state of Bi is more stable than its +5 oxidation state).

(b) What happens when
(i) White phosphorus is heated with concentrated NaOH solution in an inert atmosphere of CO2.
(ii) XeF6 undergoes partial hydrolysis.
(Give the chemical equations involved).
Answer:
(i) Phosphorus undergoes disproportionation reaction to form phosphine gas.
P4 + 3NaOH + 3H2O → PH3 + 3NaH2PO2

(ii) On partial hydrolysis, XeF6 gives oxyfluoride XeOF4 and HF.
XeF6 + H2O → XeOF4 + 2HF

OR

(a) What inspired N.Bartlett for carrying out reaction between Xe and PtF6?
Answer:
N. Bartlett first prepared a red compound O2+PtF6. He then realised that the first ionisation enthalpy of molecular oxygen was almost identical with Xenon. So he carried out reaction between Xe and PtF6.

(b) Arrange the following in the order of property indicated against each set:
(i) F2, I2, Br2, Cl2 (increasing bond dissociation enthalpy)
(ii) NH3, ASH3, SbH3, BiH3, PH3 (decreasing base strength)
Answer:
(i) I2 < F2 < Br2 < Cl2
(ii) NH3 > PH3 > ASH3 > SbH3 > BiH3

(c) Complete the following equations:
(i) Cl2 + NaOH (cold and dilute) →
(ii) Fe3+ + SO2 + H2O →
(CBSE Sample Paper 2018)
Answer:
(i) 2NaOH + Cl2 → NaCl + NaOCl + H2O
(ii) 2Fe3+ + SO2 + 2H2O → 2Fe2+ + SO42- + 4H+

Question 31.
(a) Give reasons:
(i) H3PO3 undergoes disproportionation reaction but H3PO4 does not.
(ii) When Cl2 reacts with excess of F2, ClF3 is formed and not FCl3.
(iii) Dioxygen is a gas while Sulphur is a solid at room temperature.
Answer:
(i) In +3 oxidation state phosphorus tends to disproportionate to higher and lower oxidation states / Oxidation state of P in H3PO3 is +3 so it undergoes disproportionation but in H3PO4 it is +5 which is the highest oxidation state.

(ii) F cannot show positive oxidation state as it has highest electronegativity/ Because Fluorine cannot expand its covalency / As Fluorine is a small sized atom, it cannot pack three large sized Cl atoms around it.

(iii) Oxygen has multiple bonding due to pπ-pπ bonding whereas sulphur shows catenation. Oxygen is diatomic therefore held by weak intermolecular force while sulphur is polyatomic and held by strong intermolecular forces.

(b) Draw the structures of the following:
(i) XeF4
(ii) HClO3
Answer:
(i)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 80

(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 81

OR

(a) When concentrated sulphuric acid was added to an unknown salt present in a test tube a brown gas (A) was evolved. This gas Intensified when copper turnings were added to this test tube. On cooling, the gas (A) changed into a colourless solid (B).
(i) Identify (A) and (B).
(ii) Write the structures of (A) and (B).
(iii) Why does gas (A) change to solid on cooling?
Answer:
(i) A = NO2, B = N2O4
(ii)
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 82
(iii) NO2 contains an odd electron. So it dimerises to give N2O4.

(b) Arrange the following In the decreasing order of their reducing character:
HF, HCl, HBr, HI
Answer:
HI > HBr > HCl > HF

Complete the following reaction:
XeF4 + SbF5 →, (CBSE 2018)
Answer:
(C) XeF4 + SbF5 → [XeF3] [SbF6]

Question 32.
(i) What happens when
(a) chlorine gas reacts with cold and dilute solution of NaOH?
(b) XeF2 undergoes hydrolysis?
Answer:
(a) 2NaOH + Cl2 → NaCl + NaOCl + H2O
(cold and dilute)
(b) 2XeF2(s) + 2H2O (l) → 2Xe (g) + 4HF(aq) + O2(S)

(ii) Assign suitable reasons for the following:
(a) SF6 Is inert towards hydrolysis.
(b) H3PO3 is diprotic.
(C) Out of noble gases only Xenon is known to form established chemical compounds.
Answer:
(a) Sulphur is sterically protected by six F atoms, hence does not allow the water molecules to attack.

(b) It contains only two ionisable H-atoms which are present as -OH groups, thus behaves as dibasic acid.

(c) Xe has least ionisation energy among the noble gases and hence it forms chemical compounds particularly with O2 and F2.

OR

(i) Considering the parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy, compare the oxidising power of F2 and Cl2.
Answer:
F2 is stronger oxidising agent than Cl2. This can be explained on the basis of bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
The process of oxidising behaviour may be expressed as:
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 83
The overall tendency for the change (i.e. oxidising behaviour) depends upon the net effect of three steps. As energy is required to dissociate or convert molecular halogen into atomic halogen, the enthalpy change for this step is positive. On the other hand, energy is released in steps (II) and (III), therefore, enthalpy change for these steps is negative.

Now although fluorine has less negative electron gain enthalpy, yet it is stronger oxidising agent because of low enthalpy of dissociation and very high enthalpy of hydration. In other words, large amount of energy released in step (III) and lesser amount of energy required in step (I) overweigh the smaller energy released in step (II) for fluorine. As a result, the AH overall is more negative for F2 than for Cl2.
Hence, F2 is stronger oxidising agent than Cl2.

(ii) Complete the following reactions :
(a) Cu + HNO3(dilute) →
(b) Fe3+ + SO2 + H2O →
(c) XeF4 + O2F2 → (CBSE 2018)
Answer:
(a) 3Cu + 8 HNO3 (dilute) → 3Cu(NO3)2 + 2NO + 4H3O
(b) 2Fe3+ + SO3 + 2H3O → 2 Fe2+ + SO42- + 4H+
(c) XeF4 + O2F2 → XeF6 + O2

Question 33.
A crystalline solid ‘A’ burns in air to form a gas ‘B’ which turns lime water milky. The gas is also produced as a by-product during roasting of sulphide ore. This gas decolourises acidified KMnO4 (aq.) solution and reduces Fe3+ to Fe2+. Identify ‘A’ and ‘B’ and write the reactions involved. (CBSE 2019C)
Answer:
A = S8 / Sulphur
S8 + 8 O2 → 8SO2 / S + O2 → SO2
B = SO2
Class 12 Chemistry Important Questions Chapter 7 The p-Block Elements 84
Decolourises KMnO4
2KMnO4 + 5 SO2 + 2H4O → 2H2SO4 + 2MnSO4 + K2SO4 / 2MnO4 + 5SO2 + 2H2O → 4H+ + 2Mn2+ + 5SO42-
Reduces Fe3+ to Fe2+
2Fe3+ + SO2 + 2 H2O → 2 Fe2+ + SO42- + 4H+

OR

Answer the following:
(a) Arrange the following hydrides of Group 16 elements in the decreasing order of their acidic strength:
H2O, H2S, H2Se, H2Te
Answer:
H2Te > H2Se > H2S > H2O

(b) Which one of PCl4+ and PCl4 is not likely to exist and why?
Answer:
PCl4- is not likely to exist because lone pair on P in PCl3 can be donated to Cl+ and not to Cl. Phosphorus has 10e which cannot be accommodated in sp3 orbitals.

(c) Which aliotrope of sulphur is thermally stable at room temperature?
Answer:
Rhombic sulphur is thermally stable at room temperature. Its melting point is 385.8 K. All other varieties of sulphur change into this form on standing. It has low thermal and electrical conductivity.

(d) Write the formula of a compound of phosphorus which is obtained when cone. HNO3 oxidises P4.
Answer:
HNO3 oxidises phosphorus to phosphoric acid. H3PO4 is formed
P4 + 2OHNO3 → 4H3PO4 + 2ONO2 + 4H2O

(e) Why does PCl3 fume in moisture?
Answer:
PCl3 fumes in moist air and reacts with water violently to form phosphorus acid. It hydrolyses in the presence of moisture to give fumes of HCl.
PCl3 + 3H2O → H3PO3 + 3HCl

The d-and f-Block Elements Class 12 Important Extra Questions Chemistry Chapter 8

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 8 The d-and f-Block Elements. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 8 Important Extra Questions The d-and f-Block Elements

The d-and f-Block Elements Important Extra Questions Very Short Answer Type

Question 1.
How would you account for the increasing oxidising power in the series:
V02+ < Cr207 2-< Mn04? (CBSE Sample Paper 2010)
Answer:
This is due to the increasing stability of the lower species to which they are reduced.

Question 2.
Which metal in the first transition series exhibits a +1 oxidation state most frequently and why? (CBSE Delhi 2013)
Answer:
Copper has electronic configuration 3cf104s1. It can easily lose one (4s1) electron to give stable 3d10 configuration.

Question 3.
The magnetic moments of a few transition metal ions are given below:

Metal ion Magnetic moment (BM)
Sc3+ 0.00
Cr2+ 4.90
Ni2+ 2.84
Ti3+ 1.73

(atomic no. Sc = 21, Ti = 22, Cr = 24, Ni = 28)
Which of the given metal ions:
(i) has the maximum number of unpaired electrons?
Answer:
Cr2+

(ii) forms colourless aqueous solution?
Answer:
Sc3+

(iii) exhibits the most stable +3 oxidation state? (CBSE Sample Paper 2017-18)
Answer:
Sc3+

Question 4.
Based on the data, arrange Fe2+, Mn2+ and Cr2+ in the increasing order of stability of +2 oxidation state:
Cr3+/Cr2+ = – 0.4 V, E°Mn3+/Mn2+ = 1.5 V, E°Fe3+/Fe2+ = 0.8 V (CBSE Sample Paper 2011)
Answer:
As the value of reduction potential increases, the stability of +2 oxidation state increases. Therefore, correct order of stability is Cr3+ | Cr2+ < Fe3+ | Fe2+ < Mn3+ | Mn2+.

Question 5.
(i) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
Answer:
Manganese

(ii) Name the element which shows only +3 oxidation state. (CBSEAI2013)
Answer:
Scandium

Question 6.
Identify the oxoanion of chromium which is stable in an acidic medium. (CBSE Sample Paper 2017-18)
Answer:
Cr2072-

Question 7.
Actinoid contraction is greater from element to element than lanthanoid contraction. Why? (CBSE Delhi 2015)
Answer:
This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

Question 8.
Name an important alloy that contains some of the Ianthanoid metals. (CBSE AI 2013)
Answer:
Misch metal

Question 9.
Identify the Ianthanoid element that exhibits a +4 oxidation state. (CBSE Sample Paper 2017-18)
Answer:
Cerium

Question 10.
Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number. (CBSE Delhi 2017)
Answer:
Cr2072-

The d-and f-Block Elements Important Extra Questions Short Answer Type

Question 1.
Assign reasons for the following:
(i) Cu(I) Is not known In an aqueous solution.
Answer:
Because of the Lesser hydration enthalpy of Cu(I), It Is unstable In an aqueous solution and therefore, It undergoes disproportionation.

(ii) Actinolds exhibit a greater range of oxidation states than lanthanoids. (CBSE AI 2011)
Answer:
Lanthanoids show Limited number of oxidation states, such as + 2, + 3 and + 4 + 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5, + 6 and + 7 and neptunium (Z = 94) shows oxidation states of + 3, +4, + 5, + 6 and + 7. This is because of the small energy difference between 5f and 6d orbitals.

Question 2.
Give reasons:
(a) MnO is basic whereas Mn2O7    is acidic in nature.
Answer:
When a metal is in a high oxidation state, its oxide Is acidic and when a metaL is in a low oxidation state its oxide is basic.
The oxides of manganese have the following behavior:
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 1
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 2
(b) Transition metals form alloys. (CBSE 2019C)
Answer:
The transition metals are quite similar in size and, therefore, the atoms of one metal can substitute the atoms of other metal in its crystal lattice. Thus, on cooling a mixture solution of two or more transition metals, solid alloys are formed which is shown in Fig.
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 3
Question 3.
Explain why the Eθ value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ or Fe3+/Fe2+. (CBSE AI 2008, 2010, 2018)
Answer:
Mn2+ has 3d5 electronic configuration. It is stable because of the half-filled configuration of the d-subshell. Therefore, Mn has a very high third ionization enthalpy for the change from d5 to d4 and it is responsible for a much more positive Eθ value for Mn3+/Mn2+ couple in comparison to Cr3+/Cr2+ and Fe3+/ Fe2+ couples.

Question 4.
Complete the following chemical equations:
(i) Cr2O72++ H+ + I
Answer:
Cr2O72+ + 6I + 14H+ → 2Cr3+ + 3I2 + 7H2O

(ii) MnO4+ NO2+ H+ → (CBSE Delhi 2012)
Answer:
2MnO4+ 5N02+ 6H+ → 2Mn2+ + 5NO3+ 3H2O

Question 5.
(i) Which metal in the first transition series (3d series) exhibits +1 oxidatIon state most frequently and why?
Answer:
Copper, because it has [Ar] 3d104s1 electronic configuration. After losing one electron it gets completely filled electronic configuration (3d10).

(ii) Which of the following cations are colored in aqueous solutions and why?
Sc3+, V3+, Ti4+, Mn2+
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (CBSE Delhi 2013)
Answer:
Sc3+: [Ar]  V3+: [Ar] 3d2
Ti4+: [Ar]  Mn2+: [Ar] 3d5
V3+ and Mn2+ cations are colored because they contain partially filled d-orbitals.

Question 6.
Why do transition elements exhibit higher enthalpies of atomization? (CBSE Delhi 2008, 2012)
Answer:
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Question 7.
(a) Actinoid contraction is greater than lanthanoid contraction. Give reason,
Answer:
This is due to poorer shielding by 5f electrons in actinoids as compared to shielding by 4f electrons in lanthanoids. In the Ianthanoid series, as we move from one element to another, the nuclear charge increases by one unit, and one electron are added. The new electrons are added to the same inner 4f-subshells. However, the 4f-electrons shield each other from the nuclear charge quite poorly because of the very diffused shapes of the f-orbitals.

The nuclear charge, however, increases by one at each step. Hence, with increasing atomic number and nuclear charge, the effective nuclear charge experienced by each 4f-electron increases. As a result, the whole of the 4f-electron shell contracts at each successive element, though the decrease is very small.

The actinoid contraction is due to the imperfect shielding of one 5felectron by another in the same subshell. Therefore, as we move along the series, the nuclear charge and the number of 5f-electrons increase by one unit at each step. However, due to imperfect shielding of 5f orbitals, the effective nuclear charge increases which results in contraction of the size.

(b) Out of Fe and Cu, which has a higher melting point and why? (CBSE 2019C)
Answer:
Fe has a higher melting point. The metallic bond is formed due to the interaction of electrons in the outermost orbitals. The strength of bonding is related to the number of unpaired electrons. Fe has more unpaired electrons leading to stronger metallic bonding. So, it has a higher melting point.

Question 8.
How would you account for the following:
(i) Cr2+ is reducing in nature while with the same d-orbital configuration (d4), Mn3+ is an oxidizing agent.
Answer:
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

(ii) In a transition series of metals, the metal which exhibits the greatest number of oxidation states occurs in the middle of the series. (CBSE 2011)
Answer:
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to the lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

Question 9.
When MnO2 is fused with KOH in the presence of KNO3 as an oxidizing agent, it gives a dark green compound (A). Compound (A) disproportionates in an acidic solution to give a purple compound (B). An alkaline solution of compound (B) oxidizes Kl to compound (C), whereas an acidified solution of compound (B) oxidizes Kl to (D). Identify (A), (B), (C), and (D). (CBSE Delhi 2019)
Answer:
When MnO2 is fused with KOH in the presence of KNO3, as an oxidizing agent, it gives a dark green compound (A).
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 4
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 5

Question 10.
State reasons for the following:
(i) Cu(I) is not stable in an aqueous solution.
Answer:
Cu+ ion is not stable in an aqueous solution because of its less negative enthalpy of hydration than that of Cu2+ ion.

(ii) Unlike Cr3+, Mn2+, Fe3+ and the subsequent other M2+ ions of the 3d series of elements, the Ad and the 5d series metals generally do not form stable atomic species. (CBSE 2011)
Answer:
Because of lanthanoid contraction, the expected increase in size does not occur.

Question 11.
Assign reasons for each of the following:
(i) Transition metals generally form colored compounds.
Answer:
Most of the transition metal ions are colored both in the solid-state and in aqueous solutions. The color of these ions is attributed to the presence of an incomplete (n – 1) d-subshell. The electrons in these metal ions can be easily promoted from one energy level to another in the same d-subshell. The amount of energy required to excite the electrons to higher energy states within the same d-subshell corresponds to the energy of certain colors of visible light. Therefore, when white light falls on a transition metal compound, some of its energy corresponding to a certain color is absorbed causing the promotion of d-electrons. This is known as d-d transitions. The remaining colors of white light are transmitted and the compound appears colored.

(ii) Manganese exhibits the highest oxidation state of +7 among the 3d-series of transition elements. (CBSE Delhi 2011)
Answer:
Manganese has the electronic configuration [Ar] 3d5 4s2. It can lose seven electrons due to the participation of 3d and 4s electrons and therefore, exhibits the highest oxidation state of +7 in its compounds.

Question 12.
What are the transition elements? Write two characteristics of the transition elements. (CBSE Delhi 2015)
Answer:
Transition elements are those elements that have incompletely filled (partly filled) d-subshell in their ground state or in any one of their oxidation states.

Characteristics:

  1. Transition elements show variable oxidation states.
  2. They exhibit catalytic properties.

Question 13.
Though both Cr2+ and Mn3+ have d4 configurations, yet Cr2+ is reducing while Mn3+ is oxidizing. Explain why? (CBSE AI 2008, 2012, CBSE Delhi 2012)
Answer:
E° value for Cr3+/Cr2+ is negative (-0.41V), this means that Cr3+ ions are more stable than Cr2+. Therefore, Cr2+ can readily lose electrons to undergo oxidation to form Cr3+ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn3+ Mn2+ is positive (+1.57V), this means that Mn3+ ions can be readily reduced to Mn2+ and hence Mn (III) is strongly oxidizing.

Question 14.
Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state? (CBSE Delhi 2012)
Answer:
The electronic configuration of Mn2+ is [Ar] 3d5 which is half-filled and hence it is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron1 cannot be easily removed. In the case of Fe2+, the electronic configuration is 3d6. Therefore, Fe2+ can easily lose one electron to acquire a 3d5 stable electronic configuration.

Question 15.
The E°M2+|M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior. Why? (CBSE Sample Paper 2012, CBSE AI2012)
Answer:
The E°M2+|M value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+ (aq) is not balanced by its hydration enthalpy.

Question 16.
Chromium is a typical hard metal while mercury is a liquid. (CBSE Sample Paper 2011)
Answer:
In chromium, the M-M interactions are strong due to the presence of six unpaired electrons in the 3d and 4s subshell. On the other hand, in mercury, all the electrons in the 5d and 6s subshell are paired and therefore, the M-M interactions are weak. Therefore, chromium is a typical hard metal while mercury is a liquid.

Question 17.
Silver is a transition metal but zinc is not. (CBSE Sample Paper 2011)
Answer:
According to the definition, transition elements are those which have partially filled d-subshell in their elementary state or in one of the oxidation states. Silver (Z = 47) can exhibit a +2 oxidation state in which it has an incompletely filled d-subshell (4d9 configuration). Hence, silver is regarded as a transition element.

On the other hand, zinc (Z = 30) has the configuration 3d10 4s2. It does not have partially filled d-subshells in its elementary form or in a commonly occurring oxidation state (Zn2+: 3d10). Therefore, it is not regarded as a transition element.

Question 18.
Name the ox metal anions of the first series of the transition metals in which the metal exhibits an oxidation state equal to its group number. (CBSE Delhi 2017)
Answer:
MnO4: Oxidation state of Mn = +7 (equal to its group number)
CrO42-: Oxidation state of Cr = +6 (equal to its group number)

Question 19.
Give reasons:
(a) Of the d4 species, Cr2+ is strongly reducing while Mn3+ is strongly oxidising.
Answer:
Because Cr is more stable in the +3 oxidation state due to the t2g3 configuration whereas Mn is more stable in the +2 oxidation state due to the half-filled 3d5 configuration.

E° value for Cr3+/Cr2+ is negative (-0.41 V); this means that Cr3+ ions are more stable than Cr2+. Therefore, Cr2+ can readily lose electrons to undergo oxidation to form a Cr3+ ion, and hence Cr (II) is strongly reducing. On the other hand, the E° value for Mn3+/ Mn2+ is positive (+1.57V), which means that Mn3+ ions can be readily reduced to Mn2+ and hence Mn (III) is strongly oxidizing.

(b) The d1 configuration is very unstable in ions. (CBSE 2019C)
Answer:
After the loss of ns electrons, d1 electron can easily be lost to give a stable configuration. Therefore, the elements having d1 configuration are either reducing or undergo disproportionation.

Question 20.
When chromite ore FeCr2O4 is fused, with NaOH in presence of air, a yellow-colored compound (A) is obtained, which on acidification with dilute sulphuric acid gives a compound (B). Compound (B) on reaction with KCI forms an orange colored crystalline compound (C).
(i) Write the formulae of the compounds (A), (B), and (C).
(ii) Write one use of the compound (C). (CBSE Delhi 2016)
OR
Complete the following chemical equations:
(i) 8MnO4+ 3S2O32-+ H2O →
(ii) Cr2O72-+ 3Sn2+ + 14H+
Answer:
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 6
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 7
(ii) Potassium dichromate (C) is used as a powerful oxidizing agent in redox titrations in the laboratories.
OR
(i) 8MnO4+ 3S2O32-+ H2O → 8MnO2 + 6SO42- + 2OH
(ii) Cr2O7 2+ + 3Sn2+ +14H+ → 2Cr3+ + 3Sn4+ + 7H2O

Question 21.
Why do the transition elements have higher enthalpies of atomization? In 3d series (Sc to Zn), which element has the lowest enthalpy of atomization and why? (CBSE 2015)
Answer:
The transition elements have high enthalpies of atomization because they have a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

Zinc has the lowest enthalpy of atomization because it has no unpaired electrons and hence weak metallic bonding.

The d-and f-Block Elements Important Extra Questions Long Answer Type

Question 1.
(i) Transition metals and their compounds are generally found to be good catalysts.
(ii) Metal-metal bonding is more frequent for the 4d and the 5d series of transition metals than that for the 3d series. (CBSE AI 2011)
Answer:
(i) Some transition metals and their compounds act as good catalysts for various reactions. This is due to their ability to show multiple oxidation states. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO2 to SO3, V2O5 is used as a catalyst. Solid V2O; absorbs a molecule of SO2 on the surface forming V2O4 and the oxygen is given to SO2 to form S03. The divanadium tetraoxide is then converted to V2O5 by reaction with oxygen:
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 8
(ii) The second and third transition series metals show a great tendency to form strong M-M bonds than the first transition series elements. For example, rhenium easily forms ReCl84- which contains a strong Re-Re bond. It does not have a manganese analog.

Similarly, both Niobium and tantalum form M6X12n+ species and has no vanadium analogue.

Question 2.
Explain the following observations giving an appropriate reason for each.
(i) The enthalpies of atomization of transition elements are quite high.
Answer:
The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interaction and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

(ii) There occurs much more frequent metal-metal bonding in compounds of heavy transition metals (i.e. 3rd series)
Answer:
The heavier transition metals of second and third transition series metals show a great tendency to form strong M-M bonds than first transition series elements. For example, rhenium easily forms ReCl84- which contains a strong Re-Re bond. It does not have a manganese analog. Similarly, both niobium and tantalum form M6X12n+ species and has no vanadium analog.

(iii) Mn2+ is much more resistant than Fe2+ towards oxidation. (CBSE Delhi 2012)
Answer:
Mn2+ is quite stable because it has a stable half-filled d5 electronic configuration and therefore, cannot be easily oxidized. On the other hand, Fe2+ has a less stable d6 electronic configuration and can be easily oxidized to a stable d5 configuration.

Question 3.
How would you account for the following?
(i) Transition metals exhibit variable oxidation states.
(ii) Zr (Z = 40) and Hf (Z = 72) have almost identical radii.
(iii) Transition metals and their compounds act as a catalyst.
OR
Complete the following chemical equations:
(i) Cr2072- + 6Fe2+ + 14H+
(ii) 2Cr042- + 2H+
(iii) 2Mn04 + 5C2042- + 16H+ → (CBSE Delhi 2013)
Answer:
(i) The transition elements exhibit variable oxidation states. The variable oxidation states of transition metals are due to the participation of ns and (n – 1) d-electrons. This is because of the very small difference between the energies of (n -1) d and ns orbitals. For the first five elements, the minimum oxidation state is equal to the number of electrons in the 4s orbitals and the other oxidation states are equal to the sum of 4s and some of the 3d-electrons. The highest oxidation state is equal to the sum of 4s and 3d electrons. For the remaining elements, the minimum oxidation state is equal to electrons in 4s-orbitals and the maximum oxidation state is not equal to the sum of 4s and 3d electrons.

In general, the oxidation state increases up to the middle and then decreases.

(ii) Due to lanthanoid contraction, the increase in radii from the second to third transition series vanishes. Therefore, Zr and Hf have almost the same radii.

(iii) Catalytic properties of transition metal ions. Some transition metals and their compounds act as good catalysts for various reactions. The common examples are Fe, Co, Ni, V, Cr, Mn, Pt, etc. For example, iron-molybdenum is used as a catalyst in Haber’s process for the manufacture of NH3. V205 is used for the oxidation of S02 to S03 in the Contact process for the manufacture of H2S04.

The transition metals form reaction intermediates with the substrate by using empty d-orbitals. These intermediates give reaction paths of lower activation energy and therefore, increase the rate of reaction. For example, during the conversion of SO2 to SO3, V2O5 is used as a catalyst. Solid V2O5 absorbs a molecule of SO2 on the surface forming V2O4 and the oxygen is given to SO2 to form SO3. The divanadium tetraoxide is then converted to V2O5 by reaction with oxygen:
Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 9
OR
(i) 6Fe2+ + Cr2072- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O
(ii) 2CrO42- + 2H+ → Cr2O72-+ H2O
(iii) 2MnO4 + 5C2O42 + 16H+ → 2Mn2+ + 8H2O + 10CO2

Question 4.
Give reasons for the following:
(i) Transition elements and their compounds act as catalysts.
Answer:
Transition metals and their compounds act as good catalysts because of their ability to show multiple oxidation states and form complexes, for example, Fe, Co, Ni, V, Cr, V2O5, etc.

(ii) E° value for (Mn2+|Mn) is negative whereas for (Cu2+| Cu) is positive.
Answer:
The negative value of E° (Mn2+| Mn) is due to the stability of the half-filled d-subshell in Mn2+(3d5). The positive E°(Cu2+|Cu) is due to the high ionization enthalpy and high enthalpy of atomization of copper. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(iii) Actinoids show irregularities in their electronic configuration. (CBSE Delhi 2019)
Answer:
Actinoids have irregularities in the electronic configuration because of almost equal energy of 5f, 6d and 7s orbitals. Therefore, there are some irregularities in the filling of 5f, 6d, and 7s orbitals. The electron may enter either of these orbitals.

Question 5.
(a) Complete the following chemical reactions:
(i) Na2Cr2O7 + KCl →
(ii) 2MnO4+ 5 S032- + 6 H+
Answer:
(i) Na2Cr2O7 + KCl → K2Cr2O7 + 2 NaCl
(ii) 5 S032- + 2MnO4 + 6 H+ → 2Mn2+ + 3H2O + 5SO42-

(b) How does the colour of Cr2O72- change when treated with an alkali? (CBSE2019C)
Answer:
The orange color of Cr2O72- change to yellow due to the formation of chromate ion.
Cr2O72- + 2OH → 2Cr2O72-  + H2O (Yellow)

Question 6.
(a) Write chemical equations involved in the preparation of KMn04 from Mn02.
Answer:
2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3MnO42-+ 4H+ → 2MnO+ MnO2 + 2H2O (or any other correct equation of preparation)

(b) Actinoids show wide range of oxidation states. Why? (CBSE 2019C)
Answer:
Lanthanoids show limited number of oxidation states, such as + 2, + 3 and + 4 (+ 3 is the principal oxidation state). This is because of large energy gap between 5d and 4f subshells. On the other hand, actinoids also show principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7.

This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 7.
How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
Answer:
The metallic radii of the third (5d) series of transition metals are virtually the same as those of corresponding group members of the (4d) series due to the phenomenon of lanthanoid contraction. The steady decrease in atomic and ionic sizes of lanthanoid elements with increasing atomic numbers is known as lanthanoid contraction.

(iii) Lanthanoids form primarily +3 ions, white the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical. (CBSE Delhi 2012)
Answer:
Lanthanoids form primarily +3 ions, while actinoids usually have higher oxidation states, +4 or even +6 in their compounds because in actinoids the 5f, 6d, and 7s energy levels are of comparable energies.

Therefore, all these three subshells can participate.

Question 8.
How would you account for the following:
(i) Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
Answer:
All lanthanoids exhibit a common stable oxidation state of +3. In addition, some lanthanoids show +2 and +4 oxidation states also in solution or in solid compounds. These are shown by these elements which by doing so attain the stable f (empty f-subshell), f (half-filled f-subshell), and f4 (filled f-subshell) configurations. For example,
(a) Ce and Tb exhibit +4 oxidation states. Cerium (Ce) and terbium (Tb) attain f and f configurations respectively when they get +4 oxidation state, as shown below:
Ce4+: [Xe] 4f°
Tb4+: [Xe] 4 f°

(b) Eu and Yb exhibit + 2 oxidation states.
Europium and ytterbium get f and f4 configurations in +2 oxidation state as shown below:
EU2+ : [Xe] 4 f7
Yb2+: [Xe] 4f14

(ii) The E°M2+/M for copper is positive (0.34 V). Copper is the only metal in the first series of transition elements showing this behavior.
Answer:
The E°(M2+/M) value for copper is positive and this shows that it is the least reactive metal out of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(iii) The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series. (CBSE 2012)
Answer:
The metallic radii of the third transition series elements are virtually the same as those of corresponding members of the second transition series because of the lanthanoid contraction. Due to the presence of lanthanoids between the second and third transition series, the expected increase in radii vanishes. Consequently, the pairs of elements Zr-Hf, Nb-Ta, Mo-W, etc have almost similar sizes.

Question 9.
Explain the following observations:
(i) Many of the transition elements are known to form interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(ii) There is a general increase in density from titanium (Z = 22) to copper (Z = 29).
Answer:
There is a gradual increase in density from Ti to Cu because as we move along a transition series from left to right, the atomic radii decrease due to an increase in effective nuclear charge. Therefore, the atomic volume decreases but at the same time atomic mass increases. Therefore, density (mass/volume) increases.

(iii) The members of the actinoid series exhibit a larger number of oxidation states than the corresponding members of the lanthanoid series. (CBSE 2012)
Answer:
Lanthanoids show limited a number of oxidation states, such as + 2, + 3, and + 4 (+ 3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of + 3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of + 3, + 4, + 5 and + 6 and neptunium (Z = 93) shows oxidation states of + 3, + 4, + 5, + 6 and + 7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 10.
Following are the transition metal ions of 3d series.
Ti4+, V2+, Mn3+, Cr3+
(Atomic numbers: Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(i) Which ion is most stable in an aqueous solution and why?
(ii) Which ion is a strong oxidizing agent and why?
(iii) Which ion is colorless and why? (CBSE AI 2017)
Answer:
(i) Cr3+ because of the half-filled t2g3 configuration.
(ii) Mn3+ due to stable d5 configuration of Mn2+
(iii) Ti4+ because it has no unpaired electrons.

Question 11.
Write chemical equations for the following reactions:
(i) Oxidation of nitrite ion by MnO4 in acidic medium.
(ii) Acidification of potassium chromate solution.
(iii) Disproportionation of manganese
(vi) in acidic solution. (CBSE Sample Paper 2011)
Answer:
(i) 5NO2– + 2MnO4+ 6H+ → 2Mn2+ + 3H2O + 5NO3
(ii) 2K2CrO44 + 2H+ → K2Cr2O7 + 2K+ + H2O
(iii) 3MnO42- + 4H+ → 2MnO4 + MnO2 + 2H2O

Question 12.
The actinoids exhibit a larger number of oxidation states than the corresponding members in the lanthanoid series. (CBSE AI 2012, CBSE Delhi 2012)
Answer:
Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4f subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 93) and plutonium (Z = 94) show oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f, 6d, and 7s orbitals.

Question 13.
Explain the following:
(i) Out of Sc3+, Co2+, and Cr3+ ions, only Sc3+ is colorless in aqueous solutions.
(Atomic no.: Co = 27; Sc = 21 and Cr = 24)
(iI) The E°Cu2+/Cu for copper metal is positive (+0.34), unlike the remaining members of the first transition series
(iiI) La(OH)3 is more basic than Lu(OH)3. (CBSE Sample paper 2018)
Answer:
(i) Co2+: [Ar]3d7, Sc3+: [Ar]3d°
Cr3+: [Ar]3d3
Co2+ and Cr3+ have unpaired electrons. Thus, they are colored in an aqueous solution. Sc3+ has no unpaired electron. Thus it is colorless.
(ii) Metal copper has a high enthalpy of atomization and enthalpy of ionization. Therefore the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.
(iii) Due to lanthanoid contraction the size of lanthanoid ion decreases regularly with the increase in atomic size. Thus covalent character between lanthanoid ion and OH’ increases from La3+ to Lu3+. Thus the basic character of hydroxides decreases from La(OH)3 to Lu(OH)3

Question 14.
(a) Complete the following chemical reactions:
(i) 2MnO4+ 5 NO2+ 6 H+
(ii) 3Mn042- + 4 H+
Answer:
(i) 5NO2– + 2MnO4 + 6H+ → 2Mn2+ + 5NO3 + 3H2O
(ii) 3MnO42-+ 4H+ → 2MnO4+ MnO2 + 2H2O

(b) Name a member of the lanthanoid series which shows a +4 oxidation state. (CBSE 2019C)
Answer:
Cerium / Ce

Question 15.
Give reasons:
(i) E° value for Mn3+/Mn2+ couple is much more positive than that for Fe3+/Fe2+.
Answer:
The comparatively high value for Mn shows that Mn2+(d5) is particularly stable/Much larger third ionization energy of Mn (where the required change is from d5 to d4).

(ii) Iron has a higher enthalpy of atomization than copper.
Answer:
Due to the higher number of unpaired electrons.

(iii) Sc3+ is colorless in an aqueous solution whereas Ti3+ is colored. (CBSE 2018)
Answer:
The absence of unpaired d-electron in Sc3+ whereas in Ti3+ there is one unpaired electron or Ti3+ shows the d-d transition.

Question 16.
(i) Complete the following chemical equations for reactions:
(a) MnO4(aq) + S2O32- (aq) + H2O (I) →
(b) Cr2O7(aq) + H2S(g) + H+(aq) →
Answer:
(a) 8MnO4 (aq) + 3S2O32-  (aq) + H2O (l) → 8MnO2 (s) + 6SO42- (aq) + 2OH- (aq)
(b) Cr2O72- (aq) + 3H2S(g) + 8H+ (aq) → 2Cr3+(aq) + 3S(s) + 7H2O

(ii) Give an explanation for each of the following observations:
(a) The gradual decrease in size (actinoid contraction) from element to element is greater among the actinoids than that among the lanthanoids (lanthanoid contraction).
Answer:
(a) This is because of relatively poor shielding by 5f electrons in actinoids in comparison with shielding of 4f electrons in lanthanoids.

(b) The greatest number of oxidation states are exhibited by the members in the middle of a transition series.
Answer:
This is because, in the middle of the transition series, the maximum number of electrons are available for sharing with others. The small number of oxidation states at the extreme left side is due to a lesser number of electrons to lose or share. On the other hand, at the extreme right-hand side, due to a large number of electrons, only a few orbitals are available in which the electrons can share with others for higher valence.

(c) With the same d-orbital configuration (d4) Cr2+ ion is a reducing agent but the Mn3+ ion is an oxidizing agent. (CBSE Delhi 2009)
Answer:
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

Question 17.
(i) Complete the following chemical reaction equations:
(a) Fe2+ (aq) + MnO4 (aq) + H+ (aq) →
(b) Cr2O72- (aq) + I (aq) + H+ (aq) →
Answer:
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(l)
(b) Cr2O72- (aq) + 6I (aq) + 14H+ (aq) → 2Cr3+(aq) + 3I2 (s) + 7H2O(l)

(ii) Explain the following observations:
(a) Transition elements are known to form many interstitial compounds.
Answer:
Transition metals form interstitial compounds. Transition metals have a unique character to form interstitial compounds with small non-metallic elements such as hydrogen, boron, carbon, and nitrogen. The small atoms of these non-metallic elements (H, B, C, N, etc.) fit into the vacant spaces of the lattices of the transition metal atoms. As a result of the filling up of the interstitial spaces, the transition metals become rigid and hard.

These interstitial compounds have similar chemical properties as the parent metals but have different physical properties, particularly, density, hardness, and conductivity. For example, steel and cast iron are hard because of the formation of interstitial compounds with carbon. These interstitial compounds have a variable composition and cannot be expressed by a simple formula. Therefore, these are called nonstoichiometric compounds.

(b) With the same d4 d-orbital configuration Cr2+ ion is reducing while the Mn3+ ion is oxidizing.
Answer:
Cr2+ is reducing as its configuration changes from d4 to d3, the latter having a half-filled t2g level. On the other hand, the change from Mn3+ to Mn2+ results in the half-filled (d5) configuration which has extra stability.

(c) The enthalpies of atomization of the transition elements are quite high. (CBSE Delhi 2009)
Answer:
The transition metals have strong interactions of electrons in the outermost orbitals and therefore, have high melting and boiling points. These suggest that the atoms of these elements are held together by strong forces and have high enthalpy of atomization.

Question 18.
(i) Complete the following chemical equations:
(a) MnO4(aq) + S2O32- (aq) + H2O(l) →
(b) Cr2O72- (aq) + Fe2+ (aq) + H+ (aq) →
Answer:
(a) 8MnO4 + 3S2O32- + H2O → 8MnO2 + 6SO42-+ 2OH-
(b) Cr2O72- + 14H+ + 6Fe2+ → 2Cr3+ +6Fe3+ + 7H2O

(ii) Explain the following observations:
(a) La3+ (Z = 57) and Lu3+ (Z = 71) do not show any colour in solutions.
Answer:
In La3+ there are no f-electrons and in Lu3+ the 4f sub-shell is complete (4/14). Therefore, there are no unpaired electrons and consequently, d-d transitions are not possible. Hence, La3+ and LU3+ do not show any color in solutions.

(b) Among the divalent cations in the first series of transition elements, manganese exhibits the maximum paramagnetism.
Answer:
Among the divalent cations in the first
transition series, Mn (Z = 25) exhibits the maximum paramagnetism because it has the maximum number of unpaired electrons (3d5): five unpaired electrons.

(c) Cu+ ion is not known in aqueous solutions. (CBSE 2010)
Answer:
In an aqueous solution, Cu+ undergoes disproportionation changing to Cu2+ ion.
2Cu+ → Cu2+ + Cu

Question 19.
(i) Give reasons for the following:
(a) Mn3+ is a good oxidizing agent.
(b) E°M2+/M values are not regular for first-row transition metals (3d series).
(c) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the highest oxide is Mn207.
(ii) Complete the following equations:
(a) 2CrO42- + 2H+
(b) KMnO4 Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 10
OR
(i) (a) Why do transition elements show variable oxidation states?
(b) Name the element showing a maximum number of oxidation states among the first series of transition metals from Sc(Z = 21) to Zn (Z = 30).
(c) Name the element which shows only the +3 oxidation state.
(ii) What is lanthanoid contraction? Name an important alloy that contains some of the lanthanoid metals. (CBSE2013)
Answer:
(i) (a) Mn3+ (3d4) on changing to Mn2+ (3d5) becomes stable, half-filled configuration has extra stability. Therefore, Mn3+ can be easily reduced and acts as a good oxidizing agent.

(b) E°(M2+/M) values are not regular in the first transition series metals because of irregular variation of ionization enthalpies (IE1 + IE2) and the sublimation energies.

(c) Among transition elements, the bonds formed in +2 and +3 oxidation states are mostly ionic. The compounds formed in higher oxidation states are generally formed by sharing of d-electrons. Therefore, Mn can form MnO4- which has multiple bonds also, while fluorine cannot form multiple bonds.

(ii) (a) 2CrO42- + 2H+ → Cr2O72- + H2O
(b) 2KMnO4 Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 10 K2MnO4 + MnO2 + O2
OR
(i) (a) The transition elements show variable oxidation states because their atoms can lose a different number of electrons. This is due to the participation of inner (n -1). d-electrons in addition to outer ns electrons because the energies of the ns and (n – 1) d-subshells are almost equal.

(b) Manganese

(c) Scandium

(ii) The steady decrease in atomic and ionic sizes of lanthanide elements with increasing atomic numbers is called lanthanide contraction. In the lanthanoids, there is a regular decrease in the size of atoms and ions with an increase in atomic number. For example, the ionic radii decrease from Ce3+ (111 pm) to Lu3+ (93 pm).

Cause of lanthanoid contraction. As we move through the lanthanoid series, 4f-electrons are being added, one at each step. The mutual shielding effect of electrons is very little, even smaller than that of d-electrons. This is due to the shape of f-orbitals. The nuclear charge, however, increases by one at each step. Hence, the inward pull experienced by the 4f-electrons increases. This causes a reduction in the size of the entire 4f shell. The sum of the successive reductions gives the total lanthanoid contraction. The important alloy is mischmetal.

Question 20.
(i) How do you prepare:
(a) K2MnO4 from MnO2?
(b) Na2Cr2O7 from Na2CrO4?
(ii) Account for the following:
(a) Mn2+ is more stable than Fe2+ towards oxidation to +3 state.
(b) The enthalpy of atomization is lowest for Zn in the 3d series of the transition elements.
(c) Actinoid elements show a wide range of oxidation states.
OR
(a) Name the element of 3d transition series which shows a maximum number of oxidation states. Why does it show so?
(b) Which transition metal of 3d series has a positive E°(M2+/M) value and why?
(c) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(d) Name a member of the lanthanoid series which is well known to exhibit a +2 oxidation state.
(e) Complete the following equation:
Mn04+ 8H+ +5e → (CBSE Delhi 2014)
Answer:
(i) (a) When Mn02 is fused with caustic potash (KOH) or potassium carbonate in the presence of air, a green mass of potassium manganate is formed.
2MnO2 + 4KOH + O2 Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 10 4 2K2MnO4 + 2H2O
Or
2MnO2 + 2K2CO2 + O2 → 2K2MnO4 + 2CO2.

(b) Na2Cr04 is acidified with dilute sulphuric acid to get Na2Cr2O7.
2Na2CrO4 + H2SO4 → Na2Cr2O7 + Na2SO4 + H2O

(ii) (a) Electronic configuration of Mn2+ is [Ar]Bd5 which is half-filled and hence is stable. Therefore, the third ionization enthalpy is very high, i.e. the third electron cannot be easily removed. In the case of Fe2+, the electronic configuration is 3d6. Therefore, Fe2+ can easily lose one electron to acquire a 3d5 stable configuration. Thus, Mn2+ is more stable than Fe2+ towards oxidation to + 3 states.

(b) The high enthalpies of atomization of transition elements are due to the participation of electrons (n – 1) d-orbitals in addition to ns electrons in the interatomic metallic bonding. In the case of zinc, no electrons from 3d-orbitals are involved in the formation of metallic bonds. On the other hand, in all other metals of 3d series electrons from d-orbitals are always involved in the formation of metallic bonds.

(c) Lanthanoids show a limited number of oxidation states, such as +2, +3, and +4 (+3 is the principal oxidation state). This is because of the large energy gap between 5d and 4/ subshells. On the other hand, actinoids also show a principal oxidation state of +3 but show a number of other oxidation states also. For example, uranium (Z = 92) exhibits oxidation states of +3, +4, +5 and +6 and neptunium (Z = 94) shows oxidation states of +3, +4, +5, +6 and +7. This is because of the small energy difference between 5f and 6d orbitals.
OR
(a) Manganese (Z = 25) shows the largest number of oxidation states because it has the maximum number of unpaired electrons. It shows oxidation states from +2 to +7, i.e. +2, +3, +4, +5, +6 and +7.

(b) The E°(M2+|M) value for copper is positive and this shows that it is the least reactive metal among the elements of the first transition series. This is because copper has a high enthalpy of atomization and enthalpy of ionization. Therefore, the high energy required to convert Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

(c) Mn3+ is a stronger oxidizing agent than Cr3+ because Mn3+(3d4) changes to stable half-filled (3d5) configuration while in the case of Cr3+ (3d5) is stable (half-filled) and it cannot be readily changed to Cr2+ (3d4).

(d) Europium.

(e) MnO4 + 8H+ + 5e → Mn2+ + 4H2O.

Question 21.
(a) Account for the following:
(i) Manganese shows the maximum number of oxidation states in 3d series.
(ii) E° value for Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+.
(iii) Ti4+ is colorless whereas V4+ is colored in an aqueous solution.
(b) Write the chemical equations for the preparation of KMn04 from Mn02. Why does the purple color of acidified permanganate solution decolorize when it oxidizes Fe2+ to Fe3+?
OR
(a) Write one difference between transition elements and p-block elements with reference to variability of oxidation states.
(b) Why do transition metals exhibit higher enthalpies of atomization?
(c) Name an element of the lanthanoid series which is well known to show a +4 oxidation state. Is it a strong oxidizing agent or a reducing agent?
(d) What is lanthanoid contraction? Write its one consequence.
(e) Write the ionic equation showing the oxidation of Fe(ll) salt by acidified dichromate solution. (CBSE Al 2019)
Answer:
(a) (i) Manganese has the electronic configuration: 3d5 4s2. It has a maximum number of unpaired electrons and hence shows maximum oxidation states.
(ii) Mn2+ has 3d5 electronic configuration. It is stable because of the half-filled configuration. Therefore, Mn3 easily gets reduced to Mn2+. Thus, E°(M3+/Mn2+) is positive. On the other hand, Cr3+ is more stable in the +3 oxidation state due to stable t2g3 configuration.
(iii) Ti4+ (3d°) does not have any d-electrons. Therefore, there are no d-d transitions whereas V4+, (3d1) has one electron in d-subshell and d-d transitions are possible and hence it is colored.

(b) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
3K2MnO4 + 4HCL → 2KMnO + MnO2 + 2H2O

When acidified KMnO4 oxidizes Fe2+ to Fe3+ its purple color gets decolorized due to the formation of Mn2+ from
MnO4– ion.
MnO4– + 5Fe2+ + 8W → Mn2+ + 5Fe3+ + 4H2O
Or
(a) Transition elements, in general, show variable oxidation states which differ by 1 unit, whereas p-block elements show variable oxidation states which differ by 2 units.

(b) Transition eLements have unpaired d-electron and therefore have strong metallic bonding between atoms. Hence, they have high enthalpies of atomization.

(c) Cerium. It is a strong oxidizing agent.

(d) The regular decrease In atomic and ionic radii of the Lanthanides with increasing atomic number is catted Lanthanoid contraction.

Consequence: 5d series elements have almost the same size as the 4d series.

(e) 6Fe2+ + Cr2O72- + 14H+ → 6Fe3+ + 2Cr3+ + 7H2O

Question 22.
(i) Account for the following:
(a) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(b) Zirconium and Hafnium exhibit similar properties.
(c) Transition metals act as catalysts.
(ii) Complete the following equations:
(a) 2MnO2 + 4KOH + O2 Class 12 Chemistry Important Questions Chapter 8 The d-and f-Block Elements 11
(b) Cr2O72- + 14H+ + 6I
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(a) Write the element which is not regarded as a transition element. Give reason.
(b) Which element has the highest m.p.?
(c) Write the element which can show an oxidation state of +1.
(d) Which element is a strong oxidizing agent in the +3 oxidation state and why? (CBSE 2016)
Answer:
(i) (a) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation state.
(b) Due to lanthanoid contraction, Zr and Hf show similar properties.
(c) The transition metals act as catalysts. This is due to their ability to show multiple oxidation states.

(ii) (a) 2MnO2 + 4KOH + O2 → 2K2MnO4 + 2H2O
(b) Cr2O72- + 14H+ + 6I → 2Cr3+ + 7H2O + 3I2
Or
(a) Zn because of not having partially filled d-orbitals in its ground state or ionic state.
(b) Chromium, Cr
(c) Copper, Cu
(d) Mn, because Mn2+ has extra stability due to half-filled d-subshell.
Mn: [Ar] 3d5 4s2; Mn2+: [Ar]3d5
Mn3+ has four electrons (3d4) in 3d subshell and requires only one electron to get a half-filled 3d subshell and therefore, acts as a strong oxidizing agent.

Haloalkanes and Haloarenes Class 12 Important Extra Questions Chemistry Chapter 10

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 10 Haloalkanes and Haloarenes.  Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 10 Important Extra Questions Haloalkanes and Haloarenes

Haloalkanes and Haloarenes Important Extra Questions Very Short Answer Type

Question 1.
Define ambident nucleophile with an example. (CBSE Delhi 2019)
Answer:
The nucleophile which has two sites through which it can attack is called ambident nucleophile, for example, nitrite ion (NO2). It can attack either through 0 (-O -N = O) or through N
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 1
Question 2.
What happens when bromine attacks CH2 = CH – CH2 -C ≡ CH? (CBSE AI2012)
Answer:
The Colour of bromine gets discharged.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 2
Question 3.
What happens when CH3 – Br is treated with KCN? (CBSE Delhi 2013)
Answer:
Ethane nitrile is formed.
CH3Br + KCN → CH3C ≡ N + KBr (Ethane nitrile)

Question 4.
Write the IUPAC name of
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 3
Answer:
4-Bromo-4-methylpent-2-ene

Question 5.
What happens when ethyl chloride is treated with aqueous KOH? (CBSE Delhi 2013)
Answer:
Ethyl alcohol is formed.
CH3CH2Cl + KOH(aq) → CH3CH2OH + KCl Ethyl alcohol

Question 6.
Write the IUPAC name of (CH3)2CH.CH(Cl)CH3. (CBSE Delhi 2013)
Answer:
2-Chloro-3-methyl butane

Question 7.
Which compound in the following pair undergoes a faster SN1 reaction? (CBSE Delhi 2013)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 4
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 5

Question 8.
Write the IUPAC name of the following compound: (CBSE AI 2013)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 6
Answer:
3-Chloro-2, 2-dimethylbutane

Question 9.
Which aerosol depletes the ozone layer? (CBSE AI 2013)
Answer:
Chlorofluoro compounds of methane (CCl2F2) called freons.

Question 10.
Write the IUPAC name of the following compound: (CBSE AI 2013)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 7
Answer:
2-Bromo-4-chloropentane

Question 11.
Why is CH2 = CH – CH2 – Cl more easily hydrolysed than CH3 – CH2 – CH2 – Cl? (CBSE AI 2019)
Answer:
CH2 = CH – CH2 – Cl gives allylic carbocation as intermediate, which is resonance stabilised.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 8
Therefore, CH2 = CH – CH2 – Cl is more readily hydrolysed.

Question 12.
Write the IUPAC name of the following compound:
CH2 = CHCH2Br (CBSE 2011)
Answer:
3-Bromoprop-1-ene

Question 13.
Write the IUPAC name of the following compound: (CH2 )3CCH2Br (CBSE 2011)
Answer:
1 -Bromo 2,2-dimethylpropane

Question 14.
Write the IUPAC name of the following: (CBSE 2012)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 9
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 10
3-Bromo- 2-methylpropene

Question 15.
Identify the chiral molecule in the following pair: (CBSE 2014)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 11
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 12
2-Chlorobutane is a chiral molecule.

Question 16.
Which would undergo SN2 reaction faster in the following pair and why? (CBSE Delhi 2015)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 13
Answer:
CH3CH2Br would undergo SN2 reaction faster because of less steric hindrance.

Question 17.
Out Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 14andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 15 which is more reactive towards SN1 reaction and why? (CBSE Delhi 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 16 is more reactive.
The SN1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 17
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 18
Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the SN1 reaction.

Question 18.
Write the structure of the product formed when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether. (CBSE Al 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 19

Question 19.
Which of the following two reactions is SN2 and why? (CBSE AI2016)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 20
Answer:
Reaction (i) is SN2 because it proceeds by inversion of configuration.

Question 20.
Why is chloroform kept in dark coloured bottles? (CBSE AI 2019)
Answer:
Chloroform reacts with atmospheric oxygen in presence of light to produce an extremely poisonous gas called phosgene.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 21
Chloroform is stored in dark coloured bottles so that it is not exposed to light and the formation of phosgene is prevented.

Question 21.
Write one stereochemical difference between SN1 and SN2 reactions. (CBSE Delhi 2019)
Answer:
SN1 reaction forms a racemic mixture while SN2 reaction results in inversion of configuration.

Question 22.
A solution of KOH hydrolyses CH3CHClCH2CH3 and CH3CH2CH2CH2Cl. Which one of these is more easily hydrolysed? (CBSE Delhi 2010)
Answer:
CH3CHClCH2CH3

Question 23.
Write the structure of 2, 4-dinitrochlorobenzene. (CBSE Delhi 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 22

Question 24.
Write the structure of 1-Bromo-4- chlorobut-2-ene. (CBSE Delhi 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 23

Question 25.
Write the structure of 3-Bromo-2- methylprop-1 -ene. (CBSE Delhi 2017)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 24

Question 26.
Write IUPAC name of the given compound: (CBSE Delhi 2019)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 25
Answer:
4-Chlorobenzene sulphonic acid

Question 27.
Out of chlorobenzene and benzyl chloride, which one gets easily hydrolysed by aqueous NaOH and why? (CBSE Sample paper 2018)
Answer:
Benzyl chloride;
Due to resonance, stable benzyl carbocation is formed.

Haloalkanes and Haloarenes Important Extra Questions Short Answer Type

Question 1.
Which one of the following compounds will undergo hydrolysis at a faster rate by SN1 mechanism? Justify. (CBSE Sample Paper 2019)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 26
Answer:
The following compound will undergo SN1 faster:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 27
Greater the stability of the carbocation, greater will be its ease of formation from the corresponding halide and faster will be the rate of reaction. The benzylic carbocation formed gets stabilised through resonance.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 28
CH3CH2CH2Cl forms 1° carbocation, which is less stable than benzylic carbocation.

Question 2.
How would you differentiate between SN1 and SN2 mechanisms of substitution reactions? Give one example of each. (CBSE 2010)
Answer:
The SN1 reaction occurs in two steps and the reaction is of the first order.
The SN2 reaction occurs in one step and the reaction is of second order. These can also be distinguished as:

In SN1 reaction retention of the configuration takes place while in SN2 reaction inversion of the configuration takes place.

SN2 reaction:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 29

SN1 reaction:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 30

Question 3.
(i) State one use each of DDT and iodoform.
Answer:
DDT is used as an insecticide for killing mosquitoes and other insects. chloroform is used as an antiseptic.

(ii) Which compound in the following couples will react faster in SN2 displacement and why?
(a) I -Bromopentane or 2-bromopentane
Answer:
1-Bromopentane (1°) will react faster than 2-bromopentane (2°) by SN2 displacement because It is Less stericaLly hindered in the transition state.

(b) I-Bromo-2-methyl butane or 2-Bromo-2-methyl butane (CBSE Delhi 2010)
Answer:
1 -Bromo-2-methyl butane (1°) reacts faster than 2-bron,o-2-methyl butane (2°) because being primary, it will have Less steric hindrance.

Question 4.
Although chlorine Is an electron-withdrawing group, yet It Is ortho-, para-directing in electrophilic aromatic substitution reactions. Explain why Is It so? (CBSE 2012)
Answer:
Chlorine is an electron-withdrawing group, yet it is ortho-, para-directing in eLectrophiLic aromatic substitution reactions due to eLectron reLeasing resonance effect (+ R-effect).
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 31
As can be seen from the above resonating structures that -Ct increases electron density at ortho- and para-positions due to the + R-effect. Therefore, the attack of electrophile takes pLace at ortho- and paro- positions and not at meta- positions.

Question 5.
(i) Which alkyl halide from the following pair Is chiral and undergoes a faster SN2 reaction? (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 32
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 33 is chiral

(ii) Out of SN1 and SN2 which reaction occurs with
(a) Inversion of configuration
Answer:
SN2

(b) Racemisation
Answer:
SN1

Question 6.
Draw the structure of major moon halo product In each of the following reactions: (CBSE Delhi 2014)
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 34
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 35

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 36
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 37

Question 7.
Propose the mechanism of the reaction taking place when
(i) (-) – 2 – Bromooctane reacts with sodium hydroxide to form (+) – octane-2-ol.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 38

(ii) 2-Bromopentane is heated with (ale.) KOH to form alkenes. (CBSE Sample Paper 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 39

Question 8.
Out of Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 40 andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 41 which is more reactive towards SN1 reaction
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 42 is more reactive.
The SN1 reaction proceeds through the formation of a carbocation. The compound which forms a more stable carbocation will be more reactive.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 43
Since 2° carbocation is more stable than 1° carbocation, 2-Chlorobutane will be more reactive towards the SN1 reaction.

Haloalkanes and Haloarenes Important Extra Questions Long Answer Type

Question 1.
(i) Out of (CH3)3C-Br, and (CH3)3C-I, which one is more reactive towards SN1 and why?
Answer:
(CH3)2 C-I, because it can readily form stable carbocation because the C – l bond is weaker than the C – Br bond.

(ii) Write the product formed when p-nitro chlorobenzene is heated with aqueous NaOH at 443 K followed by acidification.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 44

(iii) Why are dextro- and laevorotatory isomers of Butan-2-ol difficult to separate by fractional distillation? (CBSE Delhi 2019)
Answer:
Dextro- and laevorotatory isomers of butan-2-ol are enantiomers of each other and both have the same boiling point. Hence, these cannot be separated by fractional distillation.

Question 2.
Give reasons for the following:
(i) Ethyl iodide undergoes SN2 reaction faster than ethyl bromide.
Answer:
Ethyl iodide undergoes SN2 reaction faster than ethyl bromide because I- ion is a better leaving group than Br- ion.

(ii) (±) 2 – Butanol is optically inactive.
Answer:
(±) 2 – Butanol represents a racemic mixture of (+) 2-butanol and (-) 2-butanol which rotate the plane polarized light in different directions but to an equal extent. Therefore, the (±) compound is optically inactive.

(iii) C—X bond length in halobenzene is smaller than C—X bond length in CH3 — X. (CBSE 2013)
Answer:
In halobenzene, there is delocalisation of electrons due to resonance. For example, chlorobenzene is considered to be a resonance hybrid of the following structures:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 45
It is evident that the contribution of structures III, IV and V imparts a partial double bond character to the carbon-chlorine bond. Therefore, the C-X bond length in halobenzene is less than the C-X bond length in CH3-X, which has only a single C-X bond.

Question 3.
(i) Draw the structures of major moon halo products in each of the following reactions:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 46
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 47

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 48
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 49

(ii) Which halogen compound in each of the following pairs will react faster in SN2 reaction:
(a) CH3Br or CH3I
Answer:
CH3I

(b) (CH3)3C – Cl or CH3 – Cl (CBSE 2014)
Answer:
CH3Cl

Question 4.
How would you convert the following:
(i) Prop-1-ene to 1-fluoropropane
(ii) Chlorobenzene to 2-chlorotoluene
(iii) Ethanol to propanenitrile
OR
Write the main products when
(i) n-butyl chloride is treated with alcoholic KOH.
(ii) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
(iii) methyl chloride is treated with AgCN.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 50
OR
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 51

Question 5.
Give reasons:
(i) C – Cl bond length in chlorobenzene is shorter than C — Cl bond length in CH3 — Cl.
Answer:
In chlorobenzene, there is delocalisation of electrons due to resonance.
It is considered to be a resonance hybrid of the following structures.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 52
The contribution of structure III, IV and V imparts a partial double bond character to the C – Cl bond.

This is confirmed by X-ray analysis which shows that the C – Cl bond length in chlorobenzene is 1.69 whereas the C – Cl bond length in ethyl chloride is 1.82 A. There is no resonance in CH3Cl and hence no partial double bond character.

(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
Answer:
In chlorobenzene, the C of C-Cl bond is sp2-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 53
Therefore, the sp2-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp3-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp3 hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride. Thus, chlorobenzene is less polar than cyclohexyl chloride.

(iii) SN1 reactions are accompanied by racemisation in optically active alkyl halides. (CBSE 2016)
Answer:
The SN1 reactions proceed through the formation of carbocations. In the case of optically active alkyl halides, the product formed is a racemic mixture. This is because the intermediate carbocation is planar species, therefore, the attack of the nucleophile, OFT ion can take place from both the faces (front and rear) with equal ease. As a result, a 50: 50 mixture of the two enantiomers (laevo and dextro) is formed. Thus, the product formed is a racemic mixture (±) which is optically inactive.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 54
For example, hydrolysis of optically active 2-bromobutane results in the formation of a racemic mixture (±)-butan-2-ol.

Question 6.
The following compounds are given to you:
2-Bromopentane, 2-Bromo-2-methyl butane, 1-Bromopentane
(i) Write the compound which is most reactive towards the SN2 reaction.
Answer:
1-Bromopentane

(ii) Write the compound which is optically active.
Answer:
2-Bromopentane

(iii) Write the compound which is most reactive towards the SN2 elimination reaction.
Answer:
2-Bromo-2-methyl butane

Question 7.
Explain why
(i) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride. (CBSE Delhi 2016)
Answer:
In chlorobenzene, the C of C-Cl bond is sp2-hybridised while the C of C-Cl bond in cyclohexyl chloride is sp3-hybridised.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 55
Therefore, the sp2-hybridised C of chlorobenzene has more s-character and hence more electronegative than the sp3-hybrid C of cyclohexyl chloride. As a result, the sp2 hybrid of the C-Cl bond in chlorobenzene has less tendency to release electrons to Cl than the sp3 hybrid carbon of cyclohexyl chloride. As a result, the C-Cl bond in chlorobenzene is less polar than in cyclohexyl chloride.

Thus, chlorobenzene is less polar than cyclohexyl chloride. In other words, the magnitude of negative charge (δ-) is less on the Cl atom of chlorobenzene than in cyclohexyl chloride. Further, due to the delocalisation of the lone pair of electrons of the Cl atom over the benzene ring due to resonance, the C-Cl bond in chlorobenzene acquires some double-bond character. On the other hand, the C-Cl bond in cyclohexyl chloride is a pure single bond.

Since dipole moment is a product of charge and distance, therefore, chlorobenzene has a lower dipole moment than cyclohexyl chloride due to the lower magnitude of charge (δ-) on Cl atom and small C-Cl distance.

(ii) alkyl halides, though polar, are immiscible with water,
Answer:
Alkyl halides are polar molecules and therefore, their molecules are held together by dipole-dipole forces. On the other hand, the molecules of H20 are held together by hydrogen bonds. When alkyl halides are added to water, the new forces of attraction between water and alkyl halide molecules are weaker than the forces of attraction already existing between alkyl halide-alkyl halide molecules and water-water molecules. Hence, alkyl halides are immiscible (not soluble) in water.

(iii) Grignard reagents should be prepared under anhydrous conditions? (CBSE Sample Paper 2011)
Answer:
Grignard reagents are very reactive. They react with the moisture present in the apparatus or the starting materials (RX or Mg).
RMgX + HOH → R-H + Mg(OH)X
Therefore, Grignard reagents must be prepared in anhydrous conditions.

Question 8.
Which one of the following compounds will undergo a faster hydrolysis reaction by the SN1 mechanism? Justify your answer.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 56
OR
A compound is formed by the substitution of two chlorine atoms for two hydrogen atoms in propane. Write the structures of the isomers possible. Give the IUPAC name of the isomer which can exhibit enantiomerism. (CBSE Sample paper 2018-19)
Answer:
C6 H5CH2Cl will undergo SN1 reaction faster.

The carbocation formed by C6 H5CH2Cl gets stabilised through resonance.
Greater the stability of carbocation, greater will be its ease of formation from the respective halide.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 57
OR
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 58
The following isomer will exhibit enantiomerism:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 59
IUPAC name: 1,2-Dichloropropane.

Question 9.
(i) Identify the chiral molecule in the following pair:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 60 andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 62
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 61

(ii) Write the structure of the product when chlorobenzene is treated with methyl chloride in the presence of sodium metal and dry ether.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 63

(iii) Write the structure of the alkene formed by dehydrohalogenation of 1-Bromo-1- methylcyclohexane with alcoholic KOH. (CBSE 2018)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 64

Question 10.
(a) Out ofClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 65 andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 66 which one is more reactive towards SN2 reaction and why?
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 67 because it is a primary halide and has less steric hindrance and therefore, undergoes SN2 action faster.

(b) Out ofClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 68 andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 69 which one is more reactive towards nucleophilic substitution reaction and why?
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 70 is more reactive because of the presence of electron-withdrawing – NO2 group.

(c) Out ofClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 71 andClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 72 which one is optically active and why? (CBSE AI 2019)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 73 is optically active because of the presence of chiral carbon. OH

Question 11.
Write the products of the following reactions:
(i) CH3 CH2 -CH = CH2 + HCl →
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 74

(ii) Me2CHCH2OH Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 75
Answer:
Me2CHCH2OH Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 75 MeCHCH2Cl

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 76
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 77

(iv) Me2C = CMe2 Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 78
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 79

(v) Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 80 (CBSE AI 2016)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 81

Question 12.
(i) In the following pairs of the halogen compounds, which would undergo SN2 faster?
Answer:
SN2 reaction proceeds through the formation of a transition state involving the bonding of carbon to five atoms or groups. The reactivity is decided by the stability of the transition state on the basis of steric hindrance. The reactivity follows the order: CH3 > 1° > 2° > 3° halide.
Therefore,

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 82
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 83 is primary alkyl halide and hence undergoes SN2 reaction faster.

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 84
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 85 I will undergo SN2 reaction faster because iodine is a better leaving group due to its large size and hence it will be released at a faster rate in the presence of incoming nucleophile.

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 86
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 87 (1 -Bromo-2, 2-dimethylpentane) reacts faster because it is 1 ° alkyl halide.

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 88 (CBSE AI 2009)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 89 (2-Methyl-1-bromopropane) reacts faster because it has lesser steric hindrance in the transition state.

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 90
Answer:
CH3 CH2 Br would undergo SN2 reaction faster because of less steric hindrance.

(ii) Which one of the following pairs undergoes SN1 substitution reaction faster and why?
Answer:
SN1 reaction occurs through the formation of a carbocation. Therefore, the greater the stability of the carbocation, the faster is the rate of an SN1 reaction. Since the stability of carbocation follows the order 3° > 2° > 1° > CH3, therefore,

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 91
Answer:
Out of Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 92 orClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 93 reacts faster because it invoLves the formation of 3° -carbocationClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 94 while Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 95involves the formation of V -carbocationClass 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 96

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 97(CBSE AI 2009)
Answer:
Out of Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 98or Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 99reacts faster because it forms 2° carbocation while Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 100involves 1° carbocation.

Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 101 (CBSE AI 2015)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 102 undergoes SN1 reaction faster. This is because the carbocation derived from CH3(CH3)3 C Br is 3° carbocation and is more stable than carbocation (1°) obtained from CH3 CH2 Br.

Question 13.
(i) How would you convert the following:
(a) Prop-1-ene to 1-fluoropropane
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 103

(b) Chlorobenzene to 2-chlorotoluene
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 104

(ii) Write the main products when
(a) n-butyl chloride is treated with alcoholic KOH.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 105

(b) 2, 4, 6-trinitrochlorobenzene is subjected to hydrolysis.
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 106

(c) methyl chloride is treated with AgCN. (CBSE AI2015)
Answer:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 107

Question 14.
Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The rate of this reaction depends upon the concentration of the compound ‘A’ only. When another optically active isomer ‘B’ of this compound was treated with aq. KOH solution, the rate of reaction was found to be dependent on the concentration of compound and KOH both.
(i) Write down the structural formula of both compounds ‘A’ and ‘B’.
Answer:
The molecular formulae of isomers of C4H9Br are CH3
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 108
Since the rate of reaction of the compound ‘A’ (C4H9Br) with aqueous KOH depends upon the concentration of compound ‘A’ only, therefore, the reaction occurs by SN1 mechanism and compound ‘A’ is tertiary bromide, i.e. 2-Bromo-2-methylpropane.
(CH3)3CBr + KOH(aq) → (CH3 )3COH + KBr rate = k[(CH3)3CBr]

(ii) Out of these two compounds, which one will be converted to the product with an inverted configuration. (NCERT Exemplar)
Answer:
Since compound ‘B’ is optically active and is an isomer of compound ‘A’ (C4H9Br), therefore, compound ‘B’ must be 2-bromobutane. Since the rate of reaction of compound ‘B’ with aqueous KOH depends upon the concentration of compound ‘B’ and KOH, therefore, the reaction occurs by SN2 mechanism and the product of hydrolysis will have inverted configuration.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 109
Compound ‘B’ will be converted with an inverted configuration.

Question 15.
(i) Draw other resonance structures related to the following structure and find out whether the functional group present in the molecule is ortho, para directing or meta directing.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 110
Answer:
It is clear from resonating structures that there is an increase in electron density at ortho and para positions. Therefore, the functional group ‘X’ is ortho and para directing.

(ii) Classify the following compounds as primary, secondary and tertiary halides.
(a) 1-Bromobut-2-ene
Answer:
Primary

(b) 4-Bromopent-2-ene
Answer:
secondary

(c) 2-Bromo-2-methylpropane (NCERT Exemplar)
Answer:
tertiary

Question 16.
tert-Butylbromide reacts with aq. NaOH by SN1 mechanism while n-butyl bromide reacts by an SN2 mechanism. Why?
Answer:
In general, the SN1 reaction proceeds through the formation of a carbocation. The tert-butyl bromide readily loses Br ion to form stable 3° carbocation.
Therefore, it reacts with aqueous KOH by SN1 mechanism as:
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 111
On the other hand, n-butyl bromide does not undergo ionisation to form n-butyl carbocation (1°) because it is not stable. Therefore, it prefers to undergo reaction by an SN2 mechanism, which occurs in one step through a transition state involving the nucleophilic attack of OH ion from the backside with simultaneous expulsion of Br ion from the front side.
Class 12 Chemistry Important Questions Chapter 10 Haloalkanes and Haloarenes 112
SN1 mechanism follows the reactivity order as 3° > 2° > 1° while the SN2 mechanism follows the reactivity order as 1° >2° >3°
Therefore, tert-butyl bromide (3°) reacts by SN1 mechanism while n-butyl bromide (1°) reacts by the SN2 mechanism.

Biomolecules Class 12 Important Extra Questions Chemistry Chapter 14

Here we are providing Class 12 Chemistry Important Extra Questions and Answers Chapter 14 Biomolecules. Class 12 Chemistry Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Chemistry Chapter 14 Important Extra Questions Biomolecules

Biomolecules Important Extra Questions Very Short Answer Type

Question 1.
Write the structure of the product obtained when glucose is oxidised with nitric acid. (CBSE 2012)
Answer:
Saccharic acid (or glucaric acid) is formed.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 1

Question 2.
Write a reaction that shows that all the carbon atoms in glucose are linked in a straight chain. (CBSE 2012)
Answer:
When glucose is heated with HI and red P at 100°C for a long period, it gives n-hexane.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 2
The formation of n-hexane suggests that all six carbon atoms in glucose are arranged in a straight chain.

Question 3.
What are the hydrolysis products of sucrose? (CBSE 2019C)
Answer:
Glucose and fructose
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 3

Question 4.
Write the name of the linkage joining two amino acids. (CBSE 2013)
Answer:
Peptide linkage

Question 5.
What are the hydrolysis products of lactose? (CBSE2019C)
Answer:
Lactose on hydrolysis gives β – D – galactose and β – D – glucose.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 4

Question 6.
What are three types of RNA molecules which perform different functions? (CBSE Delhi 2013)
Answer:
(i) messenger RNA: m – RNA
(ii) ribosomal RNA: r – RNA
(iii) transfer RNA: t – RNA

Question 7.
What type of bonding helps in stabilising the a-helix structure of proteins? (CBSE Delhi 2013)
Answer:
Hydrogen bonding between -NH and -C = 0 groups of peptide bonds.

Question 8.
What is a glycosidic linkage? (CBSE Delhi 2013)
Answer:
Two monosaccharide units are linked to each other by a bond called glycosidic linkage.

Question 9.
Which of the two components of starch is water-soluble? (CBSE Delhi 2014)
Answer:
Amylose

Question 10.
Which component of starch is a branched polymer of a-glucose and insoluble in water? (CBSE Delhi 2014)
Answer:
Amylopectin

Question 11.
What are the products of hydrolysis of maltose? (CBSE 2014)
Answer:
Glucose
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 5
Question 12.
What is the basic structural difference between glucose and fructose?
OR
Write the products obtained after hydrolysis of lactose. (CBSE Delhi 2019)
Answer:
Glucose contains an aldehyde group (-CHO) and is aldose, which is present at the end of the carbon chain, i.e. C1.

Fructose contains a keto group (>C = 0) and is called ketose, which is present at a carbon atom next to the terminal, i.e. C2.
OR
Hydrolysis of lactose gives β – D – glucose and β – D – galactose.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 6

Question 13.
What is the difference between a glycosidic linkage and a peptide linkage?
OR
What is the difference between Nucleotide and Nucleoside? (CBSE AI 2019)
Answer:
Glycosidic linkage is the linkage that joins two monosaccharides through an oxygen atom (-0-). Peptide linkage is the linkage that joins two amino acids through -CONH- bond.
OR
The molecules in which one of the nitrogen bases (purines or pyrimidine) is bonded with a sugar molecule (ribose or deoxyribose) is a nucleoside. The nucleoside linked with the phosphate group is called a nucleotide.

Question 14.
What is meant by the inversion of sugar? (CBSE Sample Paper 2011)
Answer:
The change of specific rotation of sugar from dextrorotatory to laevorotatory is called inversion of sugar.

Question 15.
Glucose does not give 2, 4-DNP test and Schiff’s test. Why? (CBSE Sample Paper 2011)
Answer:
Glucose has a cyclic structure in which the -CHO group is not free because it forms a hemiacetal linkage with the -OH group at C-5. Therefore, it does not give 2, 4-DNP test although it has -CHO group.

Question 16.
Write any two reactions of glucose that could not be explained by an open-chain structure of the glucose molecule.
Answer:
The open-chain structure of a glucose molecule cannot explain the following:

  1. Glucose does not react with sodium bisulphite (NaHS03) to form an addition product though it has an aldehyde group,
  2. Glucose does not give the Schiffs test and 2,4-DNP test like other aldehydes.

Question 17.
Write the product when D-glucose reacts with the cone. HNO3.
Answer:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 7

Question 18.
(i) Which vitamin deficiency causes rickets?
Answer:
Vitamin D

(ii) Name the base that is found in the nucleotide of RNA only. (CBSE Sample Paper 2017-18)
Answer:
Uracil

Question 19.
Write the Zwitter ion structure of glycine. (CBSE Sample Paper 2011)
Answer:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 8
Question 20.
Name a carbohydrate present in the liver, muscles and brain. (CBSE 2019C)
Answer:
Glycogen

Question 21.
Name the species formed when an aqueous solution of amino acid is dissolved in water. (CBSE Sample Paper 2019)
Answer:
Zwitter ion/dipolar ion

Question 22.
Name the unit formed by the attachment of a base to the 1′ position of sugar in a nucleoside. (CBSE Sample Paper 2019)
Answer:
Nucleotide

Biomolecules Important Extra Questions Short Answer Type

Question 1.
Name the bases present in RNA. Which one of these is not present in DNA? (CBSE Delhi 2011)
Answer:

  • The bases present in RNA are guanine, adenine, cytosine and uracil.
  • Uracil is not present in DNA. Instead, it contains thymine.

Question 2.
What is meant by biocatalysts? (CBSE Delhi 2012)
Answer:
Biocatalysts. Biocatalysts or enzymes are produced by living cells which catalyse the biochemical reactions occurring in living cells.

Question 3.
Name one fibrous protein and one globular protein.
Answer:
Fibrous: Keratin Globular: Albumin

Question 4.
What is the chemical name of vitamin A and which disease is caused by its deficiency?
Answer:
Retinol, deficiency disease: Xerophthalmia.

Question 5.
What is the chemical name of vitamin C and which disease is caused by its deficiency?
Answer:
Ascorbic acid, deficiency disease: Scurvy.

Question 6.
Which enzyme is present in saliva? What is its function?
Answer:
The enzyme present in the saliva is amylase. It hydrolyses starch into maltose.

Question 7.
What is the difference between DNA and RNA on the basis of the bases they contain?
Answer:
Both DNA and RNA contain two bases derived from purine: guanine and adenine and one base derived from pyrimidine: cytosine. However, they have a fourth different base; DNA contains thymine whereas RNA contains uracil.

Question 8.
Shanti, a domestic helper of Mrs Anuradha, fainted while mopping the floor. Mrs Anuradha immediately took her to the nearby hospital where she was diagnosed to be severely ‘anaemic’. The doctor prescribed an iron-rich diet and multivitamins supplement to her. Mrs Anuradha supported her financially to get the medicines. After a month, Shanti was diagnosed to be normal.

After reading the above passage, answer the following questions:
(i) What values are displayed by Mrs Anuradha?
Answer:
Mrs Anuradha has shown generosity and a caring attitude for the health and life of her domestic helper. Timely help, kindness and financial support saved her domestic helper from a serious health problem.

(ii) Name the vitamin whose deficiency causes ‘pernicious anaemia’.
Answer:
B12

(iii) Give an example of a water-soluble vitamin. (CBSE 2013)
Answer:
Vitamin C

Biomolecules Important Extra Questions Long Answer Type

Question 1.
Differentiate between the following:
(i) Amylose and Amylopectin
(ii) Peptide linkage and Glycosidic linkage
(iii) Fibrous proteins and Globular proteins
OR
Write chemical reactions to show that the open structure of D-glucose contains the following:
(i) Straight chain
(ii) Five alcohol groups
(iii) Aldehyde as carbonyl group (CBSE Delhi 2019)
Answer:
(i) Amylose and Amylopectin:

Amylose Amylopectin
1. It consists of a long chain of a-D- glucose. 1. It consists of branched polymeric chains of a-D- glucose.
2. It is water-soluble. 2. It is water-insoluble.

(ii) Peptide linkage and Glycosidic linkage:

Peptide linkage Glycosidic linkage
The linkage joins two amino acids through -CONH- bond. The linkage joins two monosaccharides through an oxygen atom(-o).

(iii) Fibrous and Globular proteins:

Fibrous Globular
1. The polypeptide chains run parallel and are held by hydrogen and sulphide bonds. 1. The polypeptide chains coil around to give a spherical shape.
2. These are insoluble in water. 2. These are soluble in water.
3. For example, Keratin in hair. 3. For example, egg albumin.

OR
(i) Glucose on prolonged heating with HI and red P at 100 °C gives n-hexane, which shows the straight-chain structure of glucose.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 9
(ii) Glucose reacts with acetic anhydride in the presence of anhydrous zinc chloride to form glucose pentaacetate. The formation of pentaacetate confirms the presence of five -OH groups in the glucose molecule.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 10
(iii) Glucose reacts with hydrogen cyanide forming glucose cyanohydrin. This shows that glucose contains a carbonyl group (>C=0).
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 11
Glucose gets oxidised to six carbon carboxylic acid (gluconic acid) on treatment with a mild oxidising agent like Br2 water. This indicates that the carbonyl group present in glucose is an aldehyde (-CHO) group.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 12
Question 2.
Define the following with a suitable example in each:
(i) Oligosaccharides
(ii) Denaturation of protein
(iii) Vitamins
OR
Write the reactions involved when D-glucose is treated with the following reagents:
(i) Br2 water
(ii) H2N-OH
(iii) (CH3CO)20 (CBSE Delhi 2019)
Answer:
(i) Oligosaccharides. These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis, for example, disaccharides such as sucrose, lactose, maltose (C12H22O16), trisaccharides such as raffinose (C18H32O16).

(ii) Deriaturatlon of proteins: A process that changes the physical and biological properties of proteins without affecting the chemical composition of the protein is called denaturation. It is caused by certain physical change like change In temperature or chemical change like change In pH presence of electrolytes, etc., for example, coagulation of albumin present in the white of an egg.

(iii) Vitamins: The organic compounds which cannot be produced by the body and must be supplied in small amounts in the diet to perform specific biological functions for the normal health, growth and maintenance of the body are called vitamins, for example, vitamin A, B, C, D, etc.
OR
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 13
Question 3.
Define the following terms with a suitable example of each:
(a) Anomers
Answer:
Anomers: Carbohydrates that differ in configuration at the glycosidic carbon (i.e. C1 in aldoses and C2 in ketoses) are called anomers. For example, α-D-glucose and β-D- glucose.

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet are called essential amino acids. For example, valine, leucine, etc.

(c) Denaturation of protein (CBSE AI 2019)
Answer:
A process that changes the native conformation of a protein is called denaturation. The denaturation can be caused by changes in pH, temperature, presence of salts of certain chemical agents. The denatured protein will lose its biological activity. During denaturation, the protein molecule uncoils from an ordered and specific conformation into a more random conformation and protein precipitates from the solution. For example, when an egg is boiled in water, the globular proteins present in it change to a rubber-like insoluble mass.

Question 4.
Define the following terms with a suitable example of each:
(a) Tertiary structure of the protein
Answer:
The tertiary structure of proteins: Tertiary structure of proteins arises because of the folding, coiling or bending of polypeptide chains producing a three-dimensional structure. This structure gives the overall shape of proteins. For example, fibrous proteins such as silk, collagen and α-keratin have large helical structures

(b) Essential amino acids
Answer:
Essential amino acids: The amino acids which cannot be synthesised in our body and must be supplied through our diet are called essential amino acids, for example, valine, leucine, etc.

(c) Disaccharides (CBSE AI 2019)
Answer:
Disaccharides: The carbohydrates which on hydrolysis give two same or different monosaccharides are called disaccharides. Sucrose is an example of a disaccharide, which on hydrolysis gives glucose and fructose.

Question 5.
Describe what do you understand by primary structure and secondary structure of proteins. (CBSE Delhi 2011)
Answer:
Primary structure:
The sequence in which the amino acids are linked in one or more polypeptide chains of a protein is called the primary structure.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 14
The primary structure is usually determined by its successive hydrolysis with enzymes or mineral acids.

Secondary structure: The secondary structure gives the manner in which the polypeptide chains are folded or arranged. Therefore, it gives the shape or conformation of the protein molecule. This arises from the plane geometry of the peptide bond and hydrogen bond between the —C=O and N—H groups of different peptide bonds. Pauling and Corey studied that there are two common types of structures:

(i) α – Helix structure: In this structure, the formation of hydrogen bonding between amide groups within the same chain causes the peptide chains to coil up into a spiral structure (Fig. 1). This is called the a-helix.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 15
(ii) β-pleated sheet structure: In this structure, all polypeptide chains are stretched out to nearly maximum extension and then laid side by side in a zigzag manner to form a flat sheet. Each chain is held to the two neighbouring chains by a hydrogen bond. These sheets are stacked one upon another to form a three-dimensional structure called
β-pleated sheet structure (Fig. 2).
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 16

Question 6.
ExplaIn what is meant by the pyranose structure of glucose. (CBSE 2011)
Answer:
The six-membered cyclic structure of glucose is catted pyranose structure similar to the pyran heterocyclic compound. For structure,
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 17

Question 7.
Write the main structural difference between RNA and DNA. Of the four bases, name those which are common to both DNA and RNA. (CBSE 2011)
Answer:
Structural differences:

DNA RNA
(i) It has a double-stranded a-helix structure in which two strands are coiled spirally in opposite directions. (i) It has a single-stranded a-helix structure.
(ii) The sugar molecule present in DNA is 2-deoxyribose. (ii) The sugar molecule present in RNA is ribose.
(iii) Nitrogenous base uracil is not present. (iii) Nitrogenous base thymine is not present.

Common bases present in both: adenine, cytosine and guanine.

Question 8.
(i) Deficiency of which vitamin causes night blindness? (CBSE DelhI 2014)
(ii) Name the base that is found in the nucleotide of RNA only.
(iii) Glucose on reaction with HI gives n-hexane. What does It suggest about the structure of glucose?
Answer:
(i) Deficiency of Vitamin A causes night blindness.
(ii) Uracil
(iii) Glucose reacts with HI to give n-hexane.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 18
This suggests that alt the six carbon atoms in glucose are arranged in a straight chain structure of glucose.

Question 9.
(i) DefIciency of which vitamin causes rickets? (CBSE Delhi 2014)
Answer:
Vitamin D.

(ii) Give an example for each of fibrous protein and globular protein.
Answer:
Fibrous protein: Keratin Globular protein: Albumin

(iii) Write the product formed on reaction of D-glucose with Br2 water.
Answer:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 19

Question 10.
(i) Deficiency of which vitamin causes scurvy? (CBSE DelhI 2014)
Answer:
Vitamin C.

(ii) What type of linkage is responsible for the formation of proteins?
Answer:
Peptide Linkage.

(iii) Write the product formed when glucose is treated with HI.
Answer:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 20
Question 11.
Define the following terms: (CBSE 2014)
(i) Gtycosidic linkage
Answer:
Glycosidic linkage: The condensation of hydroxyl groups of two monosaccharides to form a link between them is called glycosidic linkage.

(ii) Invert sugar
Answer:
Invert sugar: The sugar which on hydrolysis with dilute adds or enzymes gives mixture having specific rotation opposite to the original is called invert sugar. For example, sucrose is inverted sugar.

(iii) Oligosaccharides
Answer:
Oligosaccharides: These are the carbohydrates that give two to ten monosaccharide molecules on hydrolysis. These are further classified as disaccharides, trisaccharides, tetrasaccharides, etc. depending upon the number of monosaccharide units present in their molecules. For example, Disaccharides: Sucrose, lactose, maltose. All these have the molecular

Formula C12H22O11.
Trisaccharide: Raffinose (C18H32O16).
Tetrasaccharides: Stachyose (C24H42O21 ).

Question 12.
Define the following terms: (CBSE 2014)
(i) Nucleotide
Answer:
Nucleotide: A unit formed by the combination of a nitrogen-containing heterocyclic base, a pentose sugar and a phosphoric acid group.

(ii) Anomers
Answer:
Anomers: The anomers are the isomers formed due to the change in the configuration of the -OH group at C-1 of glucose. For example, α-and β-forms of glucose are anomers.

(iii) Essential amino acids.
Answer:
Essential amino acids: The amino acids which cannot be made by our bodies and must be supplied in our diet for the growth of the body are called essential amino acids.

Question 13.
Differentiate between the following:
(a) Fibrous protein and Globular protein
Answer:
Difference between fibrous protein and globútar protein:

Fibrous Globular
(i) The polypeptide chains run parallel and are held together by hydrogen and disulphide bonds. 1. The polypeptide chains colt around to give a spherical shape.
(ii) These are insoluble in water. 2. These are soluble in water.
(iii) For example, keratin in hair 3. For example, albumin in egg.

(b) Essential amino acids and Non-essential amino acids
Answer:
Difference between essential amino acids and non-essential amino acids:

Essential amino acids Non-essential amino acids
These are not synthesìsed in our body and must be supplied in the diet. These are synthesised in our body and not required in our diet.
For example, valine. For example, alanine

(C) Amylose and Amylopectin (CBSE AI 2019)
Answer:
Difference between amylose and amylopectin:

Amylose Amylopectin
It consists of branched polymeric chains of α – D – glucose. It consists of a long straight chain of α – D – glucose.
It is water-insoluble. It is water-soluble.

Question 14.
(i) Which one of the following is a polysaccharide: (CBSE 2015, 2014)
Starch, Maltose, Fructose, Glucose
Answer:
Starch

(ii) What is the difference between native protein and denatured protein?
Answer:
Native protein is the protein found in a biological system with a unique three-dimensional structure and biological activity.
Denatured protein is the protein which loses its biological activity by certain physical or chemical treatment such as changes in pH, temperature, presence of certain salts or chemical agent, known as denaturation.

(iii) Write the name of the vitamin responsible for the coagulation of blood.
Answer:
Vitamin K.

Question 15.
(i) Write one reaction of D-glucose which cannot be explained by its open-chain structure. (CBSE 2016)
Answer:
Despite having an aldehydic (-CHO) group, glucose does not react with NaHS03 to form an addition product.

(ii) What type of linkage is present in nucleic acids?
Answer:
Phosphodiester linkage

(iii) Give one example each for water-soluble vitamins and fat-soluble vitamins.
Answer:
Water-soluble vitamin: vitamin B Fat-soluble vitamin: vitamin A

Question 16.
What is essentially the difference between the α-form of D-glucose and β-form of D-glucose? Explain.
Answer:
α-form and β-form of glucose differ in the orientation of —H and —OH groups around the C1 atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C1 are called anomers. The structures of these two may be shown below:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 21
These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|D = +111° and the β-form of glucose has m.p. 423 K and |α|D = +19.2°.

Question 17.
(a) What happens when D-glucose is treated with the following reagents:
(i) HI
Answer:
When glucose is treated with HI, it forms n-hexane, suggesting that all the six carbon atoms are linked in a straight line.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 22
(ii) cone. HN03
Answer:
Glucose on treatment with nitric acid gives a dicarboxylic acid, saccharic acid.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 23
(b) What is the basic structural difference between starch and cellulose? (CBSE 2019C)
Answer:
Starch consists of two components: amylose and amylopectin. Amylose is a long linear polymer of 200-1000 a-D-(+)-glucose units held by C1 – C4 glycosidic linkages. It is soluble in water. Amylopectin is a branched-chain polymer of a-D-(+)-glucose linkages whereas branching occurs by C1 – C6 glycosidic linkage. It is insoluble in water.

On the other hand, cellulose is a straight-chain polysaccharide composed only of (β – D – (+)-glucose units which are formed by the glycosidic linkage between C1 of one glucose unit and C4 of the next glucose unit.

Question 18.
(1) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (CBSE Delhi 2016)
Answer:
Glucose, Galactose

(ii) Why Vitamin C cannot be stored in our body?
Answer:
Vitamin C is water-soluble and hence cannot be stored in our body.

(iii) What is the difference between a nucleoside and a nucleotide?
Answer:
A nucleoside contains only two basic components of nucleic acids, namely a pentose sugar and a nitrogenous base.

A nucleotide contains all the three basic components of nucleic acids, namely a phosphoric acid group, a pentose sugar and a nitrogenous base.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 24
Question 19.
Define the following with an example of each: (CBSE 2018)
(i) Polysaccharides
(ii) Denatured protein
(iii) Essential amino acids
OR
(i) Write the product when D-glucose reacts with the cone. HNO3.
(ii) Amino acids show amphoteric behaviour. Why?
(iii) Write one difference between a-helix and p-pleated structures of proteins.
Answer:
(i) Carbohydrates that give a large number of monosaccharide units on hydrolysis or a large number of monosaccharide units joined together by glycosidic linkage, e.g. starch, glycogen, cellulose.
(ii) Proteins that lose their biological activity or proteins in which secondary and tertiary structures are destroyed, e.g. curdling of milk.
(iii) Amino acids cannot be synthesised in the body. e.g. Valine / Leucine
OR
(i) Saccharic acid / COOH-(CHOH)4-COOH
(ii) Due to the presence of carboxyl and amino group in the same molecule or due to formation of zwitterion or dipolar ion.
(iii) a-helix has intramolecular hydrogen bonding while p-pleated has intermolecular hydrogen bonding / a-helix results due to regular coiling of polypeptide chains while in p-pleated all polypeptide chains are stretched and arranged side by side.

Question 20.
Explain the following:
(i) Amino acids behave like salts rather than simple amines or carboxylic acids. (CBSE 2018C)
Answer:
Due to the formation of zwitterion.

(ii) The two strands of DNA are complementary to each other.
Answer:
The two strands are complementary to each other because the hydrogen bonds are formed between specific pairs of bases.

(iii) Reaction of glucose indicates that the carbonyl group is present as an aldehydic group in the open structure of glucose.
Answer:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 25
Glucose gets oxidised to gluconic acid on reaction with a mild oxidising agent like Bromine water.

Question 21.
What is essentially the difference between α-glucose and β-glucose? What is meant by the pyranose structure of glucose?
Answer:
α-form and β-form of glucose differ in the orientation of -H and -OH groups around C, atom. The isomer having the -OH group on the right is called α-D-glucose while the one having -OH group on the left is called β-D-glucose. Such pairs of optical isomers which differ in the configuration only around C1 are called anomers. The structures of these two may be shown below:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 26
These two forms are crystalline and have different melting points and optical rotations. For example, α-form of glucose has m.p. 419 K and |α|D = + 111° and the β-form of glucose has m.p. 423 K and |α|D = + 19.2°.

The α-D-glucose and β-D-glucose can be drawn in a simple six-membered ring form called pyranose structures. These resemble pyran which is a six-membered heterocyclic ring containing five carbon atoms and one oxygen atom.

These are known as pyranose structures and are shown below:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 27

Question 22.
Which sugar is called invert sugar? Why is it called so?
Answer:
Sucrose is called invert sugar. The sugar obtained from sugar beet is a colourless, crystalline and sweet substance. It is very soluble in water and its aqueous solution is dextrorotatory having [α]D = + 66.5°.

On hydrolysis with dilute acids or enzyme invertase, cane sugar gives an equimolar mixture of D-(+)-glucose and D-(-)-fructose.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 28
So, sucrose is dextrorotatory but after hydrolysis, gives dextrorotatory glucose and laevorotatory fructose. D-(-)-fructose has a greater specific rotation than D-(+)- glucose. Therefore, the resultant solution upon hydrolysis is laevorotatory in nature with a specific rotation of (-39.9°). Since there is a change in the sign of rotation from Dextro before hydrolysis to Laevo after hydrolysis, the reaction is called Inversion reaction and the mixture (glucose and fructose) is called invert sugar.

Question 23.
How do you explain the presence of an aldehydic group in a glucose molecule?
Answer:
Glucose reacts with hydroxylamine to form a monoxime and adds one molecule of hydrogen cyanide to give cyanohydrin.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 29
Therefore, it contains a carbonyl group which can be an aldehyde or a ketone. On mild oxidation with bromine water, glucose gives gluconic acid which is a carboxylic acid-containing six carbon atoms.
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 30
This indicates that the carbonyl group present in glucose is an aldehydic group.

Question 24.
Describe the term D- and L-configuration used for sugars with examples.
Answer:
The sugars are divided into two families: the D-family and L-family which have definite configurations. These configurations are represented with respect to glyceraldehyde as the standard. The glyceraldehyde may be presented in two forms:
Class 12 Chemistry Important Questions Chapter 14 Biomolecules 31
The D-configuration has —OH attached to the carbon adjacent to —CH2OH on the right while L-configuration has —OH attached to the carbon adjacent to —CH2OH on left.

The sugars are calLed D- or L- depending upon whether the configuration of the molecule is related to D-glyceraldehyde or L-glyceraldehyde. It has been found that all naturally occurring sugars beLong to D-series, e.g. D-glucose, D-ribose and D-fructose.

Nuclei Class 12 Important Extra Questions Physics Chapter 13

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 13 Nuclei. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 13 Important Extra Questions Nuclei

Nuclei Important Extra Questions Very Short Answer Type

Question 1.
What will be the ratio of the radii of two nuclei of mass numbers A1 and A2?
Answer:
The ratio is \(\frac{R_{1}}{R_{2}}=\left(\frac{A_{1}}{A_{2}}\right)^{1 / 3}\)

Question 2.
Two nuclei have mass numbers in the ratio 1: 2. What is the ratio of their nuclear densities?
Answer:
The densities of both nuclei are equal as they do not depend upon mass number.

Question 3.
A nucleus of mass number A has a mass defect Δm. Give the formula, for the binding energy per nucleon of this nucleus.
Answer:
The formula is E = \(\frac{\Delta m \times c^{2}}{A}\)

Question 4.
Write the relation between half-life and decay constant of a radioactive sample.
The relation is T1/2 = \(\frac{0.693}{\lambda}\)

Question 5.
Write the nuclear decay process for β-decay of 1532P.
Answer:
The process is
Class 12 Physics Important Questions Chapter 13 Nuclei 1

Question 6.
State the relation between the mean life (τ) of a radioactive element and its decay constant λ.
Answer:
The two are related as τ = 1 / λ.

Question 7.
Write any two characteristic properties of nuclear force. (CBSE AI 2011)
Answer:

  1. Do not obey inverse square law and
  2. Spin-dependent.

Question 8.
How is the radius of a nucleus related to its mass number? (CBSE AI 2011C, AI 2013C)
Answer:
The radius R of the nucleus and mass number A is related as R = RoA1/3, where Ro is a constant.

Question 9.
A nucleus undergoes β – decay. How does
(i) the mass number,
(ii) atomic number change? (CBSE Delhi 2011C)
Answer:
During β – decay
(i) the mass number remains the same,
(ii) atomic number increases by one.

Question 10.
Define the activity of a given radioactive substance. Write its SI unit. (CBSE AI 2013)
Answer:
The rate of disintegration in a radioactive substance is known as its activity. SI unit is becquerel (Bq).

Question 11.
Why is it found experimentally difficult to detect neutrinos in nuclear β-decay? (CBSE AI 2014)
Answer:
They are very difficult to detect since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 12.
Four nuclei of an element undergo fusion to form a heavier nucleus, with the release of energy. Which of the two — the parent or the daughter nucleus – would have higher binding energy per nucleon? (CBSE AI 2018, Delhi 2018)
Answer:
Daughter nucleus.

Question 13.
Why is nuclear fusion not possible In the laboratory?
Answer:
Because temperature as high as 107 K cannot be sustained in the laboratory.

Question 14.
Why is the penetrating power of gamma rays very large?
Answer:
Because they have high energy and are neutral.

Question 15.
Can it be concluded from beta decay that electrons exist inside the nucleus?
Answer:
No, the beta particle although an electron is actually created at the instant of beta decay and ejected at once. It cannot exist inside the nucleus as its de-Broglie wavelength is much larger than the dimensions of the nucleus.

Question 16.
Why is the ionizing power of α – parties greater than that of γ-rays?
Answer:
Because α – particles are heavy particles and their speed is comparatively small, so they collide more frequently with atoms of the medium and ionize them.

Question 17.
When a nucleus undergoes alpha decay, is the product atom electrically neutral in beta decay?
Answer:
No, in alpha decay, the atomic number decreases by 2 hence the atom is left with 2 extra orbital electrons. It, therefore, has a double negative charge. In beta decay, the atom is left with a net positive charge.

Question 18.
You are given two nuclides 37X and 34Y. Are they the isotopes of the same element? Why?
Answer:
Yes, because an atomic number of both nuclides is 3.

Question 19.
The variation of the decay rate of two radioactive samples A and B with time is shown in the figure. Which of the two has a greater decay constant? (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 13 Nuclei 2
Answer:
The decay constant of A is greater than that of B but it does not always decay faster than B.

Question 20.
Does the ratio of neutrons to protons in a nucleus increase, decrease, or remain the same after the emission of an alpha particle? (NCERT Exemplar)
Answer:
The ratio of neutrons to protons in a nucleus increases after the emission of an alpha particle.

Question 21.
Which property of nuclear force explains the approximate constancy of binding energy per nucleon with mass number A for nuclei in the range 30 < A < 170? (CBSE2019C)
Answer:
The short-range nature of the nuclear force explains the approximate constancy of binding energy per nucleon with mass number A in the range 30 < A < 170.

Question 22.
Draw a graph showing the variation of decay rate with a number of active nuclei. (NCERT Exemplar)
Answer:
The graph is as shown.
Class 12 Physics Important Questions Chapter 13 Nuclei 3
Question 23.
Which sample, A or B, shown in the figure has a shorter mean-life? (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 13 Nuclei 4
Answer:
B has shorter mean life as λ is greater forB.

Question 24.
Why do stable nuclei never have more protons than neutrons? (NCERT Exemplar)
Answer:
Protons are positively charged and repel one another electrically. This repulsion becomes so great in nuclei with more than 10 protons or so that an excess of neutrons, which produce only attractive forces, is required for stability.

Question 25.
Why does the process of spontaneous nuclear fission occur in heavy nuclei? (CBSE 2019C)
Answer:
Because heavy nuclei contain a large number of protons that exert strong repulsive forces on one another.

Nuclei Important Extra Questions Short Answer Type

Question 1.
Draw the curve showing the binding energy/nucleon with a mass number of different nuclei. Briefly state, how nuclear fusion and nuclear fission can be explained on the basis of this graph.
Answer:
The diagram is as shown.
Class 12 Physics Important Questions Chapter 13 Nuclei 5
Light nuclei have a small value of binding energy per nucleon, therefore to become more stable they fuse to increase their binding energy per nucleon.

A very heavy nucleus, say A 240, has Lower binding energy per nucLeon compared to that of a nucleus with A = 120. Thus if a nucleus A = 240 breaks into two A = 120 nuclei, nucleons get more tightLy bound. This implies energy would be released in the process.

Question 2.
Define decay constant for a radioactive sample. Which of the following radiations α, β, and γ rays
(i) are similar to X-rays,
(ii) are easily absorbed by matter, and
(iii) are similar in nature to cathode rays?
Answer:
The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37 % of its original value.
(i) Gamma
(ii) Alpha
(iii) Beta

Question 3.
State the law of radioactive decay.
Plot a graph showing the number of undecayed nuclei as a function of time (t) for a given radioactive sample having a half-life T1/2.
Depict In the plot the number of undecayed nuclei at (i) t = 3T1/2 and (ii) t = 5 T1/2 (CBSE Delhi 2011)
Answer:
The number of nuclei disintegrating per second is proportional to the number of nuclei present at the time of disintegration and is independent of alt physical conditions like temperature, pressure, humidity, chemical composition, etc.

The plot is as shown.
Class 12 Physics Important Questions Chapter 13 Nuclei 6
Question 4.
Draw a plot of the potential energy of a pair of nucleons as a function of their separations. Mark the regions where the nuclear force is (i) attractive and (ii) repulsive. Write any two characteristic features of nuclear forces. (CBSE AI 2012)
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 7
For r > r0 (attraction), For r < ro (repulsion)

  1. Strong attractive force (stronger than the repulsive electric force between the protons)
  2. Are short-range forces.

Question 5.
(a) Write the relation for binding energy (BE) (in MeV) of a nucleus of mass ZAM atomic number (Z) and mass number (A) in terms of the masses of its constituents – neutrons and protons.
Answer:
The required expression is
ΔE = (Zmp + (A – Z)mn – M) × 931 MeV

(b) Draw a plot of BE/A versus mass number A for 2 ≤ A ≤ 170. Use this graph to explain the release of energy in the process of nuclear fusion of two light nuclei. (CBSE Delhi 2014C)
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 8
Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

Question 6.
If both the number of neutrons and the number of protons are conserved in each nuclear reaction, in what way is mass converted into energy (or vice versa) in a nuclear reaction? Explain. (CBSE AI2016C)
Answer:
We know that the binding energy of a nucleus gives a negative contribution to the mass of the nucleus (mass defect). Now, since proton number and neutron number are conserved in a nuclear reaction the total rest mass of neutrons and protons is the same on either side of a reaction. But the total binding energy of nuclei on the left side need not be the same as that on the right-hand side.

The difference in these binding energies appears as the energy released or absorbed in a nuclear reaction. Since binding energy contributes to mass, we say that the difference in the total mass of nuclei on the two sides gets converted into energy or vice-versa.

Question 7.
State two properties of nuclear forces. Write the relation between half-life and decay constant of a radioactive nucleus. (CBSE AI 2017C)
Answer:

  1. They are saturated forces.
  2. They are charge-independent.

The required relation is
T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 8.
(a) Draw a graph showing the variation of binding energy per nucleon (BE/A) vs mass number A for the nuclei in 20 ≤ A ≤ 170.
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 8
Since the binding energy of the smaller nuclei like hydrogen is less, therefore they fuse together to form helium in order to increase their binding energy per nucleon and become stable. This means that the final system is more tightly bound than the initial system. Again energy would be released in such a process of fusion.

(b) A nucleus of mass number 240 and having binding energy/nucleon 7.6 MeV splits into two fragments Y, 1 of mass numbers 110 and 130 respectively. If the binding energy/ nucleon of Y, 1 is equal to 8.5 MeV each, calculate the energy released in the nuclear reaction. (CBSE Al 2017C)
Answer:
Energy released per fission
= (110 + 130) × 8.5 – 240 × 7.6
= 240 × (8.5 – 7.6) MeV
= 240 × 0.9
= 216.0 MeV

Question 9.
Explain with the help of an example, whether the neutron-proton ratio in a nucleus increases or decreases due to beta decay.
Answer:
Consider the following decay
Class 12 Physics Important Questions Chapter 13 Nuclei 9
Number of neutrons before beta decay
= 234-90 = 144
Number of neutrons after beta decay
= 234-91 =143
Number of protons before beta decay
= 90
Number of protons after beta decay
= 91
Neutron-proton ratio before beta decay
= \(\frac{144}{90}\) = 1.6

Neutron-proton ratio after beta decay
= \(\frac{143}{91}\) = 1.57

Thus neutron-proton ratio decreases during beta decay.

Question 10.
How is the size of a nucleus experimentally determined? Write the relation between the radius and mass number of the nucleus. Show that the density of the nucleus is independent of its mass number. (CBSE Delhi 2011C)
Answer:
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 × 10-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = RoA1/3, where Ro = 1.2 × 10-15 m

The density of a nucleus of mass number A and radius R is given by
Class 12 Physics Important Questions Chapter 13 Nuclei 10
which is independent of the mass number A.

Question 11.
(a) What characteristic property of nuclear force explains the constancy of binding energy per nucleon (BE/A) in the range of mass number ‘A’ lying 30 < A < 170?
Answer:
The nuclear force between two nucleons falls rapidly to zero as their distance is more than a few femtometres. This leads to the saturation of forces in a medium or a large-sized nucleus, i.e. nuclei for which A is 30 < A < 170, which is the reason for the constancy of the binding energy per nucleon.

(b) Show that the density of nucleus over a wide range of nuclei is constant- independent of mass number A. (CBSE AI 2012)
Answer:
The size of the nucleus can be determined by the Rutherford experiments on alpha particles scattering. The distance of the nearest approach is approximately the size of the nucleus. Here it is assumed that only coulomb repulsive force caused scattering. With alpha rays of 5.5 MeV, the size of the nucleus was found to be less than 4 x 10-14 m. By doing scattering experiments with fast electrons bombarding targets of different elements, the size of the nuclei of various elements determined accurately.

The required relation is
R = RoA1/3, where Ro = 1.2 × 10-15 m

The density of a nucleus of mass number A and radius R is given by
Class 12 Physics Important Questions Chapter 13 Nuclei 10
which is independent of the mass number A.

Question 12.
A radioactive nucleus ‘A’ decays as given below:
Class 12 Physics Important Questions Chapter 13 Nuclei 11.
If the mass number and atomic number of A1 are 180 and 73 respectively, find the mass number and an atomic number of A and A2.
Answer:
For A : Z = 72 and A = 180
For A2: Z = 71 and A = 176

Question 13.
The sequence of stepwise decay of a radioactive nucleus isClass 12 Physics Important Questions Chapter 13 Nuclei 12. If the nucleon number and atomic number for D2 are 176 and 71 respectively, what are the corresponding values of D and D3? Justify your answer in each case.
Answer:
For D: A = 180, Z = 72
For D3: A = 172, Z = 69

During alpha decay mass number decreases by 4 and the atomic number decreases by 2, while during beta decay the mass number remains the same, and the atomic number increases by 1.

Question 14.
Write symbolically the nuclear β+ decay process of 611C. Is the decayed product X an isotope or isobar of611C?
Given the mass values m (611C) = 11.011434 u and m (X) = 11.009305 u. (CBSE AI 2015)
Estimate the Q – value in this process.
Answer:
The required equation is
Class 12 Physics Important Questions Chapter 13 Nuclei 13
X is an isobar
Mass defect = m(C) – m(X)
= (11.011434- 11.009305) u = 0.002129 u

Therefore Q = Δm × 931.5 MeV
= 0.002129 × 931.5 = 1.98 MeV

Question 15.
Two radioactive samples, X, Y have the same number of atoms at t = 0. Their half¬lives are 3 h and 4 h respectively. Compare the rates of disintegration of the two nuclei after 12 hours. (CBSE AI 2017C)
Answer:
Let N0 be the nuclei present in X and Y at t = 0. Given Tx = 3 h and Ty = 4 h, t = 12 h.
The number of nuclei present in X and Y after 12 hours is
Class 12 Physics Important Questions Chapter 13 Nuclei 14
Question 16.
A radioactive sample has the activity of 10,000 disintegrations per second after 20 hours. After the next 10 hours, its activity reduces to 5,000 dis. sec-1. Find out its half-life and initial activity. (CBSE Delhi 2017C)
Answer:
Since activity reduces to half in 10 hours from 10000 dis. sec-1 to 5000 dis. sec-1, therefore half-life of the sample will be 10 years.
Class 12 Physics Important Questions Chapter 13 Nuclei 15
Question 17.
Why is the energy of the beta particles emitted during beta decay continuous?
Answer:
The phenomenon of beta decay arises due to the conversion of a neutron in the nucleus into a proton, electron, and an anti-neutrino. Because the energy available in beta decay is shared by the electron and the anti-neutrino in all possible ratios as they come out of the nucleus, therefore the beta ray energy spectrum is continuous in nature.

Question 18.
Explain, how radioactive nuclei can emit β-particles even though atomic nuclei do not contain these particles. Hence explain why the mass number of a radioactive nuclide does not change during β-decay.
Answer:
Beta-particles (or electrons) as such are not present inside a nucleus. However, in the case of a radioactive nuclide, sometimes a neutron decays into a proton, an electron, and an antineutrino as given by the following equation:
Class 12 Physics Important Questions Chapter 13 Nuclei 16
where mass and charge of antineutrino particle is zero. Out of the particles formed, the proton remains within the nucleus itself but electron along with antineutrino comes out of the nucleus. It is this electron that is being emitted as a beta-particle.

As in the process of β-emission, one proton is produced in the nucleus at the expense of a neutron and the mass number of both is the same, hence the mass number of the nuclide remains unchanged during p-decay.

Question 19.
Consider a radioactive nucleus A which decays to a stable nucleus C through the following sequence: A → B → C. Here B is an intermediate nucleus that is also radioactive. Considering that there are No atoms of A initially, plot the graph showing the variation of the number of atoms of A and B versus time. (NCERT Exemplar)
Answer:
At t = 0, NA = No while NB = 0. As time increases, NA falls off exponentially, while the number of atoms of B increases, becomes maximum, and finally decays to zero  ∞ (following exponential decay law).

Hence the graph is as shown.
Class 12 Physics Important Questions Chapter 13 Nuclei 17

Nuclei Important Extra Questions Long Answer Type

Question 1.
Define the terms: half-life period and decay constant of a radioactive sample. Derive the relation between these terms.
Answer:
The half-life is the time required for the number of radioactive nuclei to decrease to one-half the original number.

The decay constant is defined as the reciprocal of that time duration for which the number of nuclei of the radioactive sample decays to 1 / e or 37% of its original value.

To get the relation for half life T and decay constant λ we set N = \(\frac{N_{0}}{2}\) and t = T in the equation N = No e-λt, obtaining \(\frac{1}{2}\) = e-λt

Taking the logarithm of both sides and solving for T we have
T = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)

Question 2.
(a) Draw a graph showing the variation of the potential energy of a pair of nucleons as a function of their separation. Indicate the regions in which nuclear force is (i) attractive, and (ii) repulsive.
Answer:
Graph showing the variation of potential energy U (in MeV) of a pair of nucleons as a function of their separation r (in fm) is shown here.
Class 12 Physics Important Questions Chapter 13 Nuclei 18

  1. In the graph region AD (r > ro) shows the region where nuclear force is strongly attractive.
  2. The region DE (r < ro) shows the region where nuclear force is strongly repulsive.

(b) Write two characteristic features of nuclear force which distinguish it from the Coulomb force.
Answer:
Two characteristics of nuclear forces which distinguish it from Coulomb’s force are

  • It is charge Independent.
  • It is an extremely short-range force and does not obey the inverse square law.

Question 3.
Prove that the Instantaneous rate of change of the activity of a radioactive substance is Inversely proportional to the square of Its half-life.
Answer:
The activity of a radioactive substance is
A = \(\frac{dN}{dt}\).

We know that the number of nuclei of a radioactive substance Left behind after time t is given by N = Noe-λt

Differentiating the above relation with respect to time we have
A = \(\frac{d N}{d t}=\frac{d}{d t}\)Noe-λt = – Noλe-λt

Differentiating the above equation with respect to time we have
Class 12 Physics Important Questions Chapter 13 Nuclei 19
Therefore the Instantaneous rate of change of the activity of a radioactive substance is inversely proportional to the square of Its half Life.

Question 4.
(a) Deduce the expression N = Noe-λt the law of radioactive decay.
(b) (i) Write symbolically the process expressing the β+ decay of, 1122Na Also write the basic nuclear process underlying this decay.
(ii) Is the nucleus formed in the decay of the nucleus 1122Na an Isotope or isobar? (CBSE Delhi 2014)
Answer:
(a) Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,
Class 12 Physics Important Questions Chapter 13 Nuclei 20
where λ  is a constant of proportionality.

Question 5.
(a) Complete the following nuclear reactions:
Class 12 Physics Important Questions Chapter 13 Nuclei 21
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 22

(b) Write the basic process Involved in nuclei responsible for (i) β and (ii) β+ decay.
Answer:
The basic nuclear process underlying β decay is the conversion of the neutron to proton
n → p + e + v
while for v+ decay, it is the conversion of a proton into a neutron
p → n + e+ + v

(c) Why is it found experimentally difficult to detect neutrinos? (CBSE AI 2015 C)
Answer:
Neutrinos are neutral particles with very small (possibly, even zero) mass compared to electrons. They have only weak interaction with other particles. They are, therefore, very difficult to detect, since they can penetrate a large quantity of matter (even earth) without any interaction.

Question 6.
(a) Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (B.E./A) versus the mass number A.
Answer:
For the graph
Class 12 Physics Important Questions Chapter 13 Nuclei 5
From the plot, we note that

  • During nuclear fission: A heavy nucleus in the larger mass region (A > 200) breaks into two middle-level nuclei, resulting in an increase in B.E./ nucleon. This results in the release of energy,
  • During nuclear fusion: Light nuclei in the lower mass region (A < 20) fuse to form a nucleus having higher B.E. / nucleon. Hence Energy gets released.

(b) A radioactive Isotope has a half-life of 10 years. How long will It take for the activity to reduce to 3.125%? (CBSE AI2018)
Answer;
3.125% means that the number of nuclei decays to 1/32 of its original value.
Therefore,
\(\frac{N}{N_{0}}=\frac{1}{32}=\left(\frac{1}{2}\right)^{5}\)

Now we know that
N = N0\(\left(\frac{1}{2}\right)^{t / T}\)

Therefore we have
\(\left(\frac{1}{2}\right)^{5}=\left(\frac{1}{2}\right)^{t / T}\)

Therefore
t = 5T = 5 × 10 = 50 years

Question 7.
Group the following six nuclides into three pairs of (?) isotones, (ii) isotopes, and (iii) isobars: 612C, 23He, 80198Hg, 13H, 79197Au 614C. How does the size of the nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
Answer:
(a) Isotones: 80198Hg, 79197Au
(6) Isotopes: 612C , 614C
(c) Isobars: 23He, 13H

The size of a nucleus depends upon its mass number as R = R0 A1/3

The nuclear density is given by the expression
Class 12 Physics Important Questions Chapter 13 Nuclei 23
The calculations show that the nuclear density is independent of the mass number.

Question 8.
Define the term ‘decay constant’ of a radioactive sample. The rate of disintegration of a given radioactive nucleus is 10,000 disintegrations/s and 5,000 disintegration/s after 20 hr and 30 hr respectively from start. Calculate the half-life and an initial number of nuclei at t = 0. (CBSE Delhi 2019)
Answer:
The decay constant of a radioactive element is the reciprocal of the time in which the number of its nuclei reduces to 1 /e of its original number.

We have R = λN
R(20hrs) = 100o0 = λN20
R(30hrs) = 5000 = λN30
\(\frac{N_{20}}{N_{30}}\) = 2

This means that the number of nuclei, of the given radioactive nucleus, gets halved in a time of (30 – 20) hours = 10 hours
Half-life = 10 hours

This means that in 20 hours (= 2 half-Lives), the original number of nuclei must have gone down by a factor of 4.
Hence rate of decay at t = 0
λ N0 = 4 λ N20
R0 = 4 × 10,000 = 40,000 disintegration per second
Class 12 Physics Important Questions Chapter 13 Nuclei 24
Question 9.
(a) Write the relation between half-life and an average life of a radioactive nucleus.
Answer:
The relation is τ = 1 .44T1/2

(b) In a given sample two isotopes A and B are initially present in the ratio of 1:2. Their half-lives are 60 years and 30 years respectively. How long will it take so that the sample has these isotopes in the ratio of 2:1? (CBSE Delhi 2019)
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 25
Class 12 Physics Important Questions Chapter 13 Nuclei 26
Question 10.
Distinguish between nuclear fission and fusion. Show how in both these processes energy is released. Calculate the energy release in MeV in the deuterium-tritium fusion reaction:
Class 12 Physics Important Questions Chapter 13 Nuclei 27
Using the data m(21H) = 2.014102 u,
m(13H) = 3.016949u, m(24He) = 4.002603 u,
mn = 1.008665 u,1 u = 931.5 MeV/c2 (CBSE Delhi 2015)
Answer:
The distinction is shown in the table below.

Nuclear Fission Nuclear Fusion
1. It is the splitting of a heavy nucleus into two or tighter unstable nuclei. 1. It is the combining of two light nuclei into a heavier nucleus.
2. It may or may not be a chain reaction. 2. It is always a chain reaction.
3. It is independent of temperature. 3. It is temperature-dependent.
4. It can be controlled. 4. It can’t be controlled.
5. Tremendous amount of energy is released. 5. Energy released per unit mass is seven times the energy released during fission.
6. By-products are harmful. 6. By-products are not harmful.
7. Example of reaction – The atom bomb. 7. Example of reaction – Reaction in stars, hydrogen born

In both reactions, there is a mass defect that is converted into energy.
Now energy released in the reaction
Class 12 Physics Important Questions Chapter 13 Nuclei 28

Question 11.
(a) Draw a plot showing the variation of the potential energy of a pair of nucleons as a function of their separation. Mark the regions where the nuclear force is (a) attractive and (b) repulsive.
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 30
For r > ro, the force is attractive For r < ro, the force is repulsive

(b) In the nuclear reaction
Class 12 Physics Important Questions Chapter 13 Nuclei 29
determine the values of a and b. (CBSE Delhi 2018 C)
Answer:
We have,
1 + 235 = a + 94 + 2 × 1
∴ a = 236 – 96 = 140

Also
0 + 92 = 54+ 6 + 2 × 0
∴ b = 92 – 54 = 38

Question 12.
Binding energy per nucleon versus mass number curve is as shown. ZAS, Z1A1w, Z2A2X, and Z3A3Y, are four nuclei indicated on the curve.
Class 12 Physics Important Questions Chapter 13 Nuclei 31
Based on the graph:
(а) Arrange X, W, and S in the increasing order of stability.
Answer:
(a) S, W, and X

(b) Write the relation between the relevant A and Z values for the following nuclear reaction. S → X + W
Answer:
The equation isClass 12 Physics Important Questions Chapter 13 Nuclei 32
Z = Z1 + Z2 and A = A1 + A2

(c) Explain why binding energy for heavy nuclei is low. (CBSE Sample Paper 2018-19)
Answer:
Reason for low binding energy: For heavier nuclei, the Coulomb repulsive force between protons increases considerably and offsets the attractive effects of the nuclear forces. This can result in such nuclei being unstable.

Question 13.
(a) Derive the law of radioactive decay,
viz. N = Noe-λt
Answer:
Let N0 be the number of nuclei present in a freshly separated sample of a radioactive substance. Let after time t the number of nuclei left behind be N. Let dN number of nuclei disintegrate in a small time interval dt. Then by the – decay law,
Class 12 Physics Important Questions Chapter 13 Nuclei 20
where λ is a constant of proportionality.

(b) Explain, giving necessary reactions, how energy is released during
(i) fission
Answer:
Nuclear Fission: It is a process in which a heavy nucleus splits up into two Lighter nucLei of nearly equal masses. It is found that the sum of the masses of the product nuclei and particles is less than the sum of the masses of the reactants, i.e. there is some mass defect. This mass defect appears as energy. One such fission reaction is given below:
Class 12 Physics Important Questions Chapter 13 Nuclei 33
The Q value of the above reaction is about 200 MeV. The sum of the masses of Ba, Kr, and 3 neutrons is less than the sum of the masses of U and one neutron.

(ii) fusion
Nuclear Fusion: It is the process in which two light nuclei combine together to form a heavy nucleus. For fusion very high temperature of l is required. One such fusion reaction is given below:
Class 12 Physics Important Questions Chapter 13 Nuclei 34
The Q value of this nuclear reaction is 24 MeV. It is the energy equivalent of the mass defect in the above reaction. The energy released in fusion is much less than in fission but the energy released per unit mass infusion is much greater than that released in fission.

Question 14.
(a) Distinguish between isotopes and isobars, giving one example for each.
(b) Why is the mass of a nucleus always less than the sum of the masses of its constituents? Write one example to justify your answer.
Or
(a) Classify the following six nuclides into (i) isotones, (ii) Isotopes, and (iii) isobars: (CBSEAI2019)
Class 12 Physics Important Questions Chapter 13 Nuclei 35
(b) How does the size of a nucleus depend on its mass number? Hence explain why the density of nuclear matter should be independent of the size of the nucleus.
Answer:
(a) Isotopes have the same atomic number while isobars have the same mass number
Examples of isotopes 612C, 614C
Examples of isobars 23He, 13H

(b) Mass of a nucleus is less than its constituents because in the bound state some mass is converted into binding energy which is energy equivalent of mass defect e.g., the mass of 1860 nucleus is less than the sum of masses of 8 protons and 8 neutrons
Or
(a) Isotones: 80198Hg, 79197Au
(6) Isotopes: 612C , 614C
(c) Isobars: 23He, 13H

(b) The radius of the nucleus is given by
R = RoA1/3

Volume of the nucleus \(\frac{4}{3}\)πR3 = \(\frac{4}{3}\)πRo3 A

If m is the average mass of the nucleon then the mass of the nucleus M = mA
Hence nuclear density
Class 12 Physics Important Questions Chapter 13 Nuclei 36
Which is independent of the A i.e., the size of the nucleus.

Numerical Problems:

  • Radius of the nucleus R = RoA1/3
  • Mass defect, Δm = Z mp + (A – Z) mn– M
  • Energy released ΔE = Δm × 931 MeV
    or
    ΔE = [Z mp + (A – Z) mn – M] × 931 MeV
  • Binding energy per nucleon BE/N = ΔE/A
  • Relation between original nuclei (N) and nuclei left (No) after time t N = Noe-λt
  • Relation between decay constant (λ) and half-life (T) = \(\frac{\ln 2}{\lambda}=\frac{2.303 \log 2}{\lambda}=\frac{0.693}{\lambda}\)
  • Half-life is also given by the expression N = No\(\left(\frac{1}{2}\right)^{n}\) where n = t/T
  • The average life is given by
    τ = \(\frac{1}{\lambda}=\frac{T}{\ln 2}=\frac{T}{0.693}\) = 1.44 T
  • Activity is given by A = -λN = \(\frac{0.693N}{T}\)

Question 1.
Calculate the binding energy per nucleon of Fe5626 Given mFe = 55.934939 u, mn = 1.008665 u and mp = 1.007825 u
Answer:
Number of protons Z = 26
Number of neutrons (A – Z) = 30
Now mass defect is given by
Δm = Z mp + (A – Z)mn – M
Δm = 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 0.528461 u

Therefore binding energy
BE = Δm × 931 MeV = 0.528461 × 931
= 491.99 MeV

BE/nucleon = 491.99/56 = 8.785 MeV

Question 2.
The activity of a radioactive element drops to one-sixteenth of its initial value in 32 years. Find the mean life of the sample.
Answer:
Class 12 Physics Important Questions Chapter 13 Nuclei 37
Or
32/T = 4 or 7 = 32 / 4 = 8 years.
Therefore mean life of the sample is τ = 1.44 7 = 1.44 × 8 = 11.52 years.

Question 3.
A radioactive sample contains 2.2 mg of pure 116C which has a half-life period of 1224 seconds. Calculate (i) the number of atoms present initially and (ii) the activity when 5 pg of the sample will be left.
Answer:
Mass of sample = 2.2 pg
Now 11 g of the sample contains 6.023 × 1023 nuclei, therefore the number of nuclei in 2.2 mg = 2.2 × 10-3 g are
Class 12 Physics Important Questions Chapter 13 Nuclei 38

Question 4.
The half-life of 238 92U is 4.5 × 109 years. Calculate the activity of 1 g sample of 92238U.
Answer:
Given T = 4.5 × 109 years.
Number of nuclei of U in 1 g
= N = \(\frac{6.023 \times 10^{23}}{238}\) = 2.5 × 1021

Therefore activity
Class 12 Physics Important Questions Chapter 13 Nuclei 39
Question 5.
The decay constant for a given radioactive sample is 0.3456 per day. What percentage of this sample will get decayed in a period of 4 days?
Answer:
Given λ = 0.3456 day-1
or
T1/2 = 0.693/λ = 0. 693/ 0.3456 = 2.894 days, t = 4 days.

Let N be the mass left behind, then N = Noe-λt
or
N = No e-0 3456 × 4
or
N = N0 e-1 3824 = No × 0.25

Therefore the percentage of undecayed is
Class 12 Physics Important Questions Chapter 13 Nuclei 40
Question 6.
It is observed that only 6.25 % of a given radioactive sample is left undecayed after a period of 16 days. What is the decay constant of this sample per day?
Answer:
Given N/No = 6.25 %, t = 16 days, λ = ?
Class 12 Physics Important Questions Chapter 13 Nuclei 41
Or
16/ T = 4 or T = 4 days.

Therefore λ = 1/T = 1/4 = 0.25 day-1

Question 7.
A radioactive substance decays to 1/32th of its initial value in 25 days. Calculate its half-life.
Answer:
Given t = 25 days, N = No / 32, using
Class 12 Physics Important Questions Chapter 13 Nuclei 42
Or
25/7= 5 or T= 25 / 5 = 5 days.

Question 8.
The half-life of a radioactive sample is 30 s.
Calculate
(i) the decay constant, and
Answer:
Given T1/2 = 30 s, N = 3No / 4, λ = ?, t = ?
(i) Decay constant
λ = \(\frac{0.693}{T_{1 / 2}}=\frac{0.693}{30}\) = 0.0231 s-1

(ii) time taken for the sample to decay to 3/4 th of its initial value.
Answer:
Using N = Noe-λt we have
Class 12 Physics Important Questions Chapter 13 Nuclei 43
Question 9.
The half-life of 14 6C is 5700 years. What does it mean?
Two radioactive nuclei X and Y initially contain an equal number of atoms. Their half-lives are 1 hour and 2 hours respectively. Calculate the ratio of their rates of disintegration after 2 hours.
Answer:
It means that in 5700 years the number of nuclei of carbon decay to half their original value.
Given Nox = NoY, TX = 1 h, TY = 2 h, therefore
\(\frac{\lambda_{X}}{\lambda_{Y}}=\frac{2}{1}\) = 2

Now after 2 hours X will reduce to one- fourth and Y will reduce to half their original value.
If activities at t = 2 h are Rx and Ry respectively, then
Class 12 Physics Important Questions Chapter 13 Nuclei 44
Thus their rate of disintegration after 2 hours is the same.

Question 10.
A star converts all its hydrogen to helium achieving 100% helium composition. It then converts helium to carbon via the reaction.
Class 12 Physics Important Questions Chapter 13 Nuclei 45
The mass of the star is 5 × 1032 kg and it generates energy at the rate of 5 × 1030 watt. How long will it take to convert all the helium to carbon at this rate?

As 4 × 10-3 kg of He consists of 6.023 × 1023 He nuclei so 5 × 1032 kg He will contain
\(\frac{6.023 \times 10^{23} \times 5 \times 10^{32}}{4 \times 10^{-3}}\) = 7.5 × 1058 nuclei

Now three nuclei of helium produce 7.27 × 1.6 × 10-13 J of energy
So all nuclei in the star will produce
E = \(\frac{7.27 \times 1.6 \times 10^{-13}}{3}\) × 7.5 × 1058
= 2.9 × 1046 J

As power generated is P = 5 × 1030 W, therefore time taken to convert all He nuclei into carbon is
t = \(\frac{E}{P}=\frac{2.9 \times 10^{46}}{5 \times 10^{30}}\) = 5.84 × 1015 s
or
1.85 × 108 years

Question 11.
Radioactive material is reduced to (1/16)th of its original amount in 4 days. How much material should one begin with so that 4 × 10-3 kg of the material is left after 6 days?
Answer:
N = No / 16, t = 4 days,
N = 4 × 10-3 kg,
t = 6 days

To calculate half-life of the material we have
Class 12 Physics Important Questions Chapter 13 Nuclei 46
Now using the expression 4 × 10-3 = No\(\left(\frac{1}{2}\right)^{6 / 1}\)
Solving we have No = 0.256 kg

Question 12.
Two different radioactive elements with half-lives T1 and T2 have N1 and N2 (undecayed) atoms respectively present at a given instant. Determine the ratio of their activities at this instant.
Answer:
The activity of a radioactive sample is given by the relation
A = – λN

Therefore the ratio of activity of these two radioactive elements is
\(\frac{A_{1}}{A_{2}}=\frac{-\lambda_{1} N_{1}}{-\lambda_{2} N_{2}}=\frac{T_{2} N_{1}}{T_{1} N_{2}}\)

Question 13.
Given the mass of the iron nucleus as 55.85 u and A = 56. Find the nuclear density? (NCERT)
Answer:
Given mFe = 55.85 u = 9.27 × 10-26kg
Class 12 Physics Important Questions Chapter 13 Nuclei 47
The density of matter in neutron stars (an astrophysical object) is comparable to this density. This shows that matter in these Neutron stars has been compressed to such an extent that they resemble a big nucleus.

Question 14.
We are given the following atomic masses: 92238U = 238.05079 u, 24He = 4.00260 u, 90234Th = 234.04363 u 11H = 1.00783 u, 91237Pa =237.05121 u Here the symbol Pa is for the element protactinium (Z = 91). (a) Calculate the energy released during the alpha decay of 92238U. (b) Show that cannot spontaneously emit a proton. (NCERT)
Answer:
(i) The alpha decay of 92238Uis given by
Class 12 Physics Important Questions Chapter 13 Nuclei 48
The energy released in this process is given by
Q= (Mu – MTh – MHe) × 931.5 MeV

Substituting the atomic masses as given in the data we find that
Q = (238.05079 – 234.04363 – 4.00260) × 931.5 MeV ⇒ Q = 4.25 MeV.

(ii) If 29®U spontaneously emits a proton, the decay process would be
Class 12 Physics Important Questions Chapter 13 Nuclei 49
The Q for this process to happen is Q = (Mu – Mpa – MH) × 931.5 MeV
Q = (238.05079 – 237.05121 – 1.00783) × 931.5 MeV ⇒ Q = – 7.68 MeV

Thus the Q of the process is negative and therefore it cannot proceed spontaneously. We will have to supply energy of 7.68 MeV to the 92238U nucleus to make it emit a proton.

Question 15.
The half-life of 90Sr is 28 years. What is the disintegration rate of 15 mg of this isotope? (NCERT)
Answer:
Given T1/2 = 28 years, m = 15 mg
Now the rate of disintegration is given by
Class 12 Physics Important Questions Chapter 13 Nuclei 50

Ray Optics and Optical Instruments Class 12 Important Extra Questions Physics Chapter 9

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 9 Ray Optics and Optical Instruments. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 9 Important Extra Questions Ray Optics and Optical Instruments

Ray Optics and Optical Instruments Important Extra Questions Very Short Answer Type

Question 1.
When light undergoes refraction at the surface of separation of two media, what happens to its frequency/wavelength?
Answer:
There is no change in its frequency, but its wavelength changes.

Question 2.
Define the refractive index.
Answer:
The Refractive index of a medium is defined as the ratio of the speed of light in a vacuum to the speed of light in the given medium.

Question 3.
What is the distance between the objective and eyepiece of an astronomical telescope in its normal adjustment?
Answer:
Distance between objective and eyepiece of telescope = fo + fe

Question 4.
Name the phenomenon responsible for the reddish appearance of the sun at sunrise and sunset.
Answer:
Atmospheric refraction.

Question 5.
What are the two main considerations that have to be kept in mind while designing the ‘objective’ of an astronomical telescope?
Answer:
Two main considerations are

  1. Large light gathering power
  2. Higher resolution (or resolving power)

Question 6.
Under what condition does a biconvex lens of glass having a certain refractive index act as a plane glass sheet when immersed in a liquid? (CBSE Delhi 2012)
Answer:
When the refractive index of the liquid is equal to the refractive index of a glass of which the lens is made.

Question 7.
Write the relationship between the angle of incidence ‘i’, angle of prism ‘A’ and angle of minimum deviation for a triangular prism. (CBSE Delhi 2013)
Answer:
2i = A + δm

Question 8.
Why can’t we see clearly through the fog? Name the phenomenon responsible for it. (CBSE Al 2016)
Answer:
Because it scatters light. Scattering of light.

Question 9.
How does the angle of minimum deviation of a glass prism vary if the incident violet light is replaced by red light? Give reason. (CBSE AI 2017)
Answer:
It decreases as δm ∝ \(\frac{1}{λ}\)

Question 10.
The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1: 2. Compare the resolving powers of the two telescopes. (CBSE AI 2017C)
Answer:
Same as resolving power does not depend upon the focal length of lenses.

Question 11.
An object is kept in front of a concave lens. What is the nature of the image formed? (CBSE Delhi 2017C)
Answer:
Virtual and erect.

Question 12.
The refractive index of the material of a concave lens is n1. It is immersed in a medium of refractive index n2. A parallel beam of tight is incident on the lens. Trace the path of the emergent rays when n2 > n1.
Answer:
The path of rays is as shown
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 1
Question 13.
A convex lens made of glass of refractive index μL is immersed in a medium of refractive index μm. How will the lens behave when μL < μm?
Answer:
The lens will continue to behave as a convex lens.

Question 14.
The image of an object formed by a lens on the screen is not in sharp focus. Suggest a method to get clear focusing of the image on the screen without disturbing the position of the object, the lens or the screen.
Answer:
Limit the field of view of the lens by using a blackened glass having a small circular hole in the middle.

Question 15.
In the figure given below, the path of a parallel beam of light passing through a convex lens of refractive index ng kept in a medium of refractive index nm is shown. Is (i) ng = nm or (ii) ng > nm, or (iii) ng < nm?
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 2
Answer:
As the rays of light do not suffer any deviation, therefore ng = nm.

Question 16.
In the figure path of a parallel beam of light passing through a convex lens of refractive index ng kept in a medium of refractive index, nm is shown. Is (i) ng = nm or (ii) ng > nm, or (iii) ng < nm?
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 3
Answer:
As the rays of light diverge, therefore ng < nm.

Question 17.
The refractive index of the material of a concave lens is μ1. It is immersed in a medium of refractive index μ2. A parallel beam of light is incident on the lens. Trace the path of emergent rays when μ2 < μ1.
Answer:
The path of rays is as shown below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 4
Question 18.
Suppose that the lower half of the concave mirror’s reflecting surface is covered with an opaque (non-reflective) material. What effect will this have on the image of an object placed in front of the mirror?
Answer:
The image of the whole object will be formed. However, as the area of the reflecting surface has been reduced the intensity of the image will below (in this case, half).

Question 19.
For the same value of angle of incidence, the angles of refraction in three media A, B and C are 15°, 25° and 35° respectively. In which medium would the velocity of light be minimum? (CBSE Al 2012)
Answer:
Medium A.

Question 20.
When red light passing through a convex lens is replaced by the light of blue colour, how will the focal length of the lens change? (CBSE AI 2013C)
Answer:
It will decrease.

Question 21.
A biconvex lens made of a transparent material of refractive index 1.25 is immersed in water of refractive index 1.33. Will the lens behave as a converging or a diverging lens? Give reason. (CBSE AI 2014)
Answer:
The diverging lens as its focal length will become negative.

Question 22.
A concave lens of refractive index 1.5 is immersed in a medium of refractive index 1.65. What is the nature of the lens? (CBSE Delhi 2015)
Answer:
It will behave as a convex lens.

Question 23.
Why can’t we see clearly through a fog? Name the phenomenon responsible for it. (CBSE AI 2016)
Answer:
Because it scatters light. Scattering of light.

Question 24.
How does the angle of minimum deviation of a glass prism vary if the incident violet light is replaced by red light? Give reason. (CBSE AI 2017)
Answer:
It decreases as δm ∝ \(\frac{1}{λ}\)

Question 25.
The objective lenses of two telescopes have the same apertures but their focal lengths are in the ratio 1: 2. Compare the resolving powers of the two telescopes. (CBSE AI 2017C)
Answer:
Same as resolving power does not depend upon the focal length of lenses.

Question 26.
An object is kept in front of a concave lens. What is the nature of the image formed? (CBSE Delhi 2017C)
Answer:
Virtual and erect.

Question 27.
A convex lens is held in water. What change, if any, do you expect In its focal length?
Answer:
Focal Length of the given Lens increases in accordance with tens maker’s formula
\(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

This is because wμg aμg

Question 28.
Violet light is incident on a converging lens of focal length f. State with reason, how the focal length of the lens will change 1f the violet light is replaced by a red light.
Answer:
As μr < μv hence in accordance with the relation
\(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) focal Length of Lens for red colour will be more.

Question 29.
What is the minimum value of the refractive index of the prism shown in the figure below?
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 5
Answer:
μ = \(\frac{1}{\sin i_{c}}=\frac{1}{\sin 45^{\circ}}=\frac{1}{\sqrt{2}}\)

Question 30.
How does the resolving power of a telescope change on decreasing the aperture of its objective lens? Justify your answer.
Answer:
As the resolving power of a telescope is \(\frac{D}{1.22λ}\) hence on decreasing the aperture of its objective lens, the resolving power of the telescope decreases in the same ratio.

Question 31.
What will happen to a ray of light incident normally on the interface of air and glass?
Answer:
It will pass un-deviated into the glass.

Question 32.
What is the speed of light in glass having a refractive index of 1.5?
Answer:
Speed of light in glass v = c/n = 3 × 108 / 1.5 = 2 × 108 m s-1

Question 32.
Light of wavelength 600 nm in air enters a medium of refractive index 1.5. What will be its frequency in the medium?
Answer:
Frequency does not change when light moves from one medium into another. Therefore frequency of light
v = \(\frac{c}{\lambda}=\frac{3 \times 10^{8}}{600 \times 10^{-9}}\) = 5 × 1014 Hz

Question 34.
How does the focal length of a convex lens change if monochromatic red light Is used instead of monochromatic blue
light?
Answer:
Focal Length increases, i.e. fr > fv.

Question 35.
Two thin lenses of power + 5 D and – 3 D are in contact. What Is the focal length of the combination?
Answer:
Here P = P1 + P2 = + 5 – 3 = + 2 D
Hence f = 1/P = ½ m = + 50 cm.

Question 36.
Use the maker’s formula to write an expression for the (relative) refractive index, μ, of the material in terms of its focal length, f and the radii of curvature, r1 and r2, of Its two surfaces.
Answer:
The formula is \(\frac{1}{f}\) = (μ – 1)\(\left(\frac{1}{r_{1}}-\frac{1}{r_{2}}\right)\)

Question 37.
Two thin lenses of power – 4 D and 2 D are placed in contact coaxially. Find the focal length of the combination. (CBSE Ai 2012C)
Answer:
Total power P = – 4 + 2 = – 2 D
Now f = 1/P = 1/ -2 = – 0.5 m

Question 38.
A convex lens is placed In contact with a plane mirror. A point object at a distance of 20 cm on the axis of this combination has its image coinciding with itself. What is the focal length of the lens? (CBSE Delhi 2014)
Answer:
20 cm

Question 39.
The focal length of a biconvex lens is equal to the radius of curvature of either face. What is the refractive index of the material of the lens? (CBSEAI 2015)
Answer:
1.5
Using \(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Given R1 = R
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 6
Question 40.
An unsymmetrical double convex thin lens forms the image of a point object on its axis. Will the position of the image change if the ens are reversed? (NCERT Exemplar)
Answer:
No, the reversibility of the Lens does not change the lens formula.

Ray Optics and Optical Instruments Important Extra Questions Short Answer Type

Question 1.
The aperture of the objective lens of an astronomical telescope is doubled. How does it affect
(i) the resolving power of the telescope and
(ii) the intensity of the image? (CBSE Sample Paper 2018-19)
Answer:
The resolving power of a telescope is given by the expression \(\frac{D}{1.22λ}\).

(i) When the aperture of the objective lens is increased, the resolving power of the telescope increases in the same ratio.
(ii) The intensity of the image is given by the expression β ∝ D2, thus when the aperture is doubled, the intensity of the image becomes four times.

Question 2.
How does the resolving power of a compound microscope change on (a) decreasing the wavelength of light used, and (b) decreasing the diameter of the objective lens?
Answer:
The resolving power of a microscope is given by the expression RP = \(\frac{2 n \sin \theta}{\lambda}\)

(a) If the wavelength of the incident tight is decreased, the resolving power of the microscope increases.
(b)There is no effect of the decrease in the diameter of the objective on the resolving power of the microscope.

Question 3.
The layered lens shown in the figure is made of two kinds of glass. How many and what kinds of images will be produced by this lens with a point source placed on the optic axis? Neglect the reflection of light at the boundaries between the layers.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 7
Answer:
Two images will be formed as the lens may be thought of, as two separate lenses of different focal lengths. The images will be surrounded by bright halos.

Question 4.
Monochromatic light is refracted from air into a glass of refractive index n. Find the ratio of wavelengths of the incident and refracted light.
Answer:
Using the relation λ1n1 = λ2n2 we have
\(\frac{\lambda_{1}}{\lambda_{2}}\) = n

Question 5.
Draw a labelled ray diagram to show the image formation in a compound microscope.
Answer:
The labelled diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 8
Question 6.
A ray of light while travelling from a denser to a rarer medium undergoes total internal reflection. Derive the expression for the critical angle in terms of the speed of light in the two media.
Answer:
Snell’s law can be used to find the critical angle. Now Snell’s law, when the ray moves from denser medium ‘b’ to rarer medium ‘a’, is given by
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 9
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 10
Now we know that n = \(\frac{c}{v}\) , substituting in the above relation we have
\(\frac{c}{v}=\frac{1}{\sin i_{c}}\) or sin ic = \(\frac{v}{c}\)

Question 7.
Draw a labelled diagram for a refracting type astronomical telescope. How will its magnifying power be affected by increasing for its eyepiece (a) the focal length and (,b) the aperture? Justify your answer. Write two drawbacks of refracting type telescopes. (CBSE Sample Paper 2018-19)
Answer:
The labelled diagram of the telescope is as shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 11
(a) The magnifying power of a telescope is given by M = \(\frac{f_{0}}{f_{\mathrm{e}}}\). If the focal length of the eyepiece is increased, it will decrease the magnifying power of the telescope.

(b) Magnifying power does not depend upon the aperture of the eyepiece. Therefore there is no change in the magnifying power if the aperture of the eyepiece is increased.

Drawbacks:

  • Large-sized lenses are heavy and difficult to support.
  • Large-sized lenses suffer from chromatic and spherical aberration.

Question 8.
Draw a labelled ray diagram of a Newtonian type reflecting telescope. Write any one advantage over refracting type telescope.
Answer:
The labelled diagram is shown below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 12
Due to the large aperture of the mirror as compared to a lens the image formed is much brighter than that formed by a refracting type telescope.

Question 9.
A right-angle crown glass prism with a critical angle of 41° is placed between the object PQ in two positions as shown in figures (a) and (b). Trace the path of rays from P and Q passing through the prisms in the two cases.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 13
Answer:
The path of rays is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 14
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 15
Question 10.
Write two conditions necessary for total internal reflection to take place.
Answer:
(a) The incident ray should travel from the denser to the rarer medium.
(b) The angle of incidence, in the denser medium, should be greater than the critical angle for the given pair of media.

Question 11.
(a) Explain the working of a compound microscope with the help of a labelled diagram.
Answer:
A compound microscope is an instrument used to see highly magnified images of tiny objects.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 16
Working:
Let a tiny object AB be placed in front of the objective lens at a distance more than F0. Its real and enlarged image is formed at A ‘ B ‘. The image A’ B’ acts as an object for the eyepiece and forms the final image at A” B “,

i.e. at a distance D, the least distance of distinct vision.

(b) Write the considerations that you keep in mind while choosing lenses to be used as eyepiece and objective in a compound microscope. (CBSE 2019C)
Answer:
The objective and eyepiece should have a short focal length for large magnification.

Question 12.
Explain the working of a refracting telescope with the help of a labelled diagram. What are the main limitations of this type of telescope and how are these overcome in a reflecting telescope? (CBSE 2019C)
Answer:
Refracting telescope
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 17
Working:
When the rays of light are made to incident on the objective from a distant object, the objective forms the real and inverted image at its focal plane. The lens is so adjusted that the final image is formed at least distance of distinct vision or at infinity.

Limitations:

  • Large-sized lenses are needed which are expensive.
  • Large-sized lenses suffer from spherical aberration and distortions.

Reflecting telescope (To overcome limitations):

  • Reflecting telescopes are free from chromatic aberration and spherical aberration is very small.
  • They are less heavy and easier to support.

Question 13.
Explain why the colour of the sky is blue.
Answer:
As sunlight travels through the earth’s atmosphere, it gets scattered (changes its direction) by the atmospheric particles. Light of shorter wavelengths is scattered much more than light of longer wavelengths. (The amount of scattering is inversely proportional to the fourth power of the wavelength. This is known as Rayleigh scattering.)

Hence, the bluish colour predominates in a clear sky, since blue has a shorter wavelength than red and is scattered much more strongly. In fact, violet gets scattered even more than blue, having a shorter wavelength. But since our eyes are more sensitive to blue than violet, we see the sky blue.

Question 14.
Why does the sun look reddish during sunrise and sunset?
Answer:
At sunset or sunrise, the sun’s rays have to pass through a larger distance in the atmosphere (Figure). Most of the blue and other shorter wavelengths are removed by scattering. The least scattered light reaching our eyes, therefore, the sun looks reddish. This explains the reddish appearance of the sun near the horizon.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 18
Question 15.
What are the two ways of adjusting the position of the eyepiece while observing the final image in a compound microscope? Which of these is usually preferred and why?
Answer:
The two ways
(a) Final image formed at least distance of distinct vision.
(b) Final image formed at infinity.

The second one is usually preferred as it helps the observer to observe the final image with his/her eye in a relaxed position.

Question 16.
Write the relation between the angle of incidence (i), the angle of emergence (e), the angle of the prism (A) and the angle of deviation (δ) for rays undergoing refraction through a prism. What is the relation between ‘i’ and ‘e’ for rays undergoing minimum deviation? Using this relation obtain an expression for the refractive index (μ) of the material of the prism in terms of ‘A’ and angle of minimum deviation.
Answer:
The relation is i + e = A + δm
In the minimum deviation position ∠i = ∠e
In the minimum deviation position we have A = r1 + r2 = r + r = 2r
or
r = A / 2
and i + i = A + δm
or
2i = A + δm
or
i = \(\frac{A+\delta_{m}}{2}\)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

Question 17.
Draw a ray diagram showing the formation of the image by a concave mirror of an object placed beyond its centre of curvature. If the lower half of the mirror’s reflecting surface is covered, what effect will it have on the image? (CBSE AI 2011C)
Answer:
The required ray diagram is shown below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 19
When the lower half of the mirror’s reflecting surface is covered, the intensity of the image will be reduced.

Question 18.
An object AB is kept in front of a concave mirror as shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 20
(a) Complete the ray diagram showing the image formation of the object.
(b) How will the position and intensity of the image be affected if the lower half of the mirror’s reflecting surface is painted black? (CBSE AI 2012)
Answer:
The completed ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 21
No change in the position of the image, but the intensity will decrease.

Question 19.
Draw a ray diagram to show the formation of the image by an astronomical telescope when the final image is formed at the near point. Answer the following, giving reasons
(a) Why the objective has a larger focal length and a larger aperture than the eyepiece?
(b) What would be the effect on the resolving power of the telescope if its objective lens is immersed in a transparent medium of the higher refractive index? (CBSE AI 2012C)
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 22
(a) The objective lens has a large focal length and large aperture for large magnification and resolving power.
(b) The resolving power will increase as the wavelength of light used will decrease.

Question 20.
Two convex lenses of the same focal length but of aperture A1 and A, (A, < A1), are used as the objective lenses in two astronomical telescopes having identical eyepieces. What is the ratio of their resolving power? Which telescope will you prefer and why? Give reason. (CBSE Delhi 2011)
Answer:
The resolving power is directly proportional to the aperture. Therefore the ratio of their resolving power is RP = \(\frac{A_{1}}{A}\).

Since A1 > A therefore we will prefer the telescope with aperture A1 as it will gather a larger amount of light than the telescope of aperture A.

Question 21.
Which two of the following lens L1 L2 and L3 will you select as objective and eyepiece for constructing the best possible (a) telescope and (b) microscope. Give a reason to support your answer. (CBSE Delhi 2015C)

Lens

Power

Aperture

L1 6 D 1 cm
L2 3 D 8 cm
L3 10 D 1 cm

Answer:
(a) For telescope

  • Objective lens L2: Because it has a large focal length and large aperture
  • Eye lens L3: Because it has a small focal length and small aperture

(b) For Microscope

  • Objective lens L3: Because it has a small aperture and small focal length
  • Eye lens L3: because it has a large aperture and large focal length

Question 22.
Draw a ray diagram to show how a right-angled isosceles prism may be used to bend the path of light by 90°.
Write the necessary condition in terms of the refractive index of the material of this prism for the ray to bend by 90°. (CBSE Delhi 2016C)
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 23
The angle of incidence in the denser medium should be greater than the critical angle for the given pair of media, i.e.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 24
Question 23.
The image of a candle is formed by a convex lens on a screen. The lower half of the lens is painted black to make it completely opaque. Draw the ray diagram to show the image formation. How will this image be different from the one obtained when the lens is not painted black?
Answer:
The ray diagram depicting the image is shown
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 25
The image formed will be less bright as compared to that when half of the lens is not painted black. This is because every part of the lens forms the image. When the lower half is blackened, light from this portion will be blocked, hence the intensity of light in the image will be less.

Question 24.
A magician during a show makes a glass lens with n = 1.47 disappear in a trough of liquid. What is the refractive index of the liquid? Could the liquid be water?
Answer:
The refractive index of the liquid must be equal to 1.47 in order to make the lens disappear. This means n1 = n2. This gives 1 /f = 0 or f = ∞. The lens in the liquid will act as a plain sheet of glass. No, the liquid is not water, it could be glycerine.

Question 25.
Explain with reason, how the resolving power of an astronomical telescope will change, when
(a) frequency of the incident light on the objective lens is increased,
(b) the focal length of the objective lens is increased, and
(c) the aperture of the objective lens is halved?
Answer:
Resolving power of a telescope = \(\frac{D}{1.22 \lambda}\) hence
(a) on increasing the frequency of incident light, wavelength X decreases and consequently resolving power is increased.
(b) increase in focal length of the objective lens will have no effect on the resolving power, and
(c) if the aperture of the objective lens is halved, then resolving power is also halved.

Question 26.
Explain with reason, how the resolving power of a compound microscope will change when
(a) frequency of the incident light on the objective lens is increased,
(b) the focal length of the objective lens is increased, and
(c) the aperture of the objective lens is increased.
Answer:
Resolving power of a compound microscope RP = \(\frac{2 \mu \sin \theta}{\lambda}\), hence
(a) on increasing the frequency of incident light, its wavelength X decreases and consequently resolving power increases.
(b) on increasing the focal length of the objective lens, the value of sine and hence resolving power decreases.
(c) on increasing the aperture of the objective lens, the value of sin 0 and hence resolving power increases.

Question 27.
A convex lens of the focal length is kept in contact with a concave lens of focal length f2. Find the focal length of the combination. (CBSE AI 2013)
Answer:
For a convex lens +f1 and for a concave lens – f2, using the expression
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)
or
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 26
Question 28.
(a) The relation between the angle of incidence ‘i’ and the corresponding angle of deviation (δ), for a certain optical device, is represented by the graph shown in the figure. Identify this device. Draw a ray diagram for this device and use it for obtaining an expression for the refractive index of the material of this device in terms of an angle characteristic of the device and the angle, marked as δm, in the graph. (CBSE Al 2016C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 27
Answer:
(a) The device is a prism.
The relation is i + e = A + δm
In the minimum deviation position ∠i = ∠e
In the minimum deviation position we have A = r1 + r2 = r + r = 2r
or
r = A / 2
and i + i = A + δm
or
2i = A + δm
or
i = \(\frac{A+\delta_{m}}{2}\)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

(b) The diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 28

Ray Optics and Optical Instruments Important Extra Questions Long Answer Type

Question 1.
Draw a labelled ray diagram to show the image formation in a refracting type astronomical telescope. Obtain an expression for the angular magnifying power and the length of the tube of an astronomical telescope in its ‘normal adjustment’ position. Why should the diameter of the objective of a telescope be large?
Answer:
A labelled diagram of the telescope is shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 11
The object subtends an angle at the objective and would subtend essentially the same angle at the unaided eye. Also, since the observers’ eye is placed just to the right of the focal point f’2, the angle subtended at the eye by the final image is very nearly equal to the angle β.

Therefore, M = \(\frac{\beta}{\alpha}=\frac{\tan \beta}{\tan \alpha}\) …(1)

From right triangles ABC and ABC’ as shown in figure, we have
tan α = \(\frac{\mathrm{AB}}{\mathrm{CB}}=\frac{-h}{f_{0}}\) and
tan β = \(\frac{\mathrm{AB}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{-h}{f_{\mathrm{e}}}\)

substituting the above two equations in equation (1), we have
M = \(\frac{\beta}{\alpha}=\frac{-h^{\prime}}{f_{e}} \times \frac{f_{0}}{-h^{\prime}}=\frac{f_{0}}{f_{e}}\)

The length of the telescope is the distance between the two lenses which is L = fo + fe The diameter of the objective of a telescope should be large so that it can collect more light and image of distant objects is formed clear.

Question 2.
Draw a ray diagram to show the formation of an erect image of an object kept in front of a concave mirror. Hence deduce the mirror formula. (CBSE 2019C)
Answer:
An object AB is placed between P and F. The course of rays for obtaining erect image A1B1 of object AB is shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 29
Draw DG ⊥ on the principal axis.
Triangles DGF and A1 B1C are similar
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 30
Since Point G is close to P, so GF = PF
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 31

Multiplying and dividing both sides by uvf, we get
\(\frac{1}{f}=\frac{1}{v}+\frac{1}{u}\)

Question 3.
(a) Define the term power of a lens. Write its SI units.
Answer:
The power of a lens is the reciprocal of its focal length in the metre.
i.e. P = \(\frac{1}{f(\text { in } m)}\)

S.I. unit of power of a lens is dioptre (D).

(b) Derive the expression for the power of two thin lenses placed coaxially in contact with each other. (CBSE 2019C)
Answer:
Power of two thin lenses in contact. Consider an object O placed at a distance u on the principal axis of the lens A. Rays of light starting from O forms, the image at I1
∴ \(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) …(1)

Place lens B in contact with A. The image I1 will serve as a virtual object and forms a real image I.
∴ Again apply lens formula
\(-\frac{1}{v_{1}}+\frac{1}{v}=\frac{1}{f_{2}}\) …(2)

Adding (1) and (2), we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 32
where f is the focal length of the combination
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 33
From (3), \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

The total power of the lens combination is given by
P = P1 + P2

Question 4.
A biconvex lens of refractive index μ1 focal length ‘f and radius of curvature R is immersed in a liquid of refractive index μ2. For (i) μ2 > μ1, and (ii) μ2 < μ1, draw the ray diagrams in the two cases when a beam of light coming parallel to the principal axis is incident on the lens. Also, find the focal length of the lens in terms of the original focal length and the refractive index of the glass of the lens and that of the medium. (CBSE AI 2013C)
Answer:
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 34
(b) Given: As the lens is equiconvex therefore R1 = R2 = R, aμg = μ1; aμ1 = μ2 fa = f, fL = ?

Using the expression
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 35

Question 5.
Write the conditions for observing a rainbow. Show, by drawing suitable diagrams, how one understands the formation of a rainbow. (CBSE AI 2014C)
Answer:
The conditions for observing a rainbow are that the sun should be shining in one part of the sky (say near the western horizon) while it is raining in the opposite part of the sky (say the eastern horizon). An observer can therefore see a rainbow only when his back is towards the sun.

Sunlight is first refracted as it enters a raindrop, which causes the different wavelengths (colours) of white light to separate. The longer wavelength of light (red) is bent the least while the shorter wavelength (violet) are bent the most. Next, these component rays strike the inner surface of the water drop and get internally reflected if the angle between the refracted ray and normal to the drop surface is greater than the critical angle (48°, in this case).
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 36
The reflected light is refracted again as it comes out of the drop as shown in the figure. It is found that the violet light, emerges at an angle of 40° related to the incoming sunlight and red light emerges at an angle of 42°. For other colours, angles lie in between these two values.

Question 6.
Explain the basic differences between the construction and working of a telescope and a microscope. (CBSE AI 2015)
Answer:
(a) In a telescope the objective has a large aperture and a large focal length, while in a microscope both the aperture and focal length of the objects are small (fe > fo). In a telescope, the eyepiece has a small aperture as compared to the objective, while in a microscope the eyepiece has a bigger aperture than the objective (fe < fo).

(b) A telescope increases the angle the object subtends at the eye thereby increasing its clarity by bringing it closer, while a microscope actually magnifies the object or a telescope magnifies distant objects, while a microscope magnifies nearby objects.

Question 7.
Why does white light disperse when passed through a glass prism?
Using the lens maker’s formula show how the focal length of a given lens depends upon the wavelength of light incident on it. (CBSE Delhi 2015C)
Answer:
It is because a glass prism offers a different refractive index to different wavelengths of light.
Lens maker’s formula is \(\frac{1}{f}=(n-1)\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\).

The refractive index depends upon wavelength as n ∝ \(\frac{1}{λ}\), therefore f ∝ λ.

Hence the focal length of a lens increases with the increase in the wavelength of light.

Question 8.
A thin converging lens has a focal length f in air. If it is completely immersed in a liquid, briefly explain, how the focal length of the lens will vary?
Answer:
The focal length of a converging lens is given by
\(\frac{1}{f}\) = (ang – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
where ang = refractive index of Lens with respect to air.

And In a given liquid medium the focal Length of the Lens is given by
\(\frac{1}{f_{m}}\) = (mng – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

Therefore the ratio of the totaL focal Length is
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 37
(a) If mng < ang then f is +ve and has a value greater than f, i.e. focal length of the lens increases when immersed in a given liquid medium.

(b) If mng >ang, then f is -ve, i.e. the lens will begin to behave as a diverging lens.

Question 9.
When monochromatic light is incident on a surface separating two media, the reflected and refracted light both have the same frequency as the incident frequency. Explain why?
Answer:
Reflection and refraction arise through the interaction of incident light with the atomic constituents of matter. Atoms may be viewed as oscillators that take up the frequency of the external agency (light) causing forced oscillations. The frequency of light emitted by a charged oscillator equals its frequency of oscillation. Thus the frequency of scattered light equals the frequency of incident light.

Question 10.
The image of an object formed by the combination of a convex lens (of focal length f) and a convex mirror (of the radius of curvature R), set up as shown is observed to coincide with the object. Redraw this diagram to mark on it the position of the centre of curvature of the mirror. Obtain the expression for R in terms of the distances marked as ‘a’ and ‘d’ and focal length f of the convex lens. (CBSE Delhi 2016C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 38
Answer:
The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the tens and reflection from the mirror. The (refracted) rays are, therefore, fatling normally on the mirror. It follows that the rays A B, and A’ B’, when produced, are meeting at the centre of curvature C of the mirror. Hence C is the centre of curvature of the mirror.

From the figure, we then see that for the convex lens we have

u = – a and v = + (d + R). If f is the focal length of the lens, we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 39
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 40
Question 11.
Define the magnifying power of a compound microscope when the final image is formed at infinity. Why must both the objective and the eyepiece of a compound microscope have short focal lengths? Explain. (CBSE Delhi 2017)
Answer:
It is defined as the ratio of the angle subtended on the eye by the final image when it lies at infinity to the angle subtended on the eye by the object when it lies at the least distance of distinct vision from the eye or the near point of the eye. The magnifying power of a microscope is given by
M = Me × Mo = \(\frac{L}{f_{0}} \times\left(1+\frac{D}{f_{e}}\right)\)

Since it depends inversely on the focal lengths of the objective and the eyepiece, therefore both have to be small to get a large magnification.

Question 12.
(a) Draw a ray diagram for the formation of the image by a compound microscope.
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 41
(b) You are given the following three lenses. Which two lenses will you use as an eyepiece and as an objective to construct a compound microscope?

Lenses Power(D)

Aperture(cm)

L1 3 8
L2 6 1
L3 10 1

Answer:
The objective should have a short focal length and large aperture while the eyepiece should have a short focal length and small aperture. Thus for objective lens L3 should be used and for eyepiece lens, L2 should be used.

(c) Define the resolving power of a microscope and write one factor on which it depends. (CBSE AI 2017)
Answer:
The resolving power of a microscope is given by the expression
RP = \(\frac{2 \mu \sin \theta}{1.22 \lambda}=\frac{2 \mu v \sin \theta}{1.22 c}\)

It depends upon the wavelength/ frequency of the incident light.

Question 13.
Draw a labelled ray diagram of an astronomical telescope in the near point adjustment position. (CBSE Delhi 2019)
A giant refracting telescope at an observatory has an objective lens of focal length 15 m and an eyepiece of focal length 1.0 cm. If this telescope is used to view the Moon, find the diameter of the image of the Moon formed by the objective lens. The diameter of the Moon is 3.48 × 106 m, and the radius of the lunar orbit is 3.8 × 108 m.
Answer:
A Labelled ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 42
Given fo = 15 m, fe = 1.0 cm = 0.01 m, M = ? Dm = 3.48 × 106 m, r = 3.8 × 108 m,
Using M = \(\frac{f_{0}}{f_{e}}=\frac{15}{0.01}\) = 1500

The angle subtended by the moon at the objective of the telescope
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 43

Question 14.
(a) Amobilephoneliesalongtheprincipal axis of a concave mirror. Show with the help of a suitable diagram the formation of its image. Explain why magnification is not uniform.
Answer:
The image of the mobile phone formed by the concave mirror is shown in the below figure.

The part of the mobile phone that is at C will form an image of the same size only at C. In the figure, we can see that B’C = BC. The part of the mobile phone that lies between C and F will form an enlarged image beyond C as shown in the figure. It can be observed that the magnification of each part of the mobile phone cannot be uniform on account of different locations. That is why the image formed is not uniform.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 44
(b) Suppose the lower half of the concave mirrors reflecting surface is covered with an opaque material. What effect this will have on the image of the object. Explain. (CBSE Delhi 2014)
Answer:
The intensity of the image will decrease while the whole image will be for.

Question 15.
(a) Draw a labelled ray diagram of a compound microscope, when the final image forms at the least distance of distinct vision.
(b) Why is its objective of short focal length and of short aperture compared to its eyepiece? Explain.
(c) The focal length of the objective is 4 cm while that of the eyepiece is 10 cm. The object is placed at a distance of 6 cm from the objective lens.
(i) Calculate the magnifying power of the compound microscope, if its final image is formed at the near point.
(ii) Also calculate the length of the compound microscope. (CBSE AI 2019)
Or
(a) With the help of a labelled ray diagram, explain the construction and working of a Cassegrain reflecting telescope.
(b) An amateur astronomer wishes to estimate roughly the size of the sun using his crude telescope consisting of an objective lens of focal length 200 cm and an eyepiece of focal length 10 cm. By adjusting the distance of the eyepiece from the objective, he obtains an image of the sun on a screen 40 cm behind the eyepiece. The diameter of the sun’s image is measured to be 6-0 cm. Estimate the sun’s size, given that the average earth-sun distance is 1.5 × 1011 m.
Answer:
(a) The diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 45
(b) The magnifying power of compound microscope m = m0 × me = \(\frac{L}{f_{o}}\left(1+\frac{D}{f_{e}}\right)\)

To have high magnifying power and high resolution, the focal length of the objective and its aperture should be short.

The focal length of the eyepiece is comparatively greater than the objective so an image formed by the objective lens may form within the focal length of the eyepiece and the final magnified image may be formed.

Aperture is short for higher resolution,

(c) Given uo = – 6 cm
For objective lens we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 46
(i) For magnifying power of compound microscope we have
m = \(\frac{v_{o}}{u_{o}}\left(1+\frac{D}{f_{e}}\right)=\frac{12}{6}\left(1+\frac{25}{10}\right)\) = 7

(ii) Length of the compound microscope L = v0 + ue = 12 + 7.14 = 19.14 cm
Or
(a) Cassegrain reflecting telescope consists of a large concave (primary) parabolic mirror having a hole in its centre. There is a small convex (secondary) mirror near the focus of the concave mirror. The eyepiece is placed near the hole of the concave mirror. The parallel rays from a distant object are reflected by the large concave mirror. These rays fall on the convex mirror which reflects these rays outside the hole. The final magnified image is formed at infinity.

The diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 47
(b)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 48
Magnification produced by the eye piece is
me = \(\frac{v_{e}}{u_{e}}=\frac{40}{40 / 3}\) = 3

Diameter of the image formed by the objective is
D = 6/3 = 2 cm

If D is the diameter of the sun then the angle subtended by it on the objective will be
α = \(\frac{D}{1.5 \times 10^{11}}\) rad

Now, angle subtended by the image at the objective = angle subtended by the sun
α = \(\frac{\text { size of image }}{f_{0}}=\frac{2}{200}=\frac{1}{100}\) rad

Therefore,
\(\frac{D}{1.5 \times 10^{11}}=\frac{1}{100}\)
D = 1.5 × 109

Question 16.
A convex lens is placed in contact with a plane mirror. An axial point object, at a distance of 20 cm from this combination, has its image coinciding with itself. What is the focal length of the convex lens?
Answer:
Figure (a) shows a convex lens L in contact with a plane mirror M. P is the point object, kept in front of this combination at a distance of 20 cm, from it. Since the image of the object is coinciding with the object itself, the rays from the object, after refraction from the lens, should fall normally on the mirror M, so that they retrace their path and form an image coinciding with the object itself.

This will be so, if the incident rays from P form a parallel beam perpendicular to M, after refraction from the lens. For clarity, M has been shown at a finite distance from L, in figure (b). For lens L, since the rays from P form a parallel beam after refraction, P must be at the focus of the lens. Hence the focal length of the lens is 20 cm.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 49
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 50
Question 17.
A convex lens and a convex mirror (of the radius of curvature 20 cm) are placed co-axially with the convex mirror placed at a distance of 30 cm from the lens. For a point object, at a distance of 25 cm from the lens, the final image, due to this combination, coincides with the object itself. What is the focal length of the convex lens?
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 51
The final image, formed by the combination, is coinciding with the object itself. This implies that the rays, from the object, are retracing their path, after refraction from the lens and reflection from the mirror. The (refracted) rays are, therefore, falling normally on the mirror. It follows that the rays A B, and A’ B’, when produced, are meeting at the centre of curvature C of the mirror. Hence O2C = 20 cm, the radius of curvature of the mirror. From the figure, we then see that for the convex lens u = – 25 cm and v = + (30 + 20) cm = + 50 cm. If f is the focal length of the lens, we have
\(\frac{1}{50}-\frac{1}{(-25)}=\frac{1}{f}\)
or
f = 16.67 cm

Question 18.
Derive a relation between focal length, image distance and object distance for a
concave mirror.
Answer:
Consider a concave mirror of small aperture, image distance and object distance for a Figure below shows two rays of light leaving the tip of the object. One of these rays passes through the centre of curvature C of the mirror, hitting the mirror head-on and reflecting back on itself. The second ray BP strikes the mirror at its pole P and reflects as shown obeying the laws of reflection. The image of the tip of the arrow is located at the point where these two rays intersect. Thus the image is formed between the focus and centre of curvature.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 52
Consider the two right-angled triangles BAP and B’A’P

Now from triangle BAP using trigonometry, we have
tan θ = \(\frac{AB}{PA}\) …(1)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 53
Therefore equation (7) becomes
\(\frac{\mathrm{PA}}{\mathrm{PA}^{\prime}}=\frac{\mathrm{PA}-\mathrm{PC}}{\mathrm{PC}-\mathrm{PA}^{\prime}}\)

By Cartesian sign conventions we have PA = -u, PA’ = -v and PC = – R, substituting in equation (8), we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 54
simplifying we have 2uv = vR + uR dividing both sides by uvR we have
\(\frac{1}{u}+\frac{1}{v}=\frac{2}{R}\) …(11)

but R = 2 f, where f is the focal length of the mirror. Therefore the above equation
reduces to \(\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\)
which is the required mirror formula.

Question 19.
With the help of a suitable ray diagram, derive a relation between the object distance (u) image distance v and radius of curvature (R) for a convex spherical surface, when a ray of light travels from rarer to denser medium. (CBSE Delhi 2011C)
Answer:
Let a convex spherical surface XPY separate a rarer medium of refractive index (n<sub>1</sub>) from a denser medium of refractive (n2). The real image of object 0 is formed by refraction from the convex spherical surface of radius of curvature R. The angle α β γ are shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 55
Let ∠AOP = α, ∠AIP = β and ∠ACP = γ.

From point A, drop AN perpendicular to the principal axis of the spherical refracting surface. From triangle AOC, we have
i = α + γ

Since the aperture of the spherical refracting surface is small, point A will be close to point P and hence angles α, β and γ will be small. As such, these angles may be replaced by their tangents.

Therefore, equation (1) may be written as
i = tanα + tanγ …(2)

From right angled triangles ΔALO and ΔALC, we have

tan α = \(\frac{AL}{LO}\) and tan γ = \(\frac{AL}{LC}\)

Substituting for tan α and tan γ in equation (2), we have

i = \(\frac{AL}{LO}\) + \(\frac{AL}{LC}\) …(3)

Again, as aperture of the refracting surface is small, point L will be close to point P, the pole of the refracting surface.
Therefore,
LO ≈ PO and LC ≈ PC

Therefore, equation (3) becomes
i = \(\frac{AL}{PO}\) + \(\frac{AL}{PC}\) …(4)

Now, from triangle ΔACI, γ = r + β
or
r = γ— β

Since angles γ and β are small, we have
r = tan γ— tan β …(5)

From right angled triangles ΔALC and ΔALI,
we have
tan γ = \(\frac{A L}{L C} \approx \frac{A L}{P C}\) and tan β = \(\frac{A L}{L I} \approx \frac{A L}{P I}\)

Substituting for tan β and tan γ in equation (5) we have
r = \(\frac{\mathrm{AL}}{\mathrm{PC}}-\frac{\mathrm{AL}}{\mathrm{PI}}\) …(6)

Now by Snell’s Law at point A we have
n1 sin i = n2 sin r

Since angles are small, therefore the above relation becomes
n1 i = n2 r …(7)

Substituting the values of i and r from equations (4) and (6) we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 56
Applying new Cartesian sign conventions:
PO = – u (a distance of the object is against incident Light)

PI = + v (distance of image Is along incident Light)
PC = + R (a distance of the centre of curvature is along with incident Light) we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 57
The above equation connects u, V and R to the absolute refractive indices of the material of the refracting surface and that of the rarer medium.

Question 20.
A ray of light Is an Incident on one face of a glass prism and emerges out from the other face. Trace the path of the ray and derive an expression for the refractive index of the glass prism. Also, plot a graph between the Angle of incidence and angle of deviation. (CBSE Delhi 201 IC)
Answer:
The graph is as shown
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 58
Consider a cross-section XYZ of a prism as shown in the figure. Let A be the angle of the prism. A ray PQ of monochromatic light is incident on face XY of the prism at an angle i. The ray is called the incident ray and the angle is called the angle of incidence. This ray is refracted towards the normal NQE and travels in the prism along QR. This ray is called the refracted ray at the face XY.

Let r1 be the angle of refraction at this surface. The refracted ray QR is incident at an angle r2 on the surface XZ. The ray QR again suffers refraction and emerges out of face XZ at an angle e along with RS. The ray is called the emergent ray and the angle e is called the angle of emergence. When the ray SR is extended backwards it meets the extended ray PQ at point D such that δ is the angle of deviation of the ray.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 59
As seen from figure we have in triangle QDR that ∠DQR = i – r1 and
∠DRQ = e – r2. Therefore from triangle QDR we have
δ = ∠DQR + ∠DRQ = (i – r1) + (e – r2)
or
δ = (i + e) – (r1 + r2) …(1)

Now from the quadrilateral XQER, we have
A + E = 180° …(2)

In triangle QER we have r1 + r2 + e – 180° …(3)

From equations (2) and (3) we have
A = r1 + r2 …(4)

Substituting in equation (1) we have
δ = i + e – A
or
i + e = A + δ … (5)

The deviation produced by a prism depends upon (i) the angle of incidence (ii) the angle of prism and (Hi) the refractive index of the material of the prism. It is found that as the angle of incidence changes, the angle of deviation also changes.

A graph between the angle of incidence and the angle of deviation is shown in the figure above. As the angle of incidence increases, the angle of deviation first decreases becomes a minimum for a particular angle of incidence and then increases. The minimum value of the angle of deviation is called the angle of minimum deviation. It is denoted by δm. In this position the ray of light passes symmetrically through the prism, i.e. the refracted ray QR is parallel to the base of the prism. In this position, the angle of incidence is equal to the angle of emergence, i.e. i = e. Also in this position, the angle of refraction at the faces of the prism are equal, i.e. r1 = r2.

Substituting these values in equations (4) and (5) we have
A = r1 + r2 = r + r = 2r
or
r = A/2 …(6)
and i + i = A + δm
or
2i = A + δm
or
i = \(\frac{A+\delta_{m}}{2}\) …(7)

substituting for i and r in the expression for Snell’s law we have
μ = \(\frac{\sin i}{\sin r}=\frac{\sin \left[\frac{A+\delta_{m}}{2}\right]}{\sin \left(\frac{A}{2}\right)}\)

Question 21.
(a) Draw a ray diagram showing the image formation by a compound microscope. Obtain an expression for total magnification when the image is formed at infinity.
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 60
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 61
(b) How does the resolving power of a compound microscope get affected, when
(i) the focal length of the objective is decreased.
(ii) the wavelength of light is increased? Give reasons to justify your answer. (CBSE AI 2015C)
Answer:
The resolving power of a microscope is given by the expression RP = \(\frac{2 n \sin \theta}{\lambda}\)
(i) There is no effect of the increase in the focal length of the objective on the resolving power of the microscope.
(ii) If the wavelength of the incident light is increased, the resolving power of the microscope also decreases.

Question 22.
An optical instrument uses eye-lens of power 12.5 0 and an object lens of power 50 D and has a tube length of 20 cm. Name the optical instrument and calculate its magnifying power, if it forms the final image at infinity. (CBSE Delhi 2017C)
Answer:
It is a compound microscope.
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 62
Let mo and me be the magnifications produced by the objective lens and eye lens respectively, the total magnifying power of the microscope
M = mo x me

Now mo = – \(\frac{v_{0}}{u_{0}}\)

Also, the magnifying power of the eyepiece when the final image is formed at infinity
me = \(\frac{D}{f_{e}}\)

Thus magnifying power of the microscope is
M = \(\frac{-v_{0}}{u_{0}} \times \frac{D}{f_{e}}\)

As a first approximation, ue ≈ fe and ve ≈ L = distance between objective and eyepiece (or length of microscope tube), then we have
M = \(\frac{-L}{f_{0}} \times \frac{D}{f_{e}}\)

Question 23.
Use the mirror equation to show that
(i) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
Answer:
By mirror formula we have \(\frac{1}{f}=\frac{1}{u}+\frac{1}{v}\)
Therefore \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) and for concave mirror u and f both are having negative sign, i.e. f < 0 and u < 0.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 63
It means that v > 2f, having a negative sign. Thus, the image is real and lies beyond 2f.

(ii) a convex mirror always produces a virtual image independent of the location of the object.
Answer:
(Ii) For a convex mirror \(\frac{1}{v}=\frac{1}{f}+\frac{1}{u}\) but f is positive and u is negative, i.e. f >0 but u < 0.

Therefore, it is self-evident from the above relation that irrespective of the value of u, the value of v is aLways +ve. It means that the Imaged formed by the convex mirror is always virtual independent of the location of the object.

(iii) an object placed between the pole and the focus of a concave mirror produces a virtual and enlarged image. (CBSE AI 2011)
Answer:
As for a concave mirror f and u, both are negative, hence
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 64
It means that for an object placed between the pole and principal focus of a concave mirror the image formed (as v is +ve) is virtual and magnification

m = \(\frac{1}{O}=\frac{|v|}{|u|}\) > 1, i.e. the image is an enlarged image.

Question 24.
(i) Obtain Lens makers formula using the expression \(\frac{n_{2}}{v}-\frac{n_{1}}{u}=\frac{\left(n_{2}-n_{1}\right)}{R}\), here the ray of light propagating from a rarer medium of refractive index (n1) to a denser medium of refractive index (n2) is incident on the convex side of spherical refracting surface of radius of curvature R.
(ii) Draw a ray diagram to show the image formation by a concave mirror when the object is kept between its focus and the pole. Using this diagram, derive the magnification formula for the Image formed. (CBSE DelhI 2011)
Answer:
(i) The course of rays through the Lens Is as shown. For refraction at the spherical surface XP1Y, we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 65
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 66
Consider the second surface XP2Y. Actually, the materiaL of the Lens does not extend beyond XP1Y. Therefore, before the refracted ray from A1 could meet the principal axis, it will suffer refraction at point A2 on the second face XP2Y and the Lightray will finally meet the principal axial. Such that l is the final image. Thus point lies the real Image of the virtual object I. Hence for refraction at the surface XP2Y we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 67
The above equation becomes
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 68
But n2/n1 = n, the absolute refractive index of the material of the lens, therefore the above equation takes the form
\(\frac{1}{v}-\frac{1}{u}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

But by lens formula we have \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
Therefore, from the above two equations we have
\(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) …(7)

This is the Lens maker’s equation or formula.

(ii) The ray diagram is as shown
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 69
Now from triangle BAP we have
tan θ = \(\frac{AB}{PA}\) …(1)

Also from triangle B’A’P, we have A’B’
tan θ = \(\frac{A^{\prime} B^{\prime}}{P A^{\prime}}\) …(2)

Comparing equations (1) and (2), we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 70
By Cartesian sign convention AB = h, A’B’ = h’, PA = – u and PA’= v . Substituting in equations (3) we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 71
But \(\frac{h^{\prime}}{h}\) = m ( Linear magnification),
therefore
m = – \(\frac{\text { image distance }}{\text { object distance }}\)

Question 25.
(a) Two thin convex lenses L1 and L2 of focal lengths f1 and f2 respectively, are placed coaxially in contact. An object is placed at a point beyond the focus of lens L1. Draw a ray diagram to show the image formation by the combination and hence derive the expression for the focal length of the combined system.
Answer:
Consider two lenses A and B of focal length f1 and f2 placed in contact with each other. Let the object be placed at a point O beyond the focus of the first lens L1 as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 73
The first lens L1 produces an image at l1 Since image l2 is real, it serves as a virtual object for the second lens L2, producing the final image at l. Since the lenses are thin, we assume the optical centres of the lenses to be coincident. Let this central point be denoted by P.

For the image formed by the first lens L1, we get
\(\frac{1}{v_{1}}-\frac{1}{u}=\frac{1}{f_{1}}\) …(i)

For the image formed by the second lens L2, we get
\(\frac{1}{v}-\frac{1}{v_{1}}=\frac{1}{f_{2}}\) …(ii)

Adding equations (i) and (ii) we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\) …(iii)

If the two lens-system is regarded as equivalent to a single lens of focal length f, we have
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\) …(iv)

Therefore from equations (iii) and (iv) we get
\(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{f_{2}}\)

(b) A ray PQ incident on the face AB of a prism ABC, as shown in the figure, emerges from the face AC such that AQ = AR.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 72
Draw the ray diagram showing the passage of the ray through the prism. If the angle of the prism is 60° and the refractive index of the material of the prism is \(\sqrt{3}\), determine the values of angle of incidence and angle of deviation. (CBSE AI 2015)
Answer:
The ray diagram is as shown
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 74
The prism, in this situation, is in the minimum deviation position, therefore we have
r = \(\frac{A}{2}=\frac{60}{2}\) = 30°

Hence n = \(\frac{\sin i}{\sin r}=\frac{\sin i}{\sin 30^{\circ}}\) = \(\sqrt{3}\)

This gives, i = 60°
Hence from
i = \(\frac{A+\delta_{m}}{2}\) we have
δm = 2i – A = 2 × 60° – 60° = 60°

Question 26.
(a) Plot a graph to show the variation of the angle of deviation as a function of the angle of incidence for light passing through a prism. Derive an expression for the refractive index of the prism in terms of angle of minimum deviation and angle of prism.
(b) What is the dispersion of light? What is its cause?
(c) A ray of light incident normally on one face of a right isosceles prism is totally reflected as shown in the figure. What must be the minimum value of the refractive index of glass? Give relevant calculations. (CBSE Delhi 2016)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 75
Answer:
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 76
For a prism we have
i + e = A + δ …(1)

From the graph between the angle of incidence and the angle of deviation it follows that as the angle of incidence increases, the angle of deviation first decreases becomes a minimum for a particular angle of incidence and then again increases. The minimum value of the angle of deviation is called the angle of minimum deviation. It is denoted by 5m. In this position the ray of light passes symmetrically through the prism, i.e. the refracted ray QR is parallel to the base of. the prism. In this position, the angle of incidence is equal to the angle of emergence, i.e. i = e. Also in this position, the angle of refraction at the faces of the prism are equal, i.e. r1 = r2. Substituting these values in equations (4) and (5) we have
A = r1 + r2 = r + r = 2r
or
r = A/2 …(2)
and
i + i = A + δm
or
2i = A + δm
or
i = \(\frac{A+\delta_{m}}{2}\) …(3)

Substituting for i and r in the expression for Snell’s law we have a speed of light in a vacuum
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 77
(b) The phenomenon of splitting a ray of white tight into its constituent colours (wavelengths) is catted dispersion.
Cause: It is because different wavelengths travel at different speeds In a medium other than vacuum.

(C) The diagram of the path of rays through the prism is as shown.
Here θ = 45° > ic

Now n = \(\frac{1}{\sin i_{c}}=\frac{1}{\sin 45^{\circ}}=\frac{1}{1 / \sqrt{2}}=\sqrt{2}\)
Therefore n > \(\sqrt{2}\)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 78

Numerical Problems :

Formulae for solving numerical problems

  • When Light travels from an optically denser medium to an optically rarer medium It bends away from the normal.
  • When light travels from an optically rarer medium to an optically denser medium It bends towards from the normal. speed of Light in vacuum
  • μ = \(\frac{\text { speed of light in vacuum }}{\text { speed of light in the medium }}\)
  • μ = \(\frac{1}{\sin \theta_{c}}\)
  • In minimum deviation position A = 2r or
    Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 79
  • Lens maker’s formula
    \(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)
  • MagnifyIng power of a compound microscope
    M = Me × Mo = \(\frac{L}{f_{0}} \times\left(1+\frac{D}{f_{e}}\right)\)

Question 1.
A convex lens made up of a glass of refractive index 1.5 is dipped, In turn, In
(a) a medium of refractive index 1.65,
Answer:
When dipped in the medium of refractive index 1.65, it will behave as a concave lens and when dipped in the medium of refractive index 1.33, it will behave as a convex lens.

(b) a medium of refractive index 1.33.
(i) Will it behave as a converging or a diverging lens in the two cases?
Answer:
Its focal length in another medium is given by
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 80
Thus fm = -5.5 fa, i.e. focal length increases and becomes negative.

(ii) How will Its focal length change In the two media? (CBSE AI 2011)
Answer:
Similarly
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 81
Thus fm = 3.3 fa, i.e. focal length increases.

Question 2.
A compound microscope uses an objective lens of focal length 4 cm and an eyepiece lens of focal length 10 cm. An object is placed at 6 cm from the objective lens. Calculate the magnifying power of the compound microscope. Also, calculate the length of the microscope. (CBSE Al 2011)
Answer:
fo = 4 cm, fe = 10 cm, uo = – 6 cm, M = ?, L = ?
Using
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 82
Hence angular magnification
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 83
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 84
Question 3.
A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece lens of focal length 1 cm is used, find the angular magnification of the telescope.
If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens? The diameter of the moon is 3.42 × 106 m and the radius of the lunar orbit is 3.8 × 108 m. (CBSE AI 2011, Delhi 2015)
Answer:
Given fo = 15 m, fe = 1.0 cm = 0.01 m, M = ?
Dm = 3.48 × 106 m, r = 3.8 × 108 m,

Using M = \(\frac{f_{0}}{f_{\mathrm{e}}}=\frac{15}{0.01}\) = 1500

The angle subtended by the moon at the objective of the telescope
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 85
Question 4.
A beam of light converges at a point P. A concave lens of focal length 16 cm is placed in the path of this beam 12 cm from P. Draw a ray diagram and find the location of the point at which the beam would now converge. (CBSE Delhi 2011C)
Answer:
The ray diagram is shown in the figure. In the absence of the concave lens the beam converges at point P. When the concave lens is introduced, the incident beam of light is diverged and now comes to focus at point Q. Thus for the concave lens P serves as a virtual object giving rise to a real Image at Q.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 86
Here u = + 12 cm, f = – 16 cm, v = ?, Now for a lens
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 87
Hence v = 48 cm
i. e. the point at which the beam is focused is 48 cm from the lens.

Question 5.
Two convex lenses of focal length 20 cm and 1 cm constitute a telescope. The telescope is focused on a point that is 1 m away from the objective. Calculate the magnification produced and the length of the tube, if the final image Is formed at a distance of 25 cm from the eyepiece. (CBSE Delhi 2011 C)
Answer:
Given fo = 2o cm, fe = 1 cm, u = – 100 cm, M =?, y =?

Fora lens \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
or
\(\frac{1}{v}-\frac{1}{-100}=\frac{1}{20}\)
or
v = 25cm

Since the eye lens forms the image of the virtual object at the distance of distinct vision for the eye lens
v = – 25 cm, fe = 1 cm,

Now \(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
or
\(\frac{1}{-25}-\frac{1}{u}\) = 1
or
u = – \(\frac{25}{26}\)cm

Now magnification produced by the object lens
mo = \(\frac{v}{u}=-\frac{25}{100}=-\frac{1}{4}\)

Magnification produced by the eye Lens
me = \(\frac{v}{u}=\frac{-25}{-25}\) × 26 = 26

Hence total magnification
M = mo × me = -1 /4 × 26 = – 6.5

Question 6.
(a) Under what conditions are the phenomenon of total internal reflection of light observed? Obtain the relation between the critical angle of incidence and the refractive index of the medium.
(b) Three lenses of focal lengths +10 cm, -10 cm and +30 cm are arranged coaxially as in the figure given below. Find the position of the final image formed by the combination. (CBSE Delhi 2019 C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 88
Answer:
(a) (i) Light travels from a denser medium to a rarer medium.
(ii) Angle of Incidence in the denser medium is more than the critical angle for a given pair of media.
For the grazing incidence n sin iC = l sin 90°
n = \(\frac{1}{\sin i_{c}}\)

(b) For convex lens f = + 10 cm
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 89
Object distance for concave lens u2 = 15 – 5 = 10 cm
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 90
For third lens
\(\frac{1}{f_{3}}=\frac{1}{v_{3}}-\frac{1}{\infty}\) ⇒ v3 = 30 cm

Question 7.
A ray of light incident on an equilateral glass prism (μg = \(\sqrt{3}\)) moves parallel to the baseline of the prism inside it. Find the angle of Incidence for this ray. (CBSE Delhi 2012)
Answer:
Given A = 60°, μg = \(\sqrt{3}\), i = ?

Using the expression μ = \(\frac{\sin i}{\sin A / 2}\)
or
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 91
Question 8.
Two monochromatic rays of light are incident normally on the face AB of an isosceles right-angled prism ABC. The refractive indices of the glass prism for the two rays ‘1’ and ‘2’ are respectively 1.35 and 1.45. Trace the path of these rays after entering through the prism. (CBSE AI 2014)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 92
Answer:
The critical angle for the two rays is
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 93
This shows that the angle of Incidence for ray ‘2’ Is greater than the critical angle. Hence it suffers total internal reflection, white ray ‘1’ does not. Hence the path of rays is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 94
Question 9.
A convex lens of focal length 20 cm is placed coaxially with a convex mirror of radius of curvature 20 cm. The two are kept at 15 cm from each other. A point object lies 60 cm in front of the convex lens. Draw a ray diagram to show the formation of the image by the combination. Determine the nature and position of the image formed. (CBSE AI 2014)
Answer:
The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 95
For the convex lens, we have
u1 = – 60 cm, f = + 20 cm, v = ?

Using lens formula we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 96
Had there been only the Lens, the image would have been formed at Q1, which acts as a virtual object for the convex mirror.
Therefore u2 = OQ1 – OO’ = 30 – 15 = 15 cm

Using mirror formuLa we have
\(\frac{1}{v_{2}}+\frac{1}{u_{2}}=\frac{2}{R}\)
or
\(\frac{1}{v_{2}}+\frac{1}{u_{15}}=\frac{2}{20}\)

Solving for v2 we have
v2 = 30cm

Hence the final image is formed at (Point Q) a distance of 30 cm behind the mirror.

Question 10.
A ray PQ is an incident normally on the face AB of a triangular prism refracting angle of 60°, made of a transparent material of refractive index 2 / \(\sqrt{3}\), as shown in the figure. Trace the path of the ray as it passes through the prism. Also, calculate the angle of emergence and angle deviation. (CBSE Delhi 2014C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 97
Answer:
Critical angle for glass
µ = \(\frac{1}{\sin i_{c}}\)
or
sin ic = \(\frac{1}{\mu}=\frac{\sqrt{3}}{2}\)= 0.866
or
ic = 60°

Now the ray is incident at an angle of 60° which is equal to the critical angle, therefore the ray graces the other edge of the prism

Therefore the angle of emergence is = 90°
Hence δ = 30°

This is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 98
Question 11.
An object Is placed 15 cm In front of a convex lens of focal length 10 cm. Find the nature and position of the image formed. Where should a concave mirror of radius of curvature 20 cm be placed so that the final image is formed at the position of the object itself? (CBSE AI 2015)
Answer:
Given u = -15 cm, f = + 10 cm, v = ?
For lens we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 99
Nature of image-real, magnified Final image formed will be at the object itself only if the image formed by the lens is at the position of the centre of curvature of the mirror
∴ Distance of mirror from lens, D = 30 + R = 30 + 20 cm = 50 cm

Question 12.
Define the critical angle for a pair of media. A point source of monochromatic light ‘S’ is kept at the centre of the bottom of a cylinder with the radius of 15 cm. The cylinder contains water (refractive index 4/3) to a height of 7.0 cm. Draw the ray diagram and calculate the area of the water surface through which the light emerges in the air. (CBSE Delhi 2015C)
Answer:
It the angle of incidence in the denser medium for which the angle of refraction in the rare medium is 90°.
Given h = 7 cm, A = πr² = ?, μ = 4/3 = 1.33

Due to total internal reflection light from the bulb will not come out of the entire surface of the water as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 100
The angle of the cone through which tight will spread out is twice the critical angle.

Therefore the radius of the circular patch will be
Using the relation r= \(\frac{h}{\sqrt{\mu^{2}-1}}\) we have
r = \(\frac{7}{\sqrt{(1.33)^{2}-1}}\) = 7.98 cm

Therefore area of the surface of water through which light comes out
A = πr² = 3.14 × (7.98)² = 199.95 cm²

Question 13.
A ray PQ incident on the refracting face BA Is refracted in the prism BAC as shown In the figure and emerges from the other refracting face AC as RS such that AQ = AR. If the angle of prism A = 60° and refractive Index of the material of prism is \(\sqrt{3}\), calculate angle θ. (CBSE AI 2016)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 101
Answer:
Here As AQ = AR, therefore QR is parallel to BC, hence prism is in minimum deviation position
A = 60°, θ = δm = ?, n = \(\sqrt{3}\)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 102
Solving θ = 60°

Question 14.
(i) A ray of light incident on face AB of an equilateral glass prism shows a minimum deviation of 30°. Calculate the speed of light through the prism.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 103
(ii) Find the angle of incidence at face AB so that the emergent ray grazes along with the face AC. (CBSE Delhi 2017)
Answer:
(i) Given
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 104
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 105
Question 15.
A small illuminated bulb is at the bottom of a tank, containing a liquid of refractive index p up to a height h. Find the expression for the diameter of an opaque disc, floating symmetrically on the liquid surface in order to cut off light from the bulb. (CBSE Sample Paper 2018-19)
Answer:
Let d be the diameter of the disc. The bulb shall be invisible if the incident rays from the bulb at 0 to the surface at d/2 are at the critical angle.
Let l be the angle of incidence
Then sin i = \(\frac{1}{μ}\) = tan i.
Now
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 106
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 107

Question 16.
A ray of light is incident on a glass prism of refractive index μ and refracting angle A. If it just suffers total internal reflection at the other face, obtain an expression relating to the angle of incidence, angle of prism and critical angle.
Answer:
For a ray to just suffer total internal reflection at the second face of the prism r2 = ic
Now A = r1 + ic, therefore r1 = A – ic
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 108

Now by Snells Law, we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 109
or
μ = \(\frac{\sin i}{\sin \left(A-1_{c}\right)}=\frac{1}{\sin i_{c}}\)

Question 17.
(a) Define, the refractive index of a medium.
(b) In the following ray diagram, calculate the speed of light In the liquid of unknown refractive index. (CBSE AI 2017C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 110
Answer:
(a) It is defined as the ratio of the speed of light in a vacuum to the speed of light in the given medium.
(b) The speed of Light can be found by using the formula
\(\frac{v}{c}=\frac{\sin 1}{\sin r}\)
or
v = \(\frac{\sin i}{\sin r}\) × c

From the diagram, we find that
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 111
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 112

Question 18.
A concave lens made of material of refractive index n1 is kept In a medium of refractive index n2. A parallel beam of light is incident on the lens. Complete the path of the rays of light emerging from the concave lens If (i) n1 > n2 (ii) n1 < n2 and (iii) n1 n2
Answer:
The path of rays in the three cases is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 113

Question 19.
A convex ten made of a material of refractive index n1 is kept in a medium of refractive index n2. A parallel beam of light is incident on the lens. Complete the path of rays of light emerging from the convex lens if (i) n1 > n2, (ii) n1 = n2, and (iii) n1 < n2.
Answer:
The path of the rays in the three cases is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 114
Question 20.
Write the Lens Maker’s formula and use it to obtain the range of values of μ (the refractive index of the material of the lens) for which the focal length of an equi-convex lens, kept in air, would have a greater magnitude than that of the radius of curvature of its two surface. (CBSE Delhi 2016C)
Answer:
Lens maker’s formula is given by
\(\frac{1}{f}\) = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\)

For equiconvex Lens we have R1 = + R and
R2 = – R
Therefore we have
\(\frac{1}{f}\) = (n – 1)\(\frac{2}{R}\)

For f to be greater than R
2(n – 1) < 1
2n – 2 < 1
2n < 3
n < 1.5
Hence range is 1.0< n < 1.5

Question 21.
A ray of light passing from the air through an equilateral glass prism undergoes minimum deviation when the angle of Incidence is 3/4th of the angle of the prism. Calculate the sp..d of light In the prism. (CBSE AI 2017)
Answer:
Given A = 60°, i = 3/4 × 60 = 45°,
c = 3 × 108 m s-1,

Now δm = 2i – A
= 90 – 60 = 30°

Using the formula
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 115
Question 22.
Calculate the value of the angle of Incidence when a ray of light incident on one face of an equilateral glass prism produces the emergent ray, which just grazes along the adjacent face. Refractive index of the prism is \(\sqrt{2}\). (CBSE 2017C)
Answer:
The diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 116
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 117
Question 23.
How is the working of a telescope different from that of a microscope?
The focal lengths of the objective and eyepiece of a microscope are 1.25 cm and 5 cm respectively. Find the position of the object relative to the objective in order to obtain an angular magnification of 30 in normal adjustment. (CBSE Delhi 2012)
Answer:
A microscope increases the size of the object whereas a telescope brings the object closer for better vision.
Given fo = 1.25 cm, fe = 5 cm, M = 30 cm, u = l
In normal adjustment final image is formed at infinity, therefore we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 118
Question 24.
(a) Draw a labelled ray diagram showing the image formation of a distant object by a refracting telescope. Deduce the expression for its magnifying power when the final image is formed at infinity.
(b) The sum of focal lengths of the two lenses of a refracting telescope is 105 cm. The focal length of one lens is 20 times that of the other. Determine the total magnification of the telescope when the final image is formed at infinity. (CBSE AI 2014C)
Answer:
(a) For the diagram
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 11
From right triangles ABC and ABC’ as shown in figure, we have
tan α = \(\frac{\mathrm{AB}}{\mathrm{CB}}=\frac{-h}{f_{0}}\) and
tan β = \(\frac{\mathrm{AB}}{\mathrm{C}^{\prime} \mathrm{A}}=\frac{-h}{f_{\mathrm{e}}}\)

From the above we have we have
M = \(\frac{\beta}{\alpha}=\frac{-h}{f_{e}} \times \frac{f_{0}}{-h}=\frac{f_{0}}{f_{e}}\)

(b) L = 105 cm, fo = 20fe,

Now L = fo + fe = 21fe
Or
fe = 105/21 = 5 cm
Hence fo = 20 × fe 20 × 5 = 100 cm
Hence M = fo/fe = 100/ 5 = 20

Question 25.
(a) Explain with reason, how the power of a diverging lens changes when
(i) it is kept in a medium of refractive index greater than that of the lens.
(ii) Incident red light is replaced by violet light.
(b) Three lenses L1, L2, and L3 each of focal length 30 cm are placed coaxially as shown in the figure. An object is held at 60 cm from the optic centre of L1. The final image is formed at the focus of L3. Calculate the separation between (i) L1 and L2 and (ii) L2 and L3 (CBSE AI 2017C)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 119
Answer:
(a) (i) The power of is given by the expression P = (n – 1)\(\left(\frac{1}{R_{1}}-\frac{1}{R_{2}}\right)\) medium changes the power of the lens decreases as its focal length increases.
(ii) We know that f ∝ X, therefore a decrease in the wavelength from red to violet decreases the focal length and increases the power of the lens.

(b) Given f1 = f2 = f3 = 30 cm
For lens L1, u1 = 60 cm = 2f1, therefore the image will be formed at If on the other side of the lens L1.

Since the final image for lens L3 is formed at the focus, therefore the rays of light falling on lens L3 should come from infinity. This is possible if the image of L1 lies at the focus of L2.

Thus distance L1L2 = 60 + 30 = 90 cm

Also, distance L2L3 can have any value as the rays between L2 and L3 will be parallel.

Question 26.
Calculate the radius of curvature of an equi-concave lens of refractive Index 1.5, when It Is kept In a medium of refractive index 1.4, to have a power of -5 D?
OR
An equilateral glass prism has a refractive Index of 1.6 In the air. Calculate the angle of minimum deviation of the prism, when kept In a medium of refractive Index \(\frac{4 \sqrt{2}}{5}\). (CBSE Delhi 2019)
Answer:
Given nL = 1.5, nM = 1.4, P = -5 D
Focal length f \(\frac{1}{P}=\frac{1}{-5}=\frac{-100}{5}\) = 20 cm

Using the lens makers formula and putting
R1 = – R and
R2 = +R
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 120
Solving for R, we get
R = 20/7 = 2.86cm
or
Given A = 60, n = 1.6, δm =?,nM = \(\frac{4 \sqrt{2}}{5}\)

Using the prism formula
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 121
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 122

Question 27.
A ray of light passes through an equilateral glass prism, such that the angle of Incidence Is equal to the angle of emergence. If the angle of emergence is v times the angle of the prism, calculate the refractive index of the glass prism.
Answer:
Given the angle of prism A = 60°,
the angle of incidence i = angle of emergence ‘e’ and under this condition angle of deviation is minimum.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 123
Question 28.
The magnifying power of an astronomical telescope In the normal adjustment position is 100. The distance between the objective and the eyepiece is 101 cm. Calculate the focal length of the objective and the eyepiece.
Answer:
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 124

Question 29.
A double convex lens made of glass of refractive Index 1.5 has both radii of curvature of magnItude 20 cm. An object 2 cm high is placed at 10 cm from the lens. Find the position, nature and size of the image.
Answer:
Given n = 1.5 and for a double convex Lens R1 = + 20 cm and R2 = – 20 cm
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 125
Now size of the object O = + 2 cm and u = – 1o cm
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 126
The image ¡s virtual and erect.

Question 30.
Three rays of light-red (R), green (G) and blue (B) are Incident on the face AB of a right-angled prism ABC. The refractive indices of the material of the prism for red, green and blue wavelengths are 1.39, 1.44 and 1.47, respectively. Trace the path of the rays through the prism. How will the situation change if these rays were incident normally on one of the faces of an equilateral prism?
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 127
Answer:
The course of rays through the prism is as shown below.
Critical angle for a light ray is given by
sin ic = \(\frac{1}{n}\)
or
ic = sin-1\(\left(\frac{1}{n}\right)\)

Now critical angle for red ray (R)
iR = sin-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.39}\right)\) = 46°

Critical angle for green ray (G)
iG = sin-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.44}\right)\) = 44°

and Cnticat angLe for bLue ray (B)
iB = sin-1\(\left(\frac{1}{n}\right)\) = sin-1\(\left(\frac{1}{1.47}\right)\) = 42°52′

Thus the angle of incidence for blue and green rays inside the prism will be greater than the critical angle for these two colours. Therefore blue and green rays undergo total internal reflection. The angle of incidence for red colour is smaller than its critical angle; therefore it does not undergo total internal reflection. This is depicted in the figure below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 128
For an equilateral prism, the light rays pass through the prism suffering deviation. The angle of deviation δ = (n – 1) A. Hence, the deviation for red coloured ray will be the least and that for blue ray maximum. Paths of rays are shown in the figure below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 129
Question 31.
You are given three lenses of powers 0.5 D, 4 D, 10 D. State, with reason, which two lenses will you select for constructing a good astronomical telescope.
Calculate the resolving power of this telescope, assuming the diameter of the objective lens to be 6 cm and the wavelength of light used to be 540 nm.
Answer:
The focal lengths of the three lenses are
f1 = 1 /P1 = 1 / 0.5 = 2 m = 200 cm,
f2 = 1 /P2 = 1 / 4 = 0.25 m = 25 cm,
f3 = 1 /P3 = 1 /10 = 0. 1 m = 10 cm,

For an astronomical telescope, the objective should have a large focal length and the eyepiece should have a small focal length. Therefore for the objective, we will use the lenses of power 0.5 D and for the eyepiece the lens of power 10 D.

Given D = 6 cm = 0.06 m, λ = 540 nm = 540 × 10-9 m

Now resolving power
R.P. = \(\frac{D}{1.22 \lambda}=\frac{0.06}{1.22 \times 540 \times 10^{-9}}\) = 91074.6

Question 32.
A screen is placed 90 cm from an object. The image of the object on the screen is formed by a convex lens at two different positions separated by 20 cm. Calculate the focal length of the lens.
OR
A convex lens of focal length 20 cm and a concave lens of focal length 15 cm are kept 30 cm apart with their principal axes coincident. When an object is placed 30 cm in front of the convex lens, calculate the position of the final image formed by the combination. Would this result change if the object were placed 30 cm In front of the concave tens? Give reason. (CBSE AI 2019)
Answer:
The image of an object can be obtained at two different positions of a convex lens for a fixed value of the distance between the object and the screen if values of u and v are interchanged in these two positions. This situation is as shown below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 130
Here x + 20 + x = 90 cm or x = 35cm
Therefore, u = -35 cm, v = + 55 cm, f = ?
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 131
or
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 132
For image formed by a convex Lens
f1 = + 20 cm, u = – 30 cm
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 133
Therefore, u for concave Lens = 60 – 30 = + 30 cm and f2 = – 15 cm

Now for concave lens
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 134
No, the result will not change as per the principle of reversibility.

Question 33.
For an equilateral glass prism, the minimum angle of deviation, for a parallel beam of monochromatic Incident light, is measured to be 30°. What is the refractive index of the glass used for making this prism?
If this prism were to be placed in a medium of refractive Index 1.414, how would a ray of light incident on one of its refracting faces pass through it? Draw the ray diagram for the same.
Answer:
Given A = 60°, δm = 30°, μ = ?
Using the relation
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 135
When this prism is placed in a medium of refractive index 1.414, it will behave like a plane glass piece as both the prism and the medium have the same refractive index.

The ray diagram is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 136
Question 34.
Find the position of the image formed by the lens shown in the figure.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 137
Another lens is placed in contact with this lens to shift the image further away from the lens. What is the nature of the second lens?
Answer:
f = + 10 cm, u = – 30 cm, v =?
Using the lens equation we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 138
The second lens Is concave.

Question 35.
A beam of light of wavelength 400 nm is incident normally on a right-angled prism as shown. It is observed that the light just grazes along with the surface AC after falling on it. Given that the refractive index of the material of the prism varies with the wavelength as per the relation
μ = 1.2 + \(\frac{b}{\lambda^{2}}\) calculate the value of b and the refractive index of the prism material for a wavelength λ = 500 nm. [(Given θ = Sin-1 (0.625)]
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 139
Answer:
The figure can be redrawn as shown below.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 140
By symmetry angle, θ will be the critical angle.
Therefore
μ = \(\frac{1}{\sin \theta}=\frac{1}{0.625}\) = 1.6

Now the value of the constant b is obtained as follows:
μA = 1.2 + \(\frac{b}{\lambda^{2}}\)
or
1.6 = 1.2 + \(\frac{b}{(400)^{2}}\)

Solving for b we have b = 64000 nm2
Now for a wavelength λ = 500 nm the refractive index of the material of the prism is
μ = 1.2 + \(\frac{(64000)^{2}}{(500)^{2}}\) = 1.456

Question 36.
A convex lens, of focal length 20 cm, has a point object placed on its principal axis at a distance of 40 cm from It. A plane mirror Is placed 30 cm behind the convex lens. Locate the position of the image formed by this combination.
Answer:
We first consider the effect of the lens. For the Lens, we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 141
u = – 40 cm and f = + 20 cm
Using the lens formula, we get
\(\frac{1}{v}-\frac{1}{(-40)}=\frac{1}{20}\)
or
v = + 40 cm

Had there been the lens only the image would have been formed at Q1. The plane mirror M is at a distance of 30 cm from the lens L. We can, therefore, think of a Q1 as a virtual object, located at a distance of 10 cm, behind the plane mirror M. The plane mirror, therefore, forms a real image (of this virtual object Q1) at Q1, 10 cm in front of it. This is shown in the figure.

Question 37.
(i) Calculate the distance of an object of height h from a concave mirror of radius of curvature 20 cm, so as to obtain a real image of magnification 2. Find the location of the image also.
(ii) Using the mirror formula, explain why does a convex mirror always produce a virtual image. (CBSE Delhi 2016) Answer:
Given R = – 20 cm or f = – 10 cm, m = + 2 (real image), u = ?, v = ?
Using the relation m = \(\frac{f}{f-u}\) we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 142
(b) For a convex mirror \(\frac{1}{v}=\frac{1}{f}-\frac{1}{u}\) but f is positive and u is negative i.e., f > 0 but u < 0.
Therefore, it is self-evident from the above relation that irrespective of the value of u, the value of v is always +ve. It means that the image formed by a convex mirror is always virtual Independent of the location of the object.

Question 38.
A convex lens of focal length 20 cm and a concave mirror of focal length 10 cm are placed co-axially 50 cm apart from each other. An incident beam parallel to its principal axis is incident on the convex lens. Locate the position of the (final) image formed due to this combination.
Answer:
The incident beam, on lens L, is parallel to its principal axis. Hence the lens forms an image Q1 at its focal point, i.e. at a distance OQ1 (= 20 cm) from the lens. This image, Q1, now acts as a real object for the concave mirror. For the mirror, we then have: u = – 30 cm, and f = – 10 cm,

Hence using the mirror formula, we get
\(\frac{1}{v}+\frac{1}{(-30)}=\frac{1}{(-10)}\)
or
v = – 15 cm

The lens-mirror combination, therefore, forms a real image Q at a distance of 15 cm from M. The ray diagram is as shown in
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 143
Question 39.
A convex lens of focal length 25 cm and a concave mirror of radius of curvature 20 cm are placed coaxially 40 cm apart from each other. An incident beam parallel to the principal axis is incident on the convex lens. Find the position and nature of the image formed by the combination. (CBSE Al 2016)
Answer:
The image of the object is formed at the focus of the convex lens, i.e. 25 cm. This image acts as an object for the concave mirror and ties at a distance (40 – 25) = 15 cm from the mirror.

Hence for the mirror
u = – 15 cm, v =? f = R/2 = 20/2 = 10 cm

Using mirror formula we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 144
The image is formed in front of the mirror.

Question 40.
(i) A screen is placed at a distance of 100 cm from an object. The image of the object is formed on the screen by a convex lens for two different locations of the lens separated by 20 cm. Calculate the focal length of the lens used.
(ii) A converging lens is kept coaxially in contact with a diverging lens both the lenses being of equal focal length. What is the focal length of the combination? (CBSE AI 2016)
Answer:
(i) Given u + v= 100 or v= 100 – u
Using lens formula From lens first \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) we have
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 145
Or
f = 24 cm

(ii) f = infinity. \(\frac{1}{f}=\frac{1}{f_{1}}+\frac{1}{-f_{2}}\) = 0

Question 41.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on the face of the prism. By rotating the prism, the angle of minimum deviation is measured to be 40°. What is the refractive index of the material of the prism? If the prism is placed in water (refractive index 1.33), predict the new angle of minimum deviation of a parallel beam of light. The refracting angle of the prism is 60°. (NCERT)
Answer:
Given δm = 40°, aμg = ?, aμw = 1.33, A = 60°
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 146
Solving for δm we have δm = 100°

Question 42.
A beam of light converges to a point P. A lens is placed in the path of the convergent beams 12 cm from P. At what point does the beam converge if the lens is
(i) a convex lens of focal length 20 cm,
Answer:
(i) Here u = 12 cm, f = + 20 cm and v = ?
Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\) we have
\(\frac{1}{v}=\frac{1}{12}+\frac{1}{20}=\frac{3+5}{60}=\frac{8}{60}\)
or
v = 7.5 cm

(ii) a concave lens of focal length 16 cm? (NCERT)
Answer:
Here u = 12 cm, f = – 16 cm and v =?
Using the formula \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)we have
\(\frac{1}{v}=\frac{1}{12}+\frac{1}{-16}=\frac{-3+4}{48}=\frac{1}{48}\)
or
v = 48 cm

Question 43.
A compound microscope consists of an objective lens of focal length 2.0 cm and an eyepiece of focal length 6.25 cm separated by a distance of 15 cm. How far from the objective should an object be placed in order to obtain the final image at (i) the least distance of distinct vision (25 cm), (ii) infinity? What is the magnifying power of the microscope in each case? (NCERT)
Answer:
Given f0 = 2.0 cm, fe = 6.25 cm, L = 15 cm, u0 = ?,
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 147
Now applying Lens formula for objective Lens we have \(\frac{1}{f}=\frac{1}{v}-\frac{1}{u} \text { or } \frac{1}{u_{0}}=\frac{1}{v_{0}}-\frac{1}{f_{0}}\)
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 148
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 149

Question 44.
A small telescope has an objective lens of a focal length of 144 cm and an eyepiece of a focal length of 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eyepiece? (NCERT)
Answer:
Given fo = 144 cm, fe = 6.0 cm, M = ?,
L = fo + fe = ?

Using the formula M = \(\frac{f_{0}}{f_{e}}=\frac{144}{6}\) = 24 and
L = fo + fe = 144 + 6 = 150 cm

Question 45.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eyepiece of focal length 1.0 cm is used, what is the angular magnification of the telescope?
(b) If this telescope Is used to view the moon, what is the diameter of the Image of the moon formed by the objective lens? The diameter of the moon is 3.48 × 106 m, and the radius of lunar orbit Is 3.8 × 108 m. (NCERT)
Answer:
Given fo = 15 m, fe = 1.0 cm = 0.01 m, M = ?
Dm = 3.48 × 106 m, r = 3.8 × 108m,
(a) Using M = \(\frac{f_{0}}{f_{e}}=\frac{15}{0.01}\) = 1500
(b) AngLe subtended by moon at the objective of the teLescope
α = \(\frac{D_{m}}{r}=\frac{3.48 \times 10^{6}}{3.8 \times 10^{8}}\)

Therefore angLe subtended by the image
β = M × α = 1500 × \(\frac{3.48 \times 10^{6}}{3.8 \times 10^{8}}\)

If I be the Linear size of the image then
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 150
Question 46.
At what angle should a ray of light be incident on the face of a prism of refracting angle 60° so that it just suffers total internal reflection at the other face? The refractive index of the material of the prism is 1.524. (NCERT)
Answer:
Given A = 60°, i = ?, μ = 1.524
For a ray to just suffer total internal reflection at the second face of the prism r2 = ic
Now sin ic = \(\frac{1}{\mu}=\frac{1}{1.524}\) = 0.6562
or
ic = 41°

Now A = r1 + r2, therefore r1 = A – r2
r1 = 60 – 41 = 19°

Now by Snell’s law we have
μ = \(\frac{\sin i}{\sin r_{1}}\)
or
sin i = μ × sinr1 = 1.524 × sin 19°
or
i = 29.75° = 30°

Question 47.
A small telescope has an objective lens of a focal length of 140 cm and an eyepiece of a focal length of 5.0 cm. What is the magnifying power of the telescope for viewing distant objects when
(a) the telescope is in normal adjustment (i.e. when the final image is at infinity.
Answer:
Given fo = 140 cm, fe = 5.0 cm, M = ?
For normal adjustment we have
M = \(\frac{f_{0}}{f_{e}}=\frac{140}{5}\) = 28

(b) the final image is formed at the least distance of distinct vision (25 cm) (NCERT)
Answer:
When the final image is formed at the least distance of distinct vision we have
M = \(\frac{f_{0}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)=\frac{140}{5}\left(1+\frac{5}{25}\right)\) = 33.6

Question 48.
(a) The refractive index of glass is 1.5. What is the speed of light In a glass? (Speed of light in a vacuum is 3.0 × 108 ms-11 )
(b) Is the speed of light in glass independent of the colour of light? If not, which of the two colours red and violet travels slower in a glass prism?
Answer:
(a) Given μ = 1.5, c = 3.0 × 108 m s-1
v = \(\frac{c}{\mu}=\frac{3 \times 10^{8}}{1.5}\) = 2.0 × 108 ms-1

(b) No. The refractive index, and hence the speed of light in a medium, depends on wavelength. When no particular wavelength or colour of light is specified, we may take the given refractive index to refer to yellow colour or mean light. Now we know violet colour deviates more than red in a glass prism, i.e. μv > μR. Therefore, the violet component of white light travels slower than the red component.

Question 49.
Three immiscible liquids of densities d1 > d2 > d3 and refractive indices μ1 > μ2 > μ3 are put in a beaker. The height of each liquid column is d3. A dot is made at the bottom of the beaker. For near-normal vision, find the apparent depth of the dot. (NCERT Exemplar)
Answer:
The situation is as shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 151
Let the apparent depth be O1 for the object as seen from μ2, then
Apparent depth = real depth/μ
O1 = \(\frac{h / 3}{\mu_{1} / \mu_{2}}=\frac{\mu_{2} h}{3 \mu_{1}}\)

If seen from O3 the apparent depth O2 is
Apparent depth = real depth/μ
O2 = \(\frac{\left(h / 3+O_{1}\right)}{\mu_{2} / \mu_{3}}=\frac{\mu_{3}}{\mu_{2}}\left(\frac{h}{3}+\frac{\mu_{2}}{\mu_{1}} \frac{h}{3}\right)\)

Seen from outside the apparent height is
Apparent depth = real depth/μ
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 152
Question 50.
For a glass prism (μ = \(\sqrt{3}\)) the angle of minimum deviation Is equal to the angle of the prism. Find the angle of the prism. (NCERT Exemplar)
Answer:
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 153
Question 51.
In many experimental set-ups, the source and screen are fixed at a distance say D and the lens Is movable. Show that there are two positions for the lens for which an image Is formed on the screen. Find the distance between these points and the ratio of the Image sizes for these two points. (NCERT Exemplar)
Answer:
The situation is shown.
Class 12 Physics Important Questions Chapter 9 Ray Optics and Optical Instruments 154
If there was no cut then the object would have been at a height of 0.5 cm from the principal axis OO’. Consider the image for this case by lens formuLa we have
\(\frac{1}{v}=\frac{1}{u}+\frac{1}{f}=\frac{1}{-50}+\frac{1}{25}=\frac{1}{50}\)
Or
v = 50 cm
Hence magnification of the image is.
m = \(\frac{v}{u}=\frac{-50}{50}\) = -1

Thus the image would have been formed at 50 cm from the pole and 0.5 cm below the principal axis. Hence with respect to the X-axis passing through the edge of the cut lens, the co-ordinates of the image are (50 cm, -1 cm)

Dual Nature of Radiation and Matter Class 12 Important Extra Questions Physics Chapter 11

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 11 Dual Nature of Radiation and Matter. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 11 Important Extra Questions Dual Nature of Radiation and Matter

Dual Nature of Radiation and Matter Important Extra Questions Very Short Answer Type

Question 1.
If the maximum kinetic energy of electrons emitted by photocell is 4 eV, what is the stopping potential?
Answer:
The stopping potential is 4 V.

Question 2.
Two metals A and B have a work function 4 eV and 10 eV respectively. Which metal has a higher threshold wavelength?
Answer:
The threshold wavelength is inversely proportional to the work function. Therefore metal A has a higher threshold wavelength.

Question 3.
Ultraviolet light is incident on two photo-sensitive materials having work functions W1 and W2 (W1 > W2). In which case will the kinetic energy of the emitted electrons be greater? Why?
Answer:
The kinetic energy of the emitted photoelectrons is given by 1/2 mv² = hv – W; therefore the lesser the work function for a given frequency, the more is the kinetic energy of the emitted photoelectrons. Since W2 < W1, kinetic energy will be more for the metal having work function W2.

Question 4.
Does the threshold frequency depend on the intensity of light?
Answer:
No, it does not.

Question 5.
Name the experiment which establishes the wave nature of a particle.
Answer:
Davison-Germer experiment.

Question 6.
Mention one physical process for the release of electrons from a metal surface.
Answer:
Photoelectric effect.

Question 7.
Name a phenomenon that illustrates the particle nature of light.
Answer:
Photoelectric effect.

Question 8.
Define the work function for a given metallic surface.
Answer:
The minimum amount of energy required to just eject an electron from a given metal surface is called the work function for that metal surface.

Question 9.
What is the value of stopping potential between the cathode and anode of a photocell, if the maximum kinetic energy of the electrons emitted is 5 eV?
Answer:
Stopping potential = 5 V.

Question 10.
Define the term ‘stopping potential’ in relation to the photoelectric effect. (CBSE AI 2011)
Answer:
It is the negative potential of the collector plate for which no photoelectron reaches the collector plate.

Question 11.
Write the relationship of de-Broglie wavelength associated with a particle of mass m in terms of its kinetic energy E. (CBSE Delhi 2011C)
Answer:
The required relation is λ = \(\frac{h}{\sqrt{2 m E}}\)

Question 12.
State de-Broglie hypothesis. (CBSE Delhi 2012)
Answer:
It states that moving particles should possess a wave nature.

Question 13.
Define the term “threshold frequency”, in the context of photoelectric emission. (CBSE Delhi 2019)
Answer:
It is the minimum frequency of incident radiation (light) that can cause photoemission from a given photosensitive surface.
Or
Define the term “Intensity” in the photon picture of electromagnetic radiation. (CBSE Delhi 2019)
Answer:
It is defined as the number of energy quanta (photons) per unit area per unit time.

Question 14.
Write the expression for the de-Broglie wavelength associated with a charged particle having charge ‘q’ and mass ‘m’, when it is accelerated by a potential V. (CBSE AI 2013)
Answer:
The expression is λ = \(\frac{h}{\sqrt{2 m q V}}\)

Question 15.
Define the intensity of radiation on the basis of the photon picture of light. Write its SI unit. (CBSE AI 2014)
Answer:
The intensity of radiation is defined as the number of energy quanta per unit area per unit time. It is measured in W m-2.

Question 16.
Name the phenomenon which shows the quantum nature of electromagnetic radiation. (CBSE Al 2017)
Answer:
Photoelectric effect.

Question 17.
If the distance between the source of light and the cathode of a photocell is doubled how does it affect the stopping potential applied to the photocell? (CBSE Delhi 2017C)
Answer:
No effect.

Question 18.
Draw graphs showing the variation of photoelectric current with applied voltage for two incident radiations of equal frequency and different intensities. Mark the graph for the radiation of higher intensity. (CBSE AI 2018)
Answer:
The graphs are as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 1

Question 19.
An electron is accelerated through a potential difference V. Write the expression for its final speed if it was initially at rest. (CBSE AI Delhi 2018C)
Answer:
Using eV = \(\frac{1}{2}\)mv² or v = \(\sqrt{\frac{2 e V}{m}}\)

Question 20.
Write the name given to the frequency v0, in the following graph (showing the variation of stopping potential (Vo) with the frequency (v) of the incident radiation) for a given photosensitive material. Also name the constant, for that photosensitive material, obtained by multiplying vc with Planck’s constant.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 2
Answer:
The constant obtained by multiplying v0 with Planck’s constant is called the work function of the material. This frequency is called the threshold frequency.

Question 21.
Electrons are emitted from a photosensitive surface when it is illuminated by green light but electron emission does not take place by yellow light. Will the electrons are emitted when the surface is illuminated by
(a) red light
Answer:
No electron will be emitted when illuminated by a red light.

(b) blue light?
Answer:
Electron emission takes place with blue light.

Question 22.
The frequency (v) of incident radiation is greater than the threshold frequency v0 in a photocell. How will the stopping potential vary if frequency (v) is increased keeping other factors constant?
Answer:
On increasing the frequency v of the incident light, the value of stopping potential also increases.

Question 23.
How much time is taken by a photoelectron to come out of a metal surface, when the light of wavelength less than threshold wavelength λo is incident on it?
Answer:
It is an instantaneous process. The time required is of the order of a nanosecond (10-9 s).

Question 24.
The given graphs show the variation of photoelectric current (l) with the applied voltage (V) for two different materials and for two different intensities of the incident radiations. Identify the pairs of curves that correspond to different materials but the same intensity of incident radiations.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 3
Answer:
Pairs 1 – 2 and pairs 3 – 4

Question 25.
Red light, however bright, cannot cause the emission of electrons from a clean zinc surface. But even weak ultraviolet radiations can do so. Why?
Answer:
This is because the threshold frequency of the given metal is greater than the frequency of red light.

Question 26.
If the intensity of incident radiation on metal is doubled, what happens to the kinetic energy of electrons emitted?
Answer:
The kinetic energy of emitted electrons remains unchanged.

Question 27.
If the intensity of the incident radiation in a photocell is increased, how does the stopping potential vary?
Answer:
Stopping potential remains unaffected.

Question 28.
The work function of aluminium is 4.2 eV. If two photons each of energy 2.5 eV are incident on a surface, will the emission of photoelectron take place?
Answer:
No, emission of photoelectron will not take place.

Question 29.
A photon and an electron have the same de-Broglie wavelength. Which is moving faster?
Answer:
Photon is moving faster with a speed c = 3 × 108 ms-1. An electron must have a velocity less than the speed of light.

Question 30.
Does the ‘stopping potential’ in photoelectric emission depend upon
(i) the intensity of the incident radiation in a photocell
Answer:
No

(ii) the frequency of the incident radiation?
Answer:
Yes.

Question 31.
The de-Broglie wavelengths, associated with a proton and a neutron, are found to be equal. Which of the two has a higher value for kinetic energy?
Answer:
Proton.

Question 32.
The figure shows a plot of \(\frac{1}{\sqrt{V}}\) where V is the accelerating potential vs the de-Broglie wavelength λ in case of two particles having the same charge q but different masses m1 and m2. Which line A or B represents the particle of greater mass? (CBSE AI 2013C)
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 4
Answer:
Particle A

Question 33.
Do X-rays exhibit the phenomenon of the photoelectric effect?
Answer:
Yes, they do exhibit the phenomenon of the photoelectric effect.

Question 34.
Draw a graph showing the variation of de-Broglie wavelength with the momentum of an electron. (CBSE AI 2019)
Answer:
The graph is shown below.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 5

Question 35.
The work function of two metals A and B are 2 eV and 5 eV respectively. Which of these is suitable for a photoelectric cell using visible tight?
Answer:
Metal A having a lower work function of 2 eV is suitable for use with visible light.

Question 36.
Estimate the frequency associated with a Photon of energy 2 eV. (CBSE Delhi 2019C)
Answer:
Since E = hv
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 6

Question 37.
For a given photosensitive material and with a source of the constant frequency of incident radiation, how does the photocurrent vary with the intensity of incident light? (CBSE AI 2011C)
Answer:
The photoelectric current increases linearly with the intensity of light.

Question 38.
Draw a graph showing the variation of the de-Broglie wavelength of an electron as a function of its kinetic energy. (CBSE Delhi 2015C)
Answer:
The graph is as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 7

Question 39.
(a) In the explanation of the photoelectric effect, we assume one photon of frequency v collides with an electron and transfers its energy. This leads to the equation for the maximum energy Emax of the emitted electron as Emax = hv – Φo where Φo is the work function of the metal. If an electron absorbs 2 photons (each of frequency v), what will be the maximum energy for the emitted electron?
Answer:
Emax = 2hv – Φ

(b) Why is this fact (two photons absorption) not taken into consideration in our discussion of the stopping potential? (NCERT Exemplar)
Answer:
The probability of absorbing 2 photons by the same electron is very low. Hence ’ such emissions will be negligible.

Question 40.
Do all the electrons that absorb a photon comes out as photoelectrons? (NCERT Exemplar)
Answer:
No, most electrons get scattered into the metal. Only a few come out of the surface of the metal.

Dual Nature of Radiation and Matter Important Extra Questions Short Answer Type

Question 1.
An a-particle and a proton of the same kinetic energy are in turn allowed to pass through a magnetic field B, acting normal to the direction of motion of the particles. Calculate the ratio of radii of the circular paths described by them. (CBSE Delhi 2019)
Answer:
Given qα = 2e, qp = e, Kα = Kp, mα = 4mp, rα/rp = ?
Using the expression
r = \(\frac{\sqrt{2 m K}}{q B}\) we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 8
Question 2.
How will the photoelectric current change on decreasing the wavelength of incident radiation for a given photosensitive material?
Answer:
Photoelectric current is independent of the wavelength of the incident radiation. Therefore there will be no change in the photoelectric current.

Question 3.
Estimate the ratio of the wavelengths associated with the electron orbiting around the nucleus in the ground and first excited states of a hydrogen atom. (CBSE Delhi 2019C)
Answer:
Since De Brogue’s hypothesis is related to
Bohr’s atomic model as
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 9
Question 4.
Show graphically how the stopping potential for a given photosensitive surface varies with the frequency of the incident radiation.
Answer:
The required graph is as shown
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 10
Question 5.
the de-Broglie wavelength associated with an electron accelerated through a potential difference V is λ. What will be its wavelength when accelerating potential is increased to 4 V?
Answer:
The de-BrogLie wavelength is inversely proportional to the square root of potential, therefore = \(\frac{\lambda_{2}}{\lambda_{1}}=\frac{\sqrt{V}}{\sqrt{4 V}}=\frac{1}{2}\) . Thus wavelength
wilt become half of its previous value.

Question 6.
Plot a graph showing the variation of de Brogue wavelength (λ) associated with a charged particle of mass m, versus \(\frac{1}{\sqrt{V}}\) where V is the potential difference through which the particle is accelerated. How does this graph give us information regarding the magnitude of the charge of the particle? (CBSE Dethi 2019)
Answer:
The plot is as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 11
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 12
Question 7.
X-rays of wavelength ‘λ’ fall on a photosensitive surface, emitting electrons. Assuming that the work function of the surface can be neglected, prove that the de-Broghe wavelength of the electrons emitted will be \(\sqrt{\frac{h \lambda}{2 m c}}\)
Answer:
The energy possessed by X-rays of wavelength λ is given by E=hc / λ.
Consider an electron of mass charge e to be accelerated the potential difference of V volts the velocity gained by it.

Then kinetic energy of electron is
E = \(\frac{1}{2} m v^{2}\) = eV
or
v = \(\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}=\sqrt{\frac{2 E}{m}}\)

If λ is the de-Broglie wavelength associated with an electron, then
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 13
Substituting for e, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 14
Question 8.
Explain with the help of Einstein’s photoelectric equation any two observed features in the photoelectric effect. cannot be explained by the wave theory. (CBSE Delhi 2019)
Answer:
According to Einstein’s equation, we have
\(\frac{1}{2} m v_{\max }^{2}\) = h(v – vo)

Two features
(a) Maximum energy is directly proportional to the frequency
(b) Existence of threshold frequency Explanation of two features:

  1. The energy of the photon is directly proportional to the frequency
  2. No photoelectric emission is possible if hv < hvo

Question 9.
Why is the wave theory of electromagnetic radiation not able to explain the photoelectric effect? How does the photon picture resolve this problem? (CBSE Delhi 2019)
Answer:
According to the wave theory, the more intense a beam, more is the kinetic energy it will impart to the photoelectron. This does not agree with the experimental observations (max K.E. of the emitted photoelectron is independent of intensity) on the photoelectric effect. Also according to the wave theory photoemission can occur at all frequencies.

The photon picture resolves this problem by saying that light in interaction with matter behaves as if it is made of quanta or packets of energy, each of energy hv. This picture enables us to get a correct explanation of all the observed experimental features of the photoelectric effect.

Question 10.
(a) Define the terms,
(i) threshold frequency and
(ii) stopping potential in the photoelectric effect.
(b) Plot a graph of photocurrent versus anode potential for radiation of frequency v and intensities l1 and l2. (l1 < l2). (CBSE Delhi 2019)
Answer:
(a) Threshold frequency: It is the frequency of the incident radiation below which photoelectric effect does not take place.
Stopping potential: It is the minimum negative (retarding) potential, given to the anode (collector plate) for which the photocurrent stops or becomes zero.

(b) The plot is as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 15
Question 11.
A proton and a particle are accelerated through the same potential difference. Which one of the two has
(i) greater de-Broglie wavelength, and
(ii) less kinetic energy? Justify your answer. (CBSE AI 2016)
Answer:
(i) We know that λ = \(\frac{h}{\sqrt{2 m q V}}\), therefore
we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 16
Question 12.
Plot a graph showing the variation of de-Broglie wavelength λ versus \(\frac{1}{\sqrt{V}}\) where V is the accelerating potential for two particles A and B carrying the same charge but masses m1 m2 (m1 > m2). Which one of the two represents a particle of smaller mass and why? (CBSE Delhi 2016)
Answer:
The graph is as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 17
we know that λ = \(\frac{h}{\sqrt{2 m q V}}\)

In graphs, charge of both particles is the same, but the slope of graph B is more. It means that mass of particle B is less (since slope ∝ \(\frac{1}{\sqrt{m}}\)

Question 13.
For a photosensitive surface, the threshold wavelength is λo. Does photoemission occur if the wavelength λ of the incident radiation is
(i) more than λo and
(ii) less than λo? Justify your answer.
Answer:
The photoelectric effect occurs and hence photoelectrons are ejected when the wavelength of the incident radiation is lesser than the threshold wavelength.
(i) When λ > λo, photoemission does not take place.
(ii) When λ < λo, photoemission takes place.

Question 14.
Two monochromatic radiations of frequencies v1 and v2 (v1 > v2) and having the same intensity is, in turn, an incident on a photosensitive surface to cause photoelectric emission. Explain, giving a reason, in which case (i) more number of electrons will be emitted and (ii) maximum kinetic energy of the emitted photoelectrons will be more. (CBSE Delhi 2014C)
Answer:
The number of photoelectrons emitted depends upon the intensity of radiation and the kinetic energy of photoelectrons depends upon the frequency of radiation, therefore

  1. The same number of electrons will be emitted.
  2. Photoelectrons will have more kinetic energy for radiation of frequency v1.

Question 15.
A proton and an a-particle are accelerated, using the same potential difference. How are the de-Broglie wavelengths λP and λα related to each other? (NCERT Exemplar)
Answer:
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 18
Question 16.
There are materials that absorb photons of shorter wavelength and emit photons of longer wavelength. Can there be stable substances that absorb photons of larger wavelength and emit light of shorter wavelength? (NCERT Exemplar)
Answer:
In the first case energy given out is less than the energy supplied. In the second case, the material has to supply energy as the emitted photon has more energy. This cannot happen for stable substances.

Question 17.
Two monochromatic beams A and B of equal intensity l hit a screen. The number of photons hitting the screen by beam A is twice that by beam B. Then what inference can you make about their frequencies? (NCERT Exemplar)
Answer:
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 19
The frequency of beam B is twice that of beam A.

Question 18.
Two particles A and B of de Broglie wavelengths λ1 and λ2 combine to form a particle C. The process conserves momentum. Find the de Broglie wavelength of the particle C. (The motion is one dimensional.) (NCERJExemplar)
Answer:
The motion is one dimension, therefore
(i) If the particles move in the same direction
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 20
(ii) If the particles move in the opposite direction
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 21

Question 19.
Find the frequency of light that ejects electrons from a metal surface, fully stopped by a retarding potential of 3.3 V. If photoelectric emission begins in this metal at a frequency of 8 × 1014 Hz, calculate the work function (in eV) for this metal.
Or
Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3W. Calculate the (i) energy of a photon in the light beam and (ii) the number of photons emitted on an average by the source. (CBSE Delhi 2018C)
Answer:
We have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 22

Dual Nature of Radiation and Matter Important Extra Questions Long Answer Type

Question 1.
What is the photoelectric effect? Write Einstein’s photoelectric equation and use it to explain (a) independence of maximum energy of emitted photoelectrons from the intensity of incident light and
(b) existence of a threshold frequency for the emission of photoelectrons.
Answer:
The election of photoelectrons from a metal surface when Light of suitabLe frequency is incident on it is catted photoelectric effect.

Einstein’s equation of photoelectric effect is \(\frac{1}{2}\)mv² = hv – ω0
(a) In accordance with Einstein’s equation, the kinetic energy of the photoelectrons is independent of the intensity of the incident radiation.
(b) In accordance with Einstein’s equation, the kinetic energy will be positive and hence photoelectrons will be ejected if v > v0. Thus below a certain frequency called threshold frequency, photoelectrons are not ejected from a metal surface (if v < v0).

Question 2.
An electron of mass m and charge q is accelerated from rest through a potential difference of V. Obtain the expression for the de-Broglie wavelength associated with it. If electrons and protons are moving with the same kinetic energy, which one of them will have a larger de-Broglie wavelength associated with it? Give reason.
Answer:
Consider an electron of mass m and charge e to be accelerated through a potential difference of V volts. Let v be the velocity gained by it. Then kinetic energy of the electron is
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 23
If λ is the de-Broglie wavelength associated with an electron, then
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 24
Since de-Broglie wavelength is inversely proportional to the square root of mass, the lesser the mass, the more is the de- Broglie wavelength. Since the mass of an electron is lesser than that of the proton, the electron has a greater de-Broglie wavelength than a proton.

Question 3.
Sketch the graphs showing the variation of stopping potential with the frequency of incident radiations for two photosensitive materials A and B having threshold frequencies v0 > v’0 respectively.
(a) Which of the two metals A or B has a higher work function?
(b) What information do you get from the slope of the graphs?
(c) What does the value of the intercept of graph ‘A’ on the potential axis represent?
Answer:
The graphs are as shown below.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 25
(a) The work function is directly proportional to the threshold frequency. The threshold frequency of metal A is greater than that of metal B; therefore A has a greater work function than B.
(b) The slope of the graphs gives the value of Planck’s constant.
(c) The intercept on the potential axis is negative (-W0/e) w.r.t. stopping potential, i.e. Work function = e × magnitude of the intercept on the potential axis. We may infer it to give the voltage which, when applied with opposite polarity to the stopping voltage, will just pull out electrons from the metallic atom’s outermost orbit.

Question 4.
When a given photosensitive material is irradiated with light of frequency v, the maximum speed of the emitted photoelectrons equals Vmax. The graph shown in the figure gives a plot of V²max varying with frequency v.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 26
Obtain an expression for
(a) Planck’s constant, and
Answer:
By Einstein’s photoelectric equation we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 27
(b) The work function of the given photosensitive material in terms of the parameters T, ‘n’ and the mass ‘m’ of the electron.
Answer:
The intercept on V²max axis is = \(\frac{2 \phi_{o}}{m}\) = l
Therefore, work function Φ0 = \(\frac{ml}{2}\)

(c) How is threshold frequency determined from the plot? (CBSE AI 2019)
Answer:
The threshold frequency is the intercept on the v axis i.e. v0 = n

Question 5.
X-rays fall on a photosensitive surface to cause photoelectric emission. Assuming that the work function of the surface can be neglected, find the relation between the de-Broglie wavelength (λ) of the electrons emitted to the energy (Ev) of the incident photons. Draw the nature of the graph for λ as a function of Ev. (CBSE Delhi 2014C)
Answer:
Consider an electron of mass m and charge e to be accelerated through a potential difference of V volt. Let v be the velocity gained by it. Then kinetic energy of the electron is
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 28
If λ is the de-Broglie wavelength associated with an electron, then
λ = \(\frac{h}{m v}=\frac{h}{\sqrt{2 m E_{v}}}\)

The nature of the graph is as shown.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 29

Question 6.
Light of intensity ‘l’ and frequency ‘v’ is incident on a photosensitive surface and causes photoelectric emission. What will be the effect on anode current when
(a) the intensity of light is gradually increased,
(b) the frequency of incident radiation is increased and
(c) the anode potential is increased?
In each case, all other factors remain the same. Explain giving justification in each case. (CBSE AI 2015)
Answer:
(a) Anode current will increase with the increase of intensity as the more the intensity of light, the more is the number of photons and hence more number of photoelectrons are ejected.
(b) No effect as the frequency of light affects the maximum K.E. of the emitted photoelectrons.
(c) Anode current will increase with anode potential as more anode potential will accelerate the more electrons till it attains a saturation value and gets them collected at the anode at a faster rate.

Question 7.
The graphs, drawn here, are for the phenomenon of the photoelectric effect.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 30
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 31
(a) Identify which of the two characteristics (intensity/frequency) of incident light is being kept constant in each case.
Answer:
Graph 1: Intensity, Graph 2: Frequency

(b) Name the quantity, corresponding to theClass 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 32mark, in each case.
Answer:
Graph 1: Saturation current, Graph 2: stopping potential

(c) Justify the existence of a ‘threshold frequency’ for a given photosensitive surface. (CBSE Delhi 2016C)
Answer:
The electrons require minimum energy to set themselves free. This is called the work function. As the energy of the photon depends upon its frequency, the photons must possess a minimum frequency so that their energy becomes equal to or greater than the work function. This is called threshold frequency and is given by v0 = \(\frac{\omega_{0}}{h}\)

Question 8.
Draw a graph showing the variation of de-Broglie wavelength λ of a particle of charge q and mass, with the accelerating potential V. An alpha particle and a proton have the same de-Broglie wavelength equal to 1 Å. Explain with calculations, which of the two has more kinetic energy. (CBSE Delhi 2017C)
Answer:
The graph is as shown.
The de-Broglie wavelength of a particle is given by the expression λ = \(\frac{h}{\sqrt{2 m q V}}\)

Since the alpha particle and the proton have the same de- Broglie wavelength, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 33
\(\frac{h}{\sqrt{2 m_{a} E_{\alpha}}}=\frac{h}{\sqrt{2 m_{p} E_{p}}}\)

Therefore proton has a greater value of de-Broglie wavelength.
Now kinetic energy is given by the expression
\(\frac{E_{a}}{E_{p}}=\frac{m_{p}}{m_{\alpha}}=\frac{m}{4 m}=\frac{1}{4}\)

Thus proton has more kinetic energy.

Question 9.
(a) Plot a graph showing the variation of photoelectric current with collector plate potential for different frequencies but of the same intensity of incident radiation.
Answer:
The graph between current I and plate potential.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 34
(b) Green and blue light of the same intensity are incident separately on the same photosensitive surface. If both of these cause photoelectric emission, which one will emit photoelectrons
(i) having greater kinetic energy and
Answer:
The blue light will emit photoelectrons having greater kinetic energy because the frequency and hence energy (E = hv) of blue light is more than the frequency and hence the energy of green light.

(ii) producing larger photocurrent? Justify your answer. (CBSE 2019C)
Answer:
Photoelectric current will be nearly the same for the blue and green light because the intensity of saturation current is independent of the frequency of incident light.

Question 10.
Radiation of frequency 1015 Hz is incident on two photosensitive surfaces P and Q. The following observations were recorded:
(a) Surface P: No photoemission occurs.
(b) Surface Q: Photoemission occurs but photoelectrons have zero kinetic energy. Based on Einstein’s photoelectric equation, explain the two observations.
Answer:
Einstein photoelectric equation is h(v – v0) = \(\frac{1}{2}\) mv²
(a) As in the case of surface P no photoemission takes place, we conclude that threshold frequency for P has a value greater than the frequency 1015 Hz of the given radiation.
(b) In the case of surface Q photoemission takes place but the kinetic energy of photoelectrons means that this radiation just overcomes the work function of the metal. In other words, the frequency 1015 Hz, of this radiation is equal to the threshold frequency.

Question 11.
When the light of frequency v1 is incident on a photosensitive surface, the stopping potential is V1 If the frequency of incident radiation becomes v1/2, the stopping potential changes to V2. Find out the expression for the threshold frequency for the surface in terms of V1 and V2.
If the frequency of incident radiation is doubled, will the maximum kinetic energy of the photoelectrons also be doubled? Give reason. (CBSE AI 2019)
Answer:
By Einstein’s photoelectric equation.
eV1 = hv1 – hv0 …(i).
and
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 35
If the frequency is doubled, the maximum kinetic energy will not be doubled.
Kmax = hv – hv0
Knew = 2hv – hv0 = hv + Kmax

Question 12.
(a) Define the terms
(i) threshold frequency and
Answer:
Threshold frequency (v0) is the minimum value of frequency of incident radiations below which the photoelectric emission stops, altogether.

(ii) stopping potential in the context of the photoelectric effect.
Answer:
Stopping potential is the value of retarding potential at which the photoelectric current becomes zero for the given frequency of incident radiations.

(b) Draw a graph showing the variation of stopping potential (V0) with frequency (v) of incident radiation for a given photosensitive material. (CBSE 2019C)
Answer:
The graph between v and V0
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 36
Question 13.
State Einsteins photoelectric equation explaining the symbols used.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 37
Light of frequency y is incident on a photosensitive surface. A graph of the square of the maximum speed of the electrons (v2max) vs. y is obtained as shown in the figure. Using Einstein’s photoelectric equation, obtain expressions for (i) Planck’s constant and (ii) the work function of the given photosensitive material in terms of parameters l, n and mass of the electron m. (CBSE Delhi 2018C)
Answer:
Einstein’s photoelectric equation is
hv = hv0 (= ω0) = \(\frac{1}{2}\)mv2max
v = frequency of incident Light
ω0 = threshold frequency of photosensitive A material.
ω0 = work function.
\(\frac{1}{2}\)mv2max = max. the kinetic energy of the emitted photoelectrons

(Also accept if the student writes)
hv = ω0 + eVs
ω0 = Work function of photosensitive material
Vs = stopping potential)

From Einstein’s photoelectronic equation,
we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 38
(i) Slope of the given graph = \(\frac{l}{n}\)
∴ \(\frac{2 h}{m}=\frac{l}{n}\)
or
h = \(\frac{m l}{2 n}\)

(ii) Intercept on the y-axis = – l
From equation (1)
– l = \(\frac{-2 \omega_{0}}{m}\) ⇒ ω0 = \(\frac{ml}{2}\)

Question 14.
(a) In the photoelectric effect, do all the electrons that absorb a photon comes out as photoelectrons irrespective of their location? Explain.
Answer:
No, it is not necessary that if the energy supplied to an electron is more than the work function, it will come out. The electron after receiving energy may lose energy to the metal due to collisions with the atoms of the metal. Therefore, most electrons get scattered into the metal. Only a few electrons near the surface may come out of the surface of the metal for whom the incident energy is greater than the work function of the metal.

(b) A source of light, of frequency greater than the threshold frequency, is placed at a distance be ‘d’ from the cathode of a photocell. The stopping potential is found to be V. If the distance of the light source is reduced to d/n (where n > 1), explains the changes that are likely to be observed in the (a) photoelectric current and (b) stopping potential. (CBSE Sample Paper 2018-19)
Answer:

  1. On reducing the distance, intensity increases. Photoelectric current increases with the increase in intensity.
  2. Stopping potential is independent of intensity and therefore remains unchanged.

Question 15.
Explain the laws of the photoelectric effect on the basis of Einstein’s photoelectric equation.
Answer:
The laws of the photoelectric effect can be explained on the basis of Einstein’s photoelectric equation, i.e.
\(\frac{1}{2}\)mv² = h (v – v0)
(a) From the equation we find that the kinetic energy will be positive and hence photoelectrons will be emitted if v > v0, i.e. the incident frequency is greater than the threshold frequency.

(b) It is clear from the expression that the kinetic energy of the emitted photoelectrons varies linearly with the frequency of the incident radiation.

(c) According to Einstein highly energetic radiation has a high frequency and an intense beam of radiation has more photons. Therefore when a high-intensity beam of radiation is incident on the metal surface, a large number of photoelectrons are emitted as the photoelectric effect is a one-one phenomenon, i.e. one photon can eject one photoelectron. Thus the more the intensity of the incident radiation, the more is the photoelectric current.

(d) As soon as a photon is an incident on a metal surface it is immediately absorbed by an electron, which sets itself free and comes out of the metal surface. The time lag is being negligible. Thus the photoelectric effect is instantaneous.

Question 16.
Define the terms threshold frequency and stopping potential in relation to the phenomenon of the photoelectric effect. How is the photoelectric current affected on increasing the
(i) frequency and
(ii) the intensity of the incident radiations and why?
Answer:
Stopping potential (V0): The negative potential of the plate at which no photoelectrons reach is called the stopping potential or the cut-off potential.

Threshold frequency (v0): The minimum frequency of the incident radiation, which can eject photoelectrons from a material, is known as the threshold frequency or cut-off frequency of the material.

  1. When the frequency has increased the energy absorbed by a single electron on collision with a photon also increases. E= hv, h is Planck’s constant. This raises the kinetic energy of the emitted photoelectrons but has no effect on current as the number of electrons being liberated remains the same.
  2. When the intensity of the incident radiation is increased, the number of photons crossing a unit area per second increases. These photons liberate more electrons and hence current increases.

Question 17.
Write Einstein’s photoelectric equation and mention which important features in the photoelectric effect can be explained with the help of this equation.
The maximum kinetic energy of the photoelectrons gets doubled when the wavelength of light incident on the surface changes from λ1 to λ2. Derive the expressions for the threshold wavelength λ0 and work function for the metal surface. (CBSE Delhi 2015)
Answer:
Einstein’s equation is
Emax = hv – ω or \(\frac{1}{2}\)mv²max = hv – ω

Here m is the mass of the photoelectron and vmax is maximum velocity.
(a) In accordance with Einstein’s equation, the kinetic energy of the photoelectrons is independent of the intensity of the incident radiation.
(b) In accordance with Einstein’s equation, the kinetic energy will be positive and hence photoelectrons will be ejected if v > vo. Thus below a certain frequency called the threshold frequency, photoelectrons are not ejected from a metal surface.
Given E1 = E, E2 = 2E,
By Einstein’s equation we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 39
Question 18.
Point out two distinct features observed experimentally in the photoelectric effect which cannot be explained on the basis of the wave theory of light. State how the ‘photon picture’ of light provides an explanation of these features. (CBSE AI 2016C)
Answer:
The two features are

  1. The intensity of light determines the photoelectric current and
  2. Existence of threshold frequency.

According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater is the amplitude of electric and magnetic fields. Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this picture, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with the increase in intensity.

Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons so that they exceed the minimum energy needed to escape from the metal surface. A threshold frequency, therefore, should not exist, which is against the observations. Thus the photoelectric effect cannot be explained on the basis of the wave nature of light

Question 19.
Using photon picture of light, show how Einstein’s photoelectric equation can be established. Write two features of the photoelectric effect that cannot be explained by the wave theory. (CBSE AI 2017)
Answer:
According to Einstein, the emission of a photoelectron is the interaction of quanta of light with a single electron. The photon is completely absorbed by the electron. The electron, which absorbs the photon, utilises this energy to set itself free. An amount of energy equal to the work function (ω) is required by the electron to set itself free. The remaining energy hv – ω (if hv > ω) is available to the electron as its maximum kinetic energy. Thus according to Einstein, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 40
Here m is the mass of the photoelectron and vmax is maximum velocity. Actually, most of the electrons possess kinetic energy less than the maximum kinetic energy because they lose a part of their energy due to collisions with the atoms on their way out from inside the metal. Thus electrons with different kinetic energy are emitted.

Now the work function is given by ω = hvo, therefore the above equation becomes
\(\frac{1}{2}\)mv²max = hv – hvo = h(v – vo)

(a) Dependence of kinetic energy on the frequency of the incident radiation.
(b) Existence of threshold frequency.

Question 20.
(a) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
(b) The work function of the following metals is given: Na = 2.75 eV, K = 2.3 eV, Mo = 4.17eV and Ni = 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 Å from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away? (CBSE Delhi 2017)
Answer:
(a) According to Einstein’s equation we have
\(\frac{1}{2}\)mv² = h (v – vo)

If the frequency of the incident radiation is above the threshold frequency (vo), the KE of the photo¬electrons will be positive and hence will be emitted.

Let us calculate the energy possessed by the photon of wavelength 3300 Å.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 41
As the energy of the incident photon is less than the work functions of Mo and Ni, so the metals Mo and Ni will not give photoelectric emission.

If the laser is brought closer, the intensity of radiation increases. This does not affect the result regarding Mo and Ni, but the photoelectric current will increase for Na and K with the increase in intensity.

Question 21.
When a monochromatic yellow coloured light beam is incident on a given photosensitive surface, photoelectrons are not ejected, while the same surface gives photoelectrons when exposed to the green coloured monochromatic beam. What will happen if the same surface is exposed to (i) violet and (ii) red coloured monochromatic beam of light? Justify your answer.
Answer:
Photoelectrons are ejected from a metal surface if the frequency of incident radiation is greater than the threshold frequency of that metal.

Since yellow light does not eject photoelectrons while green light does, the threshold frequency of the given metal is greater than the frequency of yellow light.
(a) Violet light has a greater frequency than yellow light, therefore when violet light is incident on the metal surface photoelectrons will be ejected.
(b) Red light has a smaller frequency than yellow light; therefore when the red light is incident on the metal surface, photoelectrons will not be ejected.

Question 22.
(a) Why photoelectric effect cannot be explained on the basis of the wave nature of light? Give reasons.
Answer:
According to the wave picture of light, the free electrons at the surface of the metal (over which the beam of radiation falls) absorb the radiant energy continuously. The greater the intensity of radiation, the greater is the amplitude of electric and magnetic fields.

Consequently, the greater the intensity, the greater should be the energy absorbed by each electron. In this theory, the maximum kinetic energy of the photoelectrons on the surface is then expected to increase with the increase in intensity.

Also, no matter what the frequency of radiation is, a sufficiently intense beam of radiation (over sufficient time) should be able to impart enough energy to the electrons so that they exceed the minimum energy needed to escape from the metal surface. A threshold frequency, therefore, should not exist, which is against the observations. Thus the photoelectric effect cannot be explained on the basis of the wave nature of light.

(b) Write the basic features of the photon picture of electromagnetic radiation on which Einstein’s photoelectric equation is based. (CBSE Delhi 2013)
Answer:
The basic features are

  1. Radiation energy is built up of discrete units.
  2. An electron absorbs the quanta of energy.

Question 23.
Obtain the expression for the wavelength of the de-Broglie wave associated with an electron accelerated from rest through a potential difference V.
The two lines A and B as shown in the graph plot the de-Broglie wavelength A. as a function of 1/\(\sqrt{V}\) (V is the accelerating potential) for two particles having the same charge. Which of the two represents the particle of heavier mass?
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 42
Answer:
Consider an electron of mass m and charge e to be accelerated through a potential difference of V volts. Let v be the velocity gained by it. Then kinetic energy of an electron is
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 43
If λ is the de-Broglie wavelength associated with an electron, then
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 44
In graphs, the charge of both particLes is the same, but the slope of graph A is Less. It means that mass of particle A is more (since slope = \(\frac{h}{\sqrt{2 m E}}\)).

Question 24.
The given graphs show the variation of the stopping potential Vs with the frequency (v) of the incident radiations for two different photosensitive materials M1 and M2.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 45
(i) What are the values of work functions for M1 and M2?
(ii) The values of the stopping potential for M1 and M2 for a frequency (v3) of the incident radiations greater than vo2 are V1 and V2 respectively. Show that the slope of the lines equals \(\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)
Answer:
(i) The values of work functions for M1 and M2 are ω1 = hvo1 and ω2 = hvo2.
(ii) The slope of the graphs is given by the variation of quantity along the Y-axis ‘ to the variation in quantity on the X-axis.

Now for M1
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 46
Therefore slope of the graphs is
\(\frac{h}{e}=\frac{V_{1}-V_{2}}{v_{02}-v_{01}}\)

Question 25.
Why are de-Broglie waves associated with a moving football not visible?
The wavelength of a photon and the de-Broglie wavelength of an electron have the same value. Show that energy of the photon is \(\frac{2λmc}{h}\) times the kinetic energy of the electron, where m, c and h have their usual meanings.
Answer:
(i) The de-Broglie wavelength of moving particles is given by the expression
λ = \(\frac{h}{mv}\). For a football the mass is large, hence the wavelength is extremely small and hence is not visible.

(ii) Now energy of a photon of wavelength λ. is E = hv = \(\frac{hc}{λ}\) ….(1)
and kinetic energy of an electron is Ke = \(\frac{1}{2}\)mv²

If de-Broglie wavelength (A) of electron be,
λ = \(\frac{h}{mv}\)
or
v = \(\frac{h}{mλ}\)

Therefore kinetic energy of electron will be
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 47
Numerical Problems:

Question 1.
What is the de Brogue wavelength associated with an electron, accelerated through a potential difference of 100 volts? (NCERT)
Answer:
Accelerating potential V = 100 V. The de-Broglie wavelength λ is
λ = \(\frac{1.227}{\sqrt{V}}\)nm = \(\frac{1.227}{\sqrt{100}}\) = 0.123 nm
The de-Broglie wavelength associated with an electron in this case is of the order of X-ray wavetengths.

Question 2.
In an experiment on the photoelectric effect, the slope of the cut-off voltage versus frequency of incident light is found to be 4.12 × 10-15 V s. Calculate the value of Planck’s constant. (NCERT)
Answer:
Given slope = 4.12 × 10-15 V s, h = 1

The slope of the voltage versus frequency graph gives the value of h/e, therefore we have
h = Slope × e = 4.12 × 10-15 × 1.6 × 10-19
Or
h = 6.59 × 10-34 J s

Question 3.
The threshold frequency for a certain metal is 3.3 × 1014 Hz. If the light of frequency 8.2 × 1014 Hz is incident on the metal, predict the cut-off voltage for the photoelectric emission. (NCERT)
Answer:
Given vo = 3.3 × 1014 Hz, v = 8.2 × 1014 Hz, Vo = ?

Using the relation h (v – vo) = eV0 we have h(v – vo)
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 48
Question 4.
The work function for a certain metal is 4.2 eV. Will this metal give photoelectric emission for incident radiation of wavelength 330 nm? (NCERT)
Answer:
Given ωo = 4.2 eV
The energy possessed by the incident radiation of wavelength 330 nm is
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{330 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 3.8 eV

Since this energy is less than the work function of the metal, no photoelectric emission will take place.

Question 5.
Light of frequency 7.21 × 1014 Hz is incident on a metal surface. Electrons with a maximum speed of 6.0 × 10 m s-1 are ejected from the surface. What is the threshold frequency for the photoemission of electrons? (NCERT)
Answer:
Given v = 7.21 × 1014 Hz, v = 6.0 × 10 m s-1, vo = ?
Using the relation h (v – vo) = \(\frac{1}{2}\)mv² We have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 49
Question 6.
Calculate the (a) momentum and (b) de Broglie wavelength of the electrons accelerated through a potential difference of 56 V. (NCERT)
Answer:
Given V = 56 V, p = ?, λ = ?
Using the relation
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 50
The de-Broglie wavelength is given by
λ = \(\frac{h}{p}=\frac{6.63 \times 10^{-34}}{4.04 \times 10^{-24}}\) = 0.164 nm

Question 7.
A particle is moving three times as fast as an electron. The ratio of the de-Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. Calculate the particle’s mass and identify the particle. (NCERT)
Answer:
Givenv = 3ve, λ/λe = 1.813 × 10-4
We know that λ = \(\frac{h}{p}=\frac{h}{m v}\)

therefore we have
m = me\(\frac{\lambda_{e} v_{e}}{\lambda v}=\frac{9.11 \times 10^{-31}}{1.813 \times 10^{-4} \times 3}\)

Solving we have m = 1.675 × 10-27 kg
This is the mass of a proton or a neutron.

Question 8.
Using the graph shown in the figure for stopping potential v/s incident frequency of photons, calculate Planck’s constant. (CBSE Delhi 2015C)
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 51
Answer:
By Einstein’s equation we have eVo = h(v – vo)

Therefore slope of the graph is \(\frac{V_{0}}{v}=\frac{h}{e}\). Now slope of graph is
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 52
Question 9.
If the light of wavelength 412.5 nm is incident on each of the metals given below, which ones will show photoelectric emission and why? (CBSE AI Delhi 2018)

Metal Work Function (eV)
Na 1.92
K 2.15
Ca 3.20
Mo 4.17

Answer:
The energy of a photon of the given light is,
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 53
E = 3.01 eV. Thus, metals having a work function less than the energy of a photon of falling light will show the photoelectric effect So, Na and K will show the photoelectric effect.

Question 10.
Two metals A and B have work functions 2 eV and 5 eV respectively. Which metal will emit electrons, when irradiated with light of wavelength 400 nm and why?
Answer:
The energy possessed by radiation of wavelength 400 nm is E = hv = hc/λ or
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 54
Since this energy is more than the work function of metal A, metal A will emit electrons.

Question 11.
Calculate the maximum kinetic energy of electrons emitted from a photosensitive surface of work function 3.2 eV, for the incident radiation of wavelength 300 nm.
Answer:
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 55
Therefore the kinetic energy of the emitted photoelectrons is
K.Emax = hv – hvo = 4.14 – 3.2 = 0.94 ev

Question 12.
The work function, for a given photosensitive surface, equals 2.5 eV. When the light of frequency y falls on this surface, the emitted photoelectron is completely stopped by applying a retarding potential of 4.1 V. What is the value of y?
Answer:
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 56
Question 13.
A nucleus of mass M Initially at rest splits into two fragments of masses M’/3 and 2M’/3 (M > M’). Find the ratio of de-Broglie wavelengths of the two fragments.
Answer:
By the principle of conservation of momentum
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 57
The ratio of de-Broglie wavelength is
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 58
Question 14.
An electron microscope uses electrons accelerated by a voltage of 50 kV. Determine the de-Broglie wavelength associated with the electrons. Taking other factors, such as numerical aperture etc. to be the same, how does the resolving power of an electron microscope compare with that of an optical microscope which uses yellow light? (CBSE AI 2014)
Answer:
Given V = 50 kV, λ = ?
Using the expression λ = \(\frac{1.227}{\sqrt{V}}\) nm we have
λ = \(\frac{1.227}{\sqrt{50 \times 10^{3}}}\) nm = 5.48 × 10-12 m

Wavelength of yellow light λy = 5.9 × 10-7 m

Now RP ∝ \(\frac{1}{λ}\)

Therefore we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 59
Question 15.
A proton and a deuteron are accelerated through the same accelerating potential. Which one of the two has
(a) the greater value of de-Broghe wavelength associated with it and
Answer:
(a) The de-Broglie wavelength is given by the expression λ = \(\frac{h}{\sqrt{2 m q V}}\). Since potential is same, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 60
Or
λp = \(\sqrt{2} \lambda_{a}\)

Thus proton has a greater value of de-Broglie wavelength.

(b) less momentum? Give reasons to justify your answer. (CBSE Delhi 2014)
Answer:
Now p = h/λ.

As the wavelength of a proton is more than that of a deuteron, the momentum of a proton is lesser than that of a deuteron. Hence, the momentum of the proton is less.

Question 16.
(a) Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W. Estimate the number of photons emitted per second on an average by the source.
Answer:
Each photon has energy
E = hv = (6.63 × 10-34) × (6.0 × 1014)
= 3.98 × 10-19 J

If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E so that P = NE, then
N = \(\frac{P}{E}=\frac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}}\)
= 5.0 × 1015 photons per second.

(b) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface. (CBSE Delhi 2014)
Answer:
With the increase of intensity of radiations, photocurrent increase.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 61

Question 17.
An electron is revolving around the nucleus with a constant speed of 2.2 × 108 m s-1. Find the de Broglie wavelength associated with it. (CBSE AI 2014C)
Answer:
Given v= 2.2 × 108 m s-1, λ = ?
Using the expression
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 62
Question 18.
Given the ground state energy E0 = – 13.6eV and Bohr radius αo = 0.53 Å. Find out how the de Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state. (CBSE AI 2015)
Answer:
The de-Broglie wavelength is given by 2πrn = nλ

In ground state, n = 1 and r0 = 0.53 Å, therefore λo = 2 × 3.14 × 0.53 = 3.33 Å
In first excited state, n = 2 and
r1 = 4 × 0.53 Å = 2.12 Å,

therefore λ1 = (2 × 3.14 × 2.12)/2 = 6.66 Å
Therefore λ1 – λo = 6.66 – 3.33 = 3.33 Å

In other words, the de-Broglie wavelength becomes double.

Question 19.
A proton and an a-particle have the same de-Broglie wavelength. Determine the ratio of (i) their accelerating potentials and (ii) their speeds. (CBSE Delhi 2015)
Answer:
(i) The de Broglie wavelength is given by λ = \(\frac{h}{\sqrt{2 m q V}}\). Since the de-Broglie waveLength is the same (or proton and alpha particle, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 63
(ii) The de-Broglie wavelength is given by λ = \(\frac{h}{mv}\). Since the de-Broglie wavelength is the same (or proton and alpha particle, we have
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 64

Question 20.
The work function (ωo), of a metal X, equals 3 × 10-19 J. Calculate the number (N) of photons, of light of wavelength 26.52 nm, whose total energy equals W. (CBSE Delhi 2016C)
Answer:
Given ωoo = 3 × 10-19 J, N = ?, λ = 26.52 nm = 26.52 × 10-9 m
The energy possessed by one photon
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 65
Therefore the number of photons
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 66
Question 21.
The KE of a beam of electrons accelerated through a potential V, equals the energy of a photon of wavelength 5460 nm. Find the de Broglie wavelength associated with this beam of electrons. (CBSE AI 2016C)
Answer:
The de-Broglie wavelength is given by
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 67

Question 22.
Calculate the kinetic energy of an electron having de Broglie wavelength of 1 A. (CBSE AI 2017C)
Answer:
Using the relation
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 68

Question 23.
Find the frequency of light that ejects electrons from a metal surface, fully stopped by a retarding potential of 3.3 V. If photoelectric emission begins in this metal at a frequency of 8 × 1014 Hz, calculate the work function (in eV) for this metal. (CBSE AI 2018C)
Answer:
The work function is given by ω0 = hv0
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 69

Question 24.
Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W. Calculate the (i) energy of a photon in the light beam and
Answer:
Each photon has an energy
E = hv = (6.63 × 10-34)(6.0 × 1014)
= 3.98 × 10-19 J

(ii) a number of photons emitted on an average by the source. (CBSE AI 2018C)
Answer:
If N is the number of photons emitted by the source per second, the power P transmitted in the beam equals N times the energy per photon E so that P = NE. Then
N = \(\frac{P}{E}=\frac{2.0 \times 10^{-3}}{3.98 \times 10^{-19}}\)
= 5.0 × 1015 photons per second.

Question 25.
The following table gives the values of work function for a few photosensitive metals.

S.No. Metal Work Function (eV)
1. Na 1.92
2. K 2.15
3. Ca 3.20
4. Mo 4.17

If each of these metals is exposed to radiations of wavelength 300 nm, which of them will not emit photoelectrons and why?
Answer:
The energy of the radiation is
E = \(\frac{h c}{\lambda}=\frac{6.63 \times 10^{-34} \times 3 \times 10^{8}}{300 \times 10^{-9} \times 1.6 \times 10^{-19}}\) = 4.14 eV

This energy is greater than the work functions of Na and K and lesser than the work function of Mo. Hence Mo will not emit photoelectrons.

Question 26.
The work function of caesium is 2.14 eV. Find
(i) the threshold frequency for caesium, and
Answer:
For the cut-off or threshold frequency the energy hv0 of the incident radiation must be equal to work function Φ0 so that
v0 = \(\frac{\phi_{0}}{h}=\frac{2.14 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}}\)
= 5.16 × 1014 Hz
Thus, for frequencies less than this threshold frequency, no photoelectrons are ejected.

(ii) the wavelength of the incident light if the photocurrent is brought to zero by a stopping potential of 0.60 V. (NCERT)
Answer:
Photocurrent reduces to zero when the maximum kinetic energy of the emitted photoelectrons equals the potential energy eV0 by the retarding potential V0.
Einstein’s Photoelectric equation is eV0 = hv – ω0 = \(\frac{hc}{λ}\) – ω0
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 70

Question 27.
An electron, an a-particle, and a proton have the same kinetic energy. Which of these particles has the shortest de-Broglie wavelength? (NCERT)
Answer:
For a particle, de-Broglie wavelength λ = h/p
Kinetic energy, K = p2/2m.
Then, λ = \(\frac{h}{\sqrt{2 m K}}\)

For the same kinetic energy K, the de-Broglie wavelength associated with the particle is inversely proportional to the square root of their masses. A proton (11H) is 1836 times massive than an electron and a particle (24He) four times that of a proton. Hence, a-particle has the shortest de-Broglie wavelength.

Question 28.
A particle is moving three times as fast as an electron. The ratio of the de Broglie wavelength of the particle to that of the electron is 1.813 × 10-4. Calculate the particle’s mass and identify the particle. (NCERT)
Answer:
the de-Broglie wavelength of a moving particle, having mass m and velocity v
λ = \(\frac{h}{p}=\frac{h}{m v}\)

Mass, m = h/λv
For an electron, mass me = h/λe ve
Now, we have v/ve = 3 and λ/λe = 1.813 × 10-4

Then the mass of the particle
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 71
Thus, the particle with this mass could be a proton or a neutron.

Question 29.
What are the (a) momentum, (b) speed, and (c) de Brogue wavelength of an electron with the kinetic energy of 120 eV? (NCERT)
Answer:
Given E = 120 eV, p =?, v =?, λ =?
Using the relation
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 72
Question 30.
Consider a metal exposed to the light of wavelength 600 nm. The maximum energy of the electron doubles when the light of wavelength 400 nm is used. Find the work function in eV. (NCERT Exemplar)
Answer:
Given λ1 = 600 nm, E1 = E, λ2 = 400 nm,
E2 = 2E, Φ =?
Now Emax = hv — Φ

According to the question (hc = 1230 eVnm)
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 73

Question 31.
A student performs an experiment on the photoelectric effect, using two materials A and B. A plot of vis given In the figure.
(a) Which material A or B has a higher work function?
(b) Given the electric charge of an electron = 1.6 × 10-19 C, find the value of h obtained from the experiment for both A and B. Comment on whether it is consistent with Einstein’s theory. (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 74
Answer:
(a) The higher the stopping potential, the higher is the work function. The stopping potential of B ís higher than that of A, therefore the work function of B is higher than that of A.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 75

Question 32.
A particle A with a mass mA is moving with a velocity y and hits a particle B (mass mB) at rest (one-dimensional motion). Find the change in the de-Broglie wavelength of particle A. Treat the collision as elastic. (NCERT Exemplar)
Answer:
Given UA = V, UB = 0
Since the collision is elastic, momentum and kinetic energy wiLt be conserved.
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 76
Class 12 Physics Important Questions Chapter 11 Dual Nature of Radiation and Matter 77

Relations and Functions Class 12 Important Extra Questions Maths Chapter 1

Here we are providing Class 12 Maths Important Extra Questions and Answers Chapter 1 Relations and Functions. Class 12 Maths Important Questions are the best resource for students which helps in Class 12 board exams.

Class 12 Maths Chapter 1 Important Extra Questions Relations and Functions

Relations and Functions Important Extra Questions Very Short Answer Type

Question 1.
If R = {(x, y) : x + 2y = 8} is a relation in N, write the range of R.
Solution:
Range of R = {1, 2, 3}.
[∵ When x = 2, then y = 3, when x = 4, then y = 2, when x = 6, then y = 1 ]

Question 2.
Show that a one-one function :
f{1, 2, 3} → {1, 2, 3} must be onto.   (N.C.E.R.T.)
Solution:
Since ‘f’ is one-one,
∴ under ‘f’, all the three elements of {1, 2, 3} should correspond to three different elements of the co-domain {1, 2, 3}.
Hence, ‘f’ is onto.

Question 3.
What is the range of the function f(x) = \(\frac{|x-1|}{x-1}\) ? (C.B.S.E. 2010)
Solution:
When x > 1,
than f(x) = \(\frac{x-1}{x-1}\) = 1.
When x< 1,
than f(x) = \(\frac{-(x-1)}{x-1}\) = -1
Hence, Rf = {-1, 1}.

Question 4.
Show that the function f : N → N given by f(x) = 2x is one-one but not onto. (N.C.E.R.T.)
Solution:
Let x1, x2 ∈ N.
Now, f(x1) = f(x2)
⇒ 2x1 = 2x2
⇒ x1 = x2
⇒ f is one-one.
Now, f is not onto.
∵ For 1 ∈ N, there does not exist any x ∈ N such that f(x) = 2x = 1.
Hence, f is ono-one but not onto.

Question 5.
If f : R → R is defined by f(x) = 3x + 2 find f(f(x)). C.B.S.E. 2011 (F))
Solution:
f(f(x)) = 3 f(x) + 2
= 3(3x + 2) + 2 = 9x + 8.

Question 6.
If f(x) = \(\frac{x}{x-1}\) , x ≠1 then find fof. (N.C.E.R.T)
Solution:
Class 12 Maths Important Questions Chapter 1 Relations and Functions 1

Question 7.
If f: R → R is defined by f(x) = (3 -x3)1/3, find fof(x)
Solution:
fof(x) = f(f(x)) = (3-(f(x))3)1/3
= (3 – ((3 – x3)1/3)3)1/3
= (3 – (3 – x3 ))1/3= (x3)1/3 = x.

Question 8.
Are f and q both necessarily onto, if gof is onto? (N.C.E.R.T.)
Solution:
Consider f: {1, 2, 3, 4} → {1, 2, 3, 4}
and g : {1, 2, 3,4} → {1,2.3} defined by:
f(1) = 1, f(2) = 2, f(3) = f(4) = 3
g (1) = 1, g (2) = 2, g (3) = g (4) = 3.
∴ gof = g (f(x)) {1, 2,3}, which is onto
But f is not onto.
[∵ 4 is not the image of any element]

Relations and Functions Important Extra Questions Short Answer Type

Question 1.
Let A be the set of all students of a Boys’ school. Show that the relation R in A given by:
R = {(a, b): a is sister of b} is an empty relation and the relation R’ given by :
R’ = {(a, b) : the difference between heights of a and b is less than 3 metres} is an universal relation. (N.C.E.R.T.)
Solution:
(i) Here R = {(a, b): a is sister of b}.
Since the school is a Boys’ school,
∴ no student of the school can be the sister of any student of the school.
Thus R = Φ Hence, R is an empty relation.

(ii) Here R’ = {(a,b): the difference between heights of a and b is less than 3 metres}.
Since the difference between heights of any two students of the school is to be less than 3 metres,
∴ R’ = A x A. Hence, R’ is a universal relation.

Question 2.
Let f : X → Y be a function. Define a relation R in X given by :
R = {(a,b):f(a) = f(b)}.
Examine, if R is an equivalence relation. (N.C.E.R.T.)
Solution:
For each a ∈ X, (a, a) ∈ R.
Thus R is reflexive. [∵ f (a) = f(a)]
Now (a, b) ∈ R
⇒ f(a) = f(b)
⇒ f(b) = f (a)
⇒ (b, a) ∈ R.
Thus R is symmetric.
And (a, b) ∈ R
and (b, c) ∈ R
⇒ f(a) = f(b)
and f(b) = f(c)
⇒ f(a)= f(c)
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 3.
Let R be the relation in the set Z of integers given by:
R = {(a, b): 2 divides a – b}.
Show that the relation R is transitive. Write the equivalence class [0].   (C.B.S.E. Sample Paper 2019-20)
Solution:
Let 2 divide (a – b) and 2 divide (b – c), where a,b,c ∈ Z
⇒ 2 divides [(a – b) + (b – c)]
⇒ 2 divides (a – c).
Hence, R is transitive.
And [0] = {0, ± 2, ± 4, ± 6,…].

Question 4.
Show that the function :
f : N → N
given by f(1) = f(2) = 1 and f(x) = x -1, for every x > 2 is onto but not one-one. (N.C.E.R.T.)
Solution:
Since f(1) = f(2) = 1,
∴ f(1) = f(2), where 1 ≠ 2.
∴ ‘f’ is not one-one.
Let y ∈ N, y ≠ 1,
we can choose x as y + 1 such that f(x) = x – 1
= y + 1 – 1 = y.
Also 1 ∈ N, f(1) = 1.
Thus ‘f ’ is onto.
Hence, ‘f ’ is onto but not one-one.

Question 5.
Find gof and fog, if:
f : R → R and g : R → R are given by f (x) = cos x and g (x) = 3x2. Show that gof ≠ fog. (N. C.E.R. T.)
Solution:
We have :
f(x) = cos x and g(x) = 3x2.
∴ gof (x) = g (f(x)) = g (cos x)
= 3 (cos x)2 = 3 cos2 x
and fog (x) = f(g (x)) = f(3x2) = cos 3x2.
Hence, gof ≠ fog.

Question 6.
If f(x) = \(\frac{4 x+3}{6 x-4}\) , x ≠ \(\frac{2}{3}\) find fof(x)
Solution:
We have: \(\frac{4 x+3}{6 x-4}\) …(1)
∴ fof(x) — f (f (x))
Class 12 Maths Important Questions Chapter 1 Relations and Functions 2

Question 7.
Let A = N x N be the set of ail ordered pairs of natural numbers and R be the relation on the set A defined by (a, b) R (c, d) iff ad = bc. Show that R is an equivalence relation.
Solution:
Given: (a, b) R (c, d) if and only if ad = bc.
(I) (a, b) R (a, b) iff ab – ba, which is true.
[∵ ab = ba ∀ a, b ∈ N]
Thus, R is reflexive.

(II) (a, b) R (c,d) ⇒ ad = bc
(c, d) R (a, b) ⇒ cb = da.
But cb = be and da = ad in N.
∴ (a, b) R (c, d) ⇒ (c, d) R (a, b).
Thus, R is symmetric.

(III) (a,b) R (c, d)
⇒ ad = bc …(1)
(c, d) R (e,f)
⇒ cf = de … (2)
Multiplying (1) and (2), (ad). (cf) – (be), (de)
⇒ af = be
⇒ (a,b) = R(e,f).
Thus, R is transitive.
Thus, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.

Question 8.
Let f: R → R be the Signum function defined as :
Class 12 Maths Important Questions Chapter 1 Relations and Functions 3
and g : R → R be the Greatest Integer Function given by g (x) = [x], where [x] is greatest integer less than or equal to x. Then does fog and gof coincide in (0,1] ?
Solution:
For x ∈ (0,1].
Class 12 Maths Important Questions Chapter 1 Relations and Functions 4
And (gof) (x) = g(f(x)) = g(1)
[∵ f(x) = 1 ∀ x > 0]
= [1] = 1
⇒ (gof) (x) = 1 ∀ x ∈ (0, 1] …(2)
From (1) and (2), (fog) and (gof) do not coincide in (0, 1].

Relations and Functions Important Extra Questions Long Answer Type 1

Question 1.
Show that the relation R on R defined as R = {(a, b):a ≤ b}, is reflexive and transitive but not symmetric.
Solution:
We have : R = {(a, b)} = a ≤ b}.
Since, a ≤ a ∀ a ∈ R,
∴ (a, a) ∈ R,
Thus, R reflexive.
Now, (a, b) ∈ R and (b, c) ∈ R
⇒ a ≤ b and b ≤ c
⇒ a ≤ c
⇒ (a, c) ∈ R.
Thus, R is transitive.
But R is not symmetric
[∵ (3, 5) ∈ R but (5, 3) ∉ R as 3 ≤ 5 but 5 > 3]

Question 2.
Prove that function f : N → N, defined by f(x) = x2 + x + 1 is one-one but not onto. Find inverse of f: N → S, where S is range of f.
Solution:
Let x1, x2 ∈ N.
Now, f(x1) = f(x2)
Class 12 Maths Important Questions Chapter 1 Relations and Functions 5
Thus, f is one-one.
Let y ∈ N, then for any x,
f(x) = y if y = x2 + x + 1
Class 12 Maths Important Questions Chapter 1 Relations and Functions 6
Class 12 Maths Important Questions Chapter 1 Relations and Functions 7
Now, for y = \(\frac{3}{4}\) , x = \(-\frac{1}{2}\) ∉ N.
Thus, f is not onto.
⇒ f(x) is not invertible.
Since, x > 0, therefore, \(\frac{\sqrt{4 y-3}-1}{2}\) > 0
⇒ \(\sqrt{4 y-3}\) > 1
⇒ 4y – 3 > 1
⇒ 4y > 4
⇒ y > 1.
Redefining, f : (0, ∞) → (1, ∞) makes
f(x) = x2 + x + 1 on onto function.
Thus, f (x) is bijection, hence f is invertible and f-1 : (1, ∞) → (1,0)
f-1(y) = \(\frac{\sqrt{4 y-3}-1}{2}\)

Question 3.
Let A = (x ∈Z : 0 ≤ x ≤ 12}.
Show that R = {(a, b) : a, b ∈ A; |a – b| is divisible by 4} is an equivalence relation. Find the set of all elements related to 1. Also write the equivalence class [2]. (C.B.S.E 2018)
Solution:
We have:
R = {(a, b): a, b ∈ A; |a – b| is divisible by 4}.
(1) Reflexive: For any a ∈ A,
∴ (a, b) ∈ R.
|a – a| = 0, which is divisible by 4.
Thus, R is reflexive.

Symmetric:
Let (a, b) ∈ R
⇒ |a – b| is divisible by 4
⇒ |b – a| is divisible by 4
Thus, R is symmetric.

Transitive: Let (a, b) ∈ R and (b, c) ∈ R
⇒ |a – b| is divisible by 4 and |b – c| is divisible by 4
⇒ |a – b| = 4λ
⇒ a – b = ±4λ ………….(1)
and |b – c| = 4µ
⇒ b – c = ± 4µ ………….(2)

Adding (1) and (2),
(a-b) + (b-c) = ±4(λ + µ)
⇒ a – c = ± 4 (λ + µ)
⇒ (a, c) ∈ R.
Thus, R is transitive.
Now, R is reflexive, symmetric and transitive.
Hence, R is an equivalence relation.

(ii) Let ‘x’ be an element of A such that (x, 1) ∈ R
⇒ |x – 1| is divisible by 4
⇒ x – 1 = 0,4, 8, 12,…
⇒ x = 1, 5, 9, 13, …
Hence, the set of all elements of A which are related to 1 is {1, 5, 9}.

(iii) Let (x, 2) ∈ R.
Thus |x – 2| = 4k, where k ≤ 3.
∴ x = 2, 6, 10.
Hence, equivalence class [2] = {2, 6, 10}.

Question 4.
Prove that the function f: [0, ∞) → R given by f(x) = 9x2 + 6x – 5 is not invertible. Modify the co-domain of the function f to make it invertible, and hence find f-1.   (C.B.S.E. Sample Paper 2018-19)
Solution:
Let y∈ R.
For any x, f(x) = y if y = 9x2 + 6x – 5
⇒ y = (9x2 + 6x + 1) – 6
= (3x + 1)2 – 6
Class 12 Maths Important Questions Chapter 1 Relations and Functions 8
For y = – 6 ∈ R, x = \(-\frac{1}{3}\) ∉ [0, ∞).
Thus, f(x) is not onto.
Hence, f(x) is not invertible.
Class 12 Maths Important Questions Chapter 1 Relations and Functions 9
We redefine,
f: [0, ∞) → [-5, ∞),
which makes f(x) = 9x2 + 6x – 5 an onto function.
Now, x1, x2 ∈ [0, ∞) such that f(x1) = f(x2)
⇒ (3x1 + 1)2 = (3x2 + 1)2
⇒[(3x1 + 1)+ (3x2 + 1 )][(3x1 + 1)- (3x2 + 1)]
⇒ [3(x1 + x2) + 2][3(x1 – x2)] = 0
⇒ x1 = x2
[∵ 3(x1 + x2) + 2 > 0]
Thus, f(x) is one-one.
∴ f(x) is bijective, hence f is invertible
and f-1: [-5, ∞) → [0, ∞)
f-1 (y) = \(\frac{\sqrt{y+6}-1}{3}\)

Question 5.
Check whether the relation R in the set R of real numbers, defined by :
R = {(a, b): 1 + ab > 0}, is reflexive, symmetric or transitive.   (C.B.S.E. Sample Paper 2018-19)
Solution:
R = {(a, b): 1 + ab> 0}.

Reflexive:
Now, 1 + a.a = 1 + a2 > 0
⇒ (a, a) ∈ R ∀ a ∈ R.
Thus, R is reflexive.

Symmetric:
Let (a, b) ∈ R.
Then 1 + ab > 0
⇒ 1 + ba > 0
⇒ (b, a) ∈ R.
Thus, R is symmetric.

Transitive:
Take a = -8,b = -1, c = \(\frac { 1 }{ 2 }\)
Now, 1 + ab = 1 + (-8) (-1)
= 9 > 0
⇒ (a, b) ∈ R
and, 1 + bc = 1 + (-1)(\(\frac { 1 }{ 2 }\) )
= 1 – \(\frac { 1 }{ 2 }\) = \(\frac { 1 }{ 2 }\) > 0
⇒ (b, c) ∈ R.
But 1 + ac – 1 + (-8) (\(\frac { 1 }{ 2 }\) )
= 1 – 4 = -3 < 0
⇒ (a, c) ∉ R.
Thus, R is not transitive.
Hence, R is reflexive, symmetric but not transitive.

Question 6.
Let T be the set of all triangles in a plane with R, a relation in T given by :
R = {(T1, T2): T1 is congruent to T2}.
Show that R is an equivalence relation. (N.C.E.R.T.)
Solution:
We have:
R = {(T1, T2): T1 is congruent to T2}.
Now (T1, T2) ∈ R.
[ ∵ Every triangle is congruent to itself]
Thus R is reflexive.
(T1, T2) ∈ R
⇒ T1 is congruent to T2
⇒ T2 is congruent to T1
(T2, T1) ∈ R.
Thus R is symmetric.

(T1, T2) ∈ R and (T2, T3) ∈ R
⇒ T1 is congruent to T2 and T2 is congruent to T3
⇒ T1 is congruent to T3
(T1, T3) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 7.
If R1, and R2 are equivalence relations in a set A, show that R1 ∩ R2 is also an equivalence relation. (N.C.E.R.T.)
Solution:
Since R1 and R2 are equivalence relations, [Given]
∴ (a, a) ∈ R1
and (a, a) ∈ R2 ∀ a ∈ A
⇒ (a, a) ∈ R1 ∩ R2 ∀ ∈ G
Thus R1 ∩ R2 is reflexive.
Now (a, b) ∈ R1 ∩ R2
⇒ (a, b) ∈ R1
and (a, b) ∈ R2
⇒ (b, a) ∈ R1
and (b, a) ∈ R2
⇒ (b, a) ∈ R1 ∩ R2

Thus R1 n R2 is symmetric.
And (a, b) ∈ R1 ∩ R2
and (b, c) ∈ R1 ∩ R2
⇒ (a, c) ∈ R1
and (a, c) ∈ R2
⇒ (a, c) ∈ R1 ∩ R2
Thus R1 ∩ R2 is transitive.
Hence, R1 ∩ R2 is an equivalence relation.

Question 8.
Show that f: R-{2} → R -{1} defined by:
f(x) = \(\frac{x}{x-2}\) is one-one.
Also, if g : R -{1} → R -{2} as g{x) = \(\frac{2 x}{x-1}\) find gof.
Solution:
We have: f(x) = \(\frac{x}{x-2}\) and g(x) = \(\frac{2 x}{x-1}\)
Class 12 Maths Important Questions Chapter 1 Relations and Functions 10

Question 9.
Let f: N → N be a function defined as :
f(x) – 9x2 + 6x – 5.
Show that f: N → S, where S is the range of f, is invertible. Find the inverse of f and hence find f-1 (43) and f-1 (163).   (C.B.S.E. 2016)
Solution:
We have : f : N → S,
fix) = 9x2 + 6x – 5.
Now f(x1) = f(x2)
⇒ 9x12 + 6x1 – 5 = 9 x22 + 6x2 – 5
⇒ 9(x12 – x22) + 6 (x1 – x2) = 0
⇒ (x1 – x2) [9x1 + 9x2+ 6] = 0
⇒ x1 = x2 [x1, x2 ∈ N ]
⇒ f is one-one.
Let y ∈ S be an arbitrary number.
Now y = f(x)
⇒ y = 9x2+ 6x – 5
⇒ y = (3x + 1)2 – 6
⇒ \(\sqrt{y+6}\) = 3x + 1
⇒ x = \(\frac{\sqrt{y+6}-1}{3}\) ∈ N
⇒ x = f-1 (y).

Since f is one-one and onto
⇒ f is invertible.
Class 12 Maths Important Questions Chapter 1 Relations and Functions 11
Class 12 Maths Important Questions Chapter 1 Relations and Functions 12

Question 10.
Let f: A → B be a function defined as f(x) = \(\frac{2 x+3}{x-3}\) where A = R – {3} and B = R – {2}.
Is the function ‘f ’one-one and onto ?
Is ‘f’ invertible? If yes, then find its inverse.
Solution:
Let x1, x2 ∈ A = R – {3}.
Now, f(x1) = f(x2)
⇒ \(\frac{2 x_{1}+3}{x_{1}-3}=\frac{2 x_{2}+3}{x_{2}-3}\)
⇒ (2x1 + 3) (x2– 3) = (2x2 + 3) (x1 – 3)
⇒ 2x1x2 – 6x1 + 3x2 – 9 = 2x1 x2 – 6x2 + 3x1 – 9
⇒ – 6x1 + 3x2 = – 6x2 + 3x1
⇒ 9x1 = 9x2
⇒ x1 = x2
Thus, ‘f’ is one-one.
Let y ∈ R- {2}.
Let y = f(x0).
Then \(\frac{2 x_{0}+3}{x_{0}-3}\) = y
⇒ 2x0 + 3 = x0y – 3y
⇒ x0(y – 2) = 3(y + 1)
⇒ x0 = \(\frac{3(y+1)}{y-2}\)
Now , y ∈ R – {2 } ⇒ \(\frac{2 x_{0}+3}{x_{0}-3}\) ∈ R – {2}
Class 12 Maths Important Questions Chapter 1 Relations and Functions 13
Thus, ‘f’ is onto.
Hence, ‘f’ is one-one onto and consequently ‘f’ is invertible.
Also y = \(\frac{2 x+3}{x-3}\)
xy – 3y = 2x + 3
x(y- 2) = 3(y +1)
x = \(\frac{3(y+1)}{y-2}\)
f-1 (y) = \(\frac{3(y+1)}{y-2}\)
Hence, f-1 (x) = \(\frac{3(x+1)}{x-2}\) for all x ∈ R – {2}.

Question 11.
Let A = R – {2} and B = R – {1}. If f: A → B is a function defined by :
f(x) = \(\frac{x-1}{x-2}\) Show that f is one one and onto.
Hence, find f-1.
Solution:
(i) One-one : Let x1 x2 ∈ R – {2} such that
f(x1) = f(x2)
⇒ \(\frac{x_{1}-1}{x_{1}-2}=\frac{x_{2}-1}{x_{2}-2}\)
⇒ x1x2 – 2x1 – x2 + 2 = x1x2 – 2x2 – x1 + 2
⇒ x1 = x2
Thus,f one-one.

Onto : Let f(x) = y.
Thus \(\frac{x-1}{x-2}\) = y ⇒ x = \(\frac{2 y-1}{y-1}\)
∴ Range of f = R – {1}
= Co-domain of B
Thus f is onto.

(ii) f-1(y) = \(\frac{2 y-1}{y-1}\)
Hence f-1(x) = \(\frac{2 x-1}{x-1}\)

Question 12.
Show that the relation S on the set:
A = {x ∈ Z : 0 ≤ x ≤ 12} given by S = {(a, b) :a,b e Z, |a – b| is divisible by 3} is an equivalence relation.
Solution:
Reflexive :
(a, a) ∈ S
⇒ |a – a| i.e., | 0 | is divisible by 3.
Thus, S is reflexive.

Symmetric:
(a, b) ∈ S ⇒ | a – b | is divisible by 3
⇒ |b – a| is divisible by 3
⇒ (b, a) ∈ S.
Thus, S is symmetric.

Transitive :
Let (a, b) ∈ S and (b, c) ∈ S.
Thus |a – b| = 3m and |b – c| = 3n
⇒ a – b = ± 3m and b – c = ± 3 n.
∴ (a – c) = 3(± m ± n)
⇒ a – c is divisible by 3
⇒ | a – c | is divisible by 3
⇒ (a, c) ∈ S.
Thus, S is transitive.
Hence, S is an equivalence relation.

Relations and Functions Important Extra Questions Long Answer Type 2

Question 1.
Let A = (1,2,3,…, 9} and R be the relation in A x A defined by (a, b) R (c, d) if: a + d = b + c for (a, b), (c, d) in A x A.
Prove that R is an equivalence relation. Also obtain the equivalence class {(2,5)}.   (C.B.S.E. 2014)
Solution:
(i) We have : (a, b) R (c, d)
⇒ a + d = b + c on the set A = {1, 2, 3,…, 9}.
(a) (a, b) R (a, b)
⇒ a + b = b + a, which is true.
[∵ a + b = b + a ∀ a, b ∈ A]
Thus R is reflexive.

(b) (a, b) R (c, d)
⇒ a + d = b + c
(c, d) R (a, b)
⇒ c + b = d + a.
But c + b = b + c
and d + a = a + d ∀ a, b, c, d ∈ A.
∴ (a, b) R(c, d) = (c, d) R (a, b).
Thus R is symmetric.

(c) (a,b)R(c,d)
⇒ a + d = b + c ∀ a, b, c, d ∈ A ……..(1)
(c,d)R(e,f)
⇒ c + f = d + e ∀ c, d, e,f ∀ A ……………. (2)

∴ Adding (1) and (2),
(a + d) + (c +f) = (b + c) + (d + e)
⇒ a + f = b + e
⇒ (a, b) R (e, f).
Thus R is transitive.
Hence, the relation R is an equivalence relation.

(ii) {(2, 5)} = {(1, 4), (2, 5), (3, 6), (4, 7), (5, 8), (6, 9)}.
[∵ 2 + 4 = 5 + 1; etc.]

Question 2.
Let N denote the set of all natural numbers and R be the relation on N x N defined by : (a, b) R(c,d) is ad(b + c) = bc(a + d).
Show that R is an equivalence relation.    (C.B.S.E. 2015)
Solution:
We have : (a, b) R (c, d)
⇒ ad (b + c) = bc (a + d) on N.
(i) {a, b) R (a, b)
⇒ ab(b + a) = ba (a + b)
⇒ ab(a + b) = ab (a + ft),
which is true.
Thus R is reflexive.

(ii) (a, b) R (c, d)
⇒ ad(b + c) = be (a + d)
⇒ bc(a + d) = ad (b + c)
⇒ cb(d + a) = da (c + b)
[∵ bc = cb and a + d = d + a;
etc. ∀ a, b, c, d ∈ N]
⇒ (cb) R (a,b).
Thus R is symmetric

(iii) Let (a , b) R (c, d) and (c, d) R ( e, f)
∴ ad (b + c) = bc (a + d)
and cf(d + e) = de(c + f)
Class 12 Maths Important Questions Chapter 1 Relations and Functions 14
Class 12 Maths Important Questions Chapter 1 Relations and Functions 15
⇒  be(a+f) = af(b+e)
⇒  af(b+e) = be(a+f)
⇒  (a, b) R (e,f).
Thus R is transitive.
Hence, R is an equivalence relation.

Question 3.
Show that the relation in the set A = {x:x ∈ W, 0 ≤ x ≤ 12} given by R = {(a, b) : (a – b) is a multiple of 4} is an equivalence relation. Also find the set of all. elements related to 2.
Solution:
We have :
R = {(a, b): (a- b) is a multiple of 4},
where a, b ∈ A = {x: x ∈ W, 0 ≤ x ≤ 12}.
(a) For any a ∈ A, we have :
(a – a) = 0, which is a multiple of 4
⇒ (a, a) ∈ R ∀ a ∈ A.
Thus R is reflexive.

(b) Let a, b ∈ A.
Now (a, b) ∈ R
⇒ (a – b) is a multiple of 4
⇒ (b – a) is a multiple of 4
⇒ (b, a) ∈ R
Thus R is symmetric.

(c) Let a, b, c ∈ A.
Now (a, b) ∈ R
and (b, c) ∈ R
⇒ (a – b) is a multiple of 4 and (b – c) is a multiple of 4
⇒ (a – b) + (b – c) is a multiple of 4
⇒ (a – c) is a multiple of 4
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.

Question 4.
Let R be the relation defined in the set A = {1,2,3,4,5,6,7} by :
R = {(a, b) : both a and b are either odd or even}.
Show that R is an equivalence relation. Further, show that all the elements of the subset:
(1, 3, 5, 7} are related to each other and all the elements of the subset {2,4,6} are related to each other, but no element of the subset {1, 3, 5, 7} is related to any element of the subset {2,4,6}. (N.C.E.R.T.)
Solution:
We have:
R = {{a, b): both a, b are either odd or even}.
(i) Let a ∈ A.
Both a and a are either odd or even.
∴ (a, a) ∈ R.
Thus R is reflexive.

(ii) Let (a, b) ∈ R
⇒ both a and b are either odd or even
⇒ both b and a are either odd or even
⇒ (b, a) ∈ R.
Thus R is symmetric.

(iii) Let {a, b) ∈ R and (b, c) ∈ R.
∴ Both a, b and both b, c are either odd or even
⇒ both a, c are either odd or even
⇒ (a, c) ∈ R.
Thus R is transitive.
Hence, R is an equivalence relation.
Further all elements of {1,3,5,7} are related to each other.
[∵ All elements of this subset are odd] Similarly all elements of {2,4,6} are related to each other.
[∵All elements of this subset are even] But no element of {1, 3, 5, 7} is related to any element of {2, 4, 6}.
[∵Elements of {1, 3, 5, 7] are odd while elements of [2,4,6] are even]

Question 5.
Show that f : N → N given by :
Class 12 Maths Important Questions Chapter 1 Relations and Functions 16
is both one-one and onto.   (A.I.C.B.S.E. 2012)
Solution:
One-One.
Here we discuss the following possible cases :
(i) When x1 is odd and x2 is even.
Here f(x1) = f(x2)
⇒ x1 + 1 = x2 – 1
⇒ x2 = x1 = 2, which is impossible.

(ii) When x1 is even and x2 is odd.
Here f(x1) =f(x2)
⇒ x1 – 1 = x2 + 1
⇒ x1 – x1 = 2, which is impossible.

(iii) When x1 and x2 are both odd.
Here f(x1) = f(x2)
⇒ x1 + 1 = x2 + 1
⇒ x1 = x2
∴ ‘f’ is one-one.

(iv) When x1 and x2 are both even.
Here f(x1) = f(x2)
⇒ x1 – 1 = x2 – 1
⇒ x1 = x2
∴ ‘ f’ is one-one.

Onto. Let ‘x’ be an arbitrary natural number. When x is an odd natural number, then there exists an even natural number (x + 1) such that: f(x + 1) = (x + 1) – 1 = x.
When x is an even natural number, then there exists an odd natural number (x – 1) such that:
f(x – 1) = (x – 1) + 1 = x.
∴ Each x ∈ N has its pre-image in N.
Thus ‘f ’ is onto.
Hence, ‘f ’ is both one-one and onto.

Question 6.
Show that the function f : R → R defined by:
\(\frac{x}{x^{2}+1}\) ∀ x ∈ R is neither one-one nor onto. Also, if g : R → R is defined g(x) = 2x -1, find fog (x). (C.B.S.E. 2018)
Solution:
We have : f(x) = \(\frac{x}{x^{2}+1}\)
(i) One-one, f(x1) = f(x2)
⇒ \(\frac{x_{1}}{x_{1}^{2}+1}=\frac{x_{2}}{x_{2}^{2}+1}\)
⇒ x1x22 + x1 = x2x12 + x2
⇒  x1x2 (x2 – x1) = x2 – x1
x1x2 = 1
x1 = \(\frac{1}{x_{2}}\)
x1 ≠ x2
Thus, f is not one-one.

Onto:
⇒ f(x) = y
⇒ \(\frac{x}{x^{2}+1}\) = y
⇒ x = yx2 + y
⇒ x2y + y – x = 0
⇒ x can not be expressed in y.
Thus, f is not onto.
Hence, f is neither one-one nor onto.

(ii) Since g (x) = 2x – 1,
∴ fog (x) = f(g (x)) = f(2x – 1)
= \(\frac{2 x-1}{(2 x-1)^{2}+1}=\frac{2 x-1}{4 x^{2}-4 x+2}\)

Question 7.
Show that the relation R on the set Z of all integers defined by (x, y) ∈ R ⇔ (x – y) is divisible by 3 is an equivalence relation.
(C.B.S.E. 2018 C)
Solution:
(x – x) = 0 is divisible by 3 for all x ∈ Z.
So, (x, x) ∈ R
:. R is reflexive.
(x – y) is divisibile by 3 implies (y – x) is divisible by 3.
So (x, y) ∈ R implies (y – x) ∈ R, x. y ∈ Z
⇒ R is symmetric.
(x – y) is divisibile by 3 and (y – z) is divisible by 3.
So (x – z) = (x – y) + (y + z) is divisible by 3.
∴ (x, z) ∈ R ⇒ R is transitive.
Hence, R is an equivalence relation.

Question 8.
Let A = R – {3} and B = R – {1}. Consider the function f: A → B defined by:
f(x) = [latyex]=\left(\frac{x-2}{x-3}\right)[/latex]. Show that ‘f’ is one-one and onto and hence find f-1. (C.B.S.E. 2012)
Solution:
Let x1, x2 ∈ R-{3}.
Now f(x1) = f(x2)
⇒ \(\frac{x_{1}-2}{x_{1}-3}=\frac{x_{2}-2}{x_{2}-3}\)
⇒ (x1 – 2) (x2 – 3) = (x1 – 3) (x2 – 2)
⇒ x1x2 – 3x1 – 2x2 + 6 = ⇒ x1x2 – 2x1 – 3x2 + 6
⇒ x1 = x2
Thus ‘f’ is one-one.
Let y ∈ R – {1}.
Let y = f(x0).
Then \(\frac{x_{0}-2}{x_{0}-3}\) = y
x0 – 2 = x0y – 3y
x0(y-1) = 3y – 2
x0 = \(\frac{3 y-2}{y-1}\)
Now y ∈ R – {1}.
Class 12 Maths Important Questions Chapter 1 Relations and Functions 17
Thus ‘f’ is onto.
Hence, ‘f’ is one-one and onto function
⇒ ‘f’ is invertible.
Also y = \(\frac{x-2}{x-3}\)
⇒ xy — 3y = x – 2
⇒ x (y — 1) = 3y – 2
x = \(\frac{3y-2}{y-1}\)
f-1 (y) = \(\frac{3 y-2}{y-1}\)
Hence, f-1 (x) = \(\frac{3 x-2}{x-1}\)– for all x ∈ R-{1}.

Question 9.
Let Y = {n2: n ∈ N} ⊂ N. Consider f : N → Y as f(n) = n2. Show that if is invertible. Find the inverse of ‘f.   (N.C.E.R.T.)
Solution:
Let y ∈ Y, where y is arbitrary.
Here y is of the form n2, for n∈ N
⇒  n = √y.
This motivates a function :
g : Y → N, defined by g (y) = √y .

Now gof(n ) = g (f(n)) = g (n2) = √n2= n
and fog (y) =f(g (y))
= f (√y) = (√y)2 = y
Thus gof = IN
and fog = IY.
Hence, ‘ f ’ is invertible with f-1 = g.

Question 10.
Let f : W → W be defined by:
Class 12 Maths Important Questions Chapter 1 Relations and Functions 18
Show that ‘f ’ is invertible. Find the inverse of ‘f ’. (Here ‘ W’ is the set of whole numbers) (A.I.C.B.S.E. 2015)
Solution:
We have : f : W → W defined by :
Class 12 Maths Important Questions Chapter 1 Relations and Functions 18
f is one-one.
When n1 and n2 are both odd,
then f(n1) = f(n2)
⇒ n1 – 1 = n2 – 1
⇒ n1 = n2
When n1 and n2 are both even, then
f(n1) = f(n2)
⇒ n1 + 1 = n2 + 1
⇒ n1= n2

∴ In both cases,
f(n1) = f(n2)
⇒ n1 = n2
When n1is odd and n2 is even,
then f(n1) =  n1 – 1, which is even
and f(n2) = n2 + 1, which is odd.
∴ (n1) ≠ (n2)
⇒ f(n1) ≠ f(n2)
Similarly when n1 is even and n2 is odd,
∴ (n1) ≠ (n2)
⇒ f(n1) ≠ f(n2)
In each case, ‘f’ is one-one.

f is onto.
When n is odd whole number, then there exists an even whole number
n – 1 ∈ W such that
f(n- 1) = (n -1) + 1 = n.
When n is even whole number, then there exists an odd whole number n + 1 ∈ W such that
f(n +1) = (n + 1) -1 = n.
Also f(1) = 0 ∈ W.
∴ each number of W has its pre-image in W.
Thus f ’ is onto.
Hence, ‘f’ is one-one onto
⇒ ‘f’ is invertible.
To obtain f-1.
Let n1, n2 ∈ W
such that f(n1) = f(n2)
n1 + 1 = n2, if n1 is even
n1 – 1 = n2, if n1 is odd.
Class 12 Maths Important Questions Chapter 1 Relations and Functions 19
Hence f = f-1

Alternating Current Class 12 Important Extra Questions Physics Chapter 7

Here we are providing Class 12 Physics Important Extra Questions and Answers Chapter 7 Alternating Current. Important Questions for Class 12 Physics with Answers are the best resource for students which helps in Class 12 board exams.

Class 12 Physics Chapter 7 Important Extra Questions Alternating Current

Alternating Current Important Extra Questions Very Short Answer Type

Question 1.
The instantaneous current flowing from an ac source is l = 5 sin 314 t. What is the rms value of current?
Answer:
The rms value of current is \(\frac{5}{\sqrt{2}}\).

Question 2.
The instantaneous emf of an ac source is given by E = 300 sin 314 t. What is the rms value of emf?
Answer:
The rms value of voltage is \(\frac{300}{\sqrt{2}}\)

Question 3.
Give the phase difference between the applied ac voltage and the current in an LCR circuit at resonance.
Answer:
The applied ac voltage and the current in an LCR circuit at resonance are in phase.
Hence phase difference = 0.

Question 4.
What is the phase difference between the voltage across the inductor and the capacitor in an LCR circuit?
Answer:
The phase difference is 180°.

Question 5.
What is the power factor of an LCR series circuit at resonance?
Answer:
The power factor is one.

Question 6.
In India, the domestic power supply is at 220 V, 50 Hz, while in the USA it is 110 V, 50 Hz. Give one advantage and one disadvantage of 220 V supply over 110 V supply.
Answer:
Advantage: less power loses
Disadvantage: more fatal.

Question 7.
Define the term ‘wattles current’. (CBSE Delhi 2011)
Answer:
It is the current at which no power is consumed.

Question 8.
In a series LCR circuit, VL = VC ≠ VR. What is the value of the power factor? (CBSE AI 2015)
Answer:
One.

Question 9.
Define capacitor reactance. Write its SI units. (CBSE Delhi 2015)
Answer:
It is the opposition offered to the flow of current by a capacitor. It is measured in ohm.

Question 10.
Define quality factor in series LCR circuit. What is its SI unit? (CBSE Delhi 2016)
Answer:
The quality factor is defined as the ratio of the voltage developed across the capacitor or inductor to the applied voltage. It does not have any unit.

Question 11.
A choke and a bulb are in series to a dc source. The bulb shines brightly. How does its brightness change when an iron core is inserted inside the choke coil?
Answer:
There is no change in the final brightness as the inductive reactance is zero for dc.

Question 12.
A solenoid with an iron core and a bulb are connected to a dc source. How does the brightness of the bulb change, when the iron core is removed from the solenoid?
Answer:
There Is no change In the finaL brightness as the inductive reactance is zero for dc.

Question 13.
In a serles LCR circult, the voltage across an inductor, capacitor and a resistor are 20 V, 20 V and 40 V, respectively. What Is the phase difference between the applied voltage and the current In the circuit?
Answer:
WhenVL = VC, then the circuit is in series resonance, therefore both current and voltage are in phase.

Question 14.
Why Is there no power consumption In an Ideal inductor connected to an ac source?
Answer:
This is because current and voltage across an ideal inductor are out of phase by 900.
Hence P = VRMS IRMS cos 90° = O

Question 15.
Can a choke be replaced by a capacitor of suitable capacitance?
Answer:
Yes, because even then the power consumed will be zero.

Question 16.
Find the inductance of the Inductor that would have a reactance of 50 ohm when used with an ac source of frequency 25/π kHz.
Answer:
Using XL = 2 πf L or L = \(\frac{X_{L}}{2 \pi f}\), therefore
L = \(\frac{50 \times \pi}{2 \pi \times 25}\) = 1 H

Question 17.
The figure gIven below shows the variation of an alternating emf with time. What Is the average value of the emf for the shaded part of the graph?
Class 12 Physics Important Questions Chapter 7 Alternating Current 1
Answer:
Average or mean value of ac over half cycle or in time T/2 is
Em = 2Eo/π = 0.637 Eo = 0.637 × 314
or Em = 200 V.

Question 18.
What is the power dissipated in an ac circuit in which the voltage and current are given by V = 230 sin(ωt + π/2) and l = 10 sin ωt.
Answer:
Since the phase difference between the voltage and current is π/2, therefore power consumed = Vrms lrms cos π/2 = 0

Question 19.
When a lamp is connected to an alternating voltage supply, it lights with the same brightness as when compared to a 12 V dc battery. What Is the peak value of alternating voltage?
Answer:
The peak value is V = 12 × \(\sqrt{2}\) = 16.97 V

Question 20.
Why is the use of a.c. voltage preferred over d.c. voltage? Give two reasons. (CBSEAI 2014)
Answer:

  1. Can be increased or decreased easily.
  2. Can easiLy be converted into dc.

Question 21.
Can the Instantaneous power output of an ac source ever be negative? Can the average power output be negative?
Answer:
Yes, No.

Alternating Current Important Extra Questions Short Answer Type

Question 1.
State the phase relationship between the current flowing and the voltage applied in an ac circuit for (i) a pure resistor (ii) a pure inductor.
Answer:

  1. Electric current and voltage applied in a pure resistor are in same phase, i.e. Φ = 0°
  2. Applied voltage leads electric current flowing through pure-inductor in an ac circuit by phase angle of π/2.

Question 2.
A light bulb is in turn connected in a series (a) across an LR circuit, (b) across an RC circuit, with an ac source. Explain, giving the necessary mathematical formula, the effect on the brightness of the bulb in case (a) and (b), when the frequency of the ac source is increased. (CBSE 2019C)
Answer:
(a) The current in LR circuit is given by
l = \(\frac{V}{\sqrt{R^{2}+\omega^{2} L^{2}}}\)

When the frequency of ac source ω increases, l decreases, and hence brightness decreases.

(b) The current in RC circuit is given by
l = \(\frac{V}{\sqrt{R^{2}+\frac{1}{\omega^{2} C^{2}}}}\)

When the frequency of ac source ω increases, l increases, and hence brightness increases.

Question 3.
An air-core solenoid is connected to an ac source and a bulb. If an iron core is inserted in the solenoid, how does the brightness of the bulb change? Give reasons for your answer.
Answer:
Insertion of an iron core in the solenoid increases its inductance. This in turn increases the value of inductive reactance. This decreases the current and hence the brightness of the bulb.

Question 4.
A bulb and a capacitor are connected in series to an ac source of variable frequency. How will the brightness of the bulb change on increasing the frequency of the ac source? Give reason.
Answer:
When the frequency of the ac is increased, it will decrease the impedance of the circuit as Z = \(\sqrt{R^{2}+(1 / 2 \pi f C)^{2}}\). As a result, the current and hence the brightness of the bulb will increase.

Question 5.
An ideal inductor is in turn put across 220 V, 50 Hz, and 220 V, 100 Hz supplies. Will the current flowing through it in the two cases be the same or different?
Answer:
The current through the inductor is given by l = \(\frac{V}{X_{L}}=\frac{V}{2 \pi f L}\). The current is inversely proportional to the frequency of applied ac.

Since the frequency is different therefore the current will also be different.

Question 6.
State the condition under which the phenomenon of resonance occurs in a series LCR circuit, Plot a graph showing the variation of current with a frequency of ac source in a series LCR circuit.
Answer:
The phenomenon occurs when the inductive reactance becomes equal to the capacitive reactance., i.e. XL – XC
⇒ ω L = \(\frac{1}{ωC}\)
⇒ ω = \(\frac{1}{\sqrt{L C}}\)

The graph is as shown below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 2
Question 7.
Give two advantages and two disadvantages of ac over dc.
Answer:
Advantages of ac:
(a) The generation and transmission of ac are more economical than dc.
(b) The alternating voltage may be easily stepped up or down as per need by using suitable transformers.

Disadvantages of ac:
(a) It is more fatal than dc.
(b) It cannot be used for electrolysis.

Question 8.
In a series, LCR circuit connected to an ac source of variable frequency and voltage v = vm sin ωt, draw a plot showing the variation of current (l) with angular frequency (ω) for two different values of resistance R1 and R2 (R1 > R2). Write the condition under which the phenomenon of resonance occurs. For which value of the resistance out of the two curves, a sharper resonance is produced? Define the Q-factor of the circuit and give its significance. (CBSE Delhi 2013C)
Answer:
The plot is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 3
Resonance occurs in an LCR circuit when
XL = XC.
The smaller the value of R sharper is the resonance. Therefore the curve will be sharper for R2. It determines the sharpness of the resonance. The larger the value of Qsharper is the resonance.

Question 9.
You are given three circuit elements X, Y, and Z. When the element X is connected across an a.c. source of a given voltage, the current and the voltage are in the same phase. When the element Y is connected in series with X across the source, voltage is ahead of the current in phase by π/2. But the current is ahead of the voltage in phase by π/2 when Z is connected in series with X across the source. Identify the circuit elements X, Y, and Z. When all the three elements are connected in series across the same source, determine the impedance of the circuit. Draw a plot of the current versus the frequency of the applied source and mention the significance of this plot. (CBSE AI 2015)
Answer:
X-Resistor, Y-Inductor, Z-Capacitor For expression of the impedance of LCR circuit see X is a resistor
Y is a capacitor
Z is an inductor

Consider the impedance triangle
Z = \(\sqrt{R^{2}+\left(X_{t}-X_{C}\right)^{2}}\)
Class 12 Physics Important Questions Chapter 7 Alternating Current 5
The plot is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 4
Significance, at ω = ω0 (resonance frequency) current, is maximum.

Question 10.
Given three elements X, Y, and Z to be connected across an ac source. With the only X connected across the ac source, voltage and current are found to be in the same phase. With only element Y in the circuit, the voltage lags behind the current in phase by π/2, while with the element Z in the circuit, the voltage leads the current in phase by π/2
(a) Identify the elements X, Y, and Z.
Answer:
X is resistor
Y is a capacitor
Z is an inductor

Consider the impedance triangle
Z = \(\sqrt{R^{2}+\left(X_{t}-X_{C}\right)^{2}}\)
Class 12 Physics Important Questions Chapter 7 Alternating Current 5
(b) When all these elements are connected in series across the same source,
(i) determine the power factor,
Answer:
Power factor = cos Φ = R/Z
= \(\frac{R}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\)

(ii) find out the condition when the circuit is in resonant state. (CBSE AI 2019)
Answer:
Circuit will be in resonance when XL – XC = 0

Question 11.
A coil with an air core and an electric bulb are connected in series across a 220 V 50 Hz ac source. The bulb glows with some brightness. How will the glow of the bulb be affected by introducing a capacitor in series with the circuit? Justify your answer.
Answer:
The impedance of an LR circuit is given by the expression
Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and

the impedance of a series LCR circuit is given by the expression
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\).

Now current flowing through the two circuits is given by V
l = \(\frac{V}{Z}\).

Since Z decreases when a capacitor is connected to an LR circuit therefore there is an increase in current through the circuit. This increases the brightness of the bulb.

Question 12.
In the given circuit, inductor L and resistor R have identical resistance. Two similar electric lamps B1 and B2 are connected as shown. When switch S is closed,
(i) which one of the lamps lights earlier,
(ii) will the lamps be equally bright after some time? Justify your answer.
Class 12 Physics Important Questions Chapter 7 Alternating Current 6
Answer:
(i) Lamp B2 connected with the resistor will light up first. This is because the current through the inductor will grow before attaining maximum value.
(ii) When the current through the inductor becomes maximum, after some time, both the lamps will be equally bright.

Question 13.
Figure (a), (b), and (c) show three ac circuits in which equal currents are flowing. If the frequency of emf be increased, how will the current be affected in these circuits? Give the reason for your answer.
Class 12 Physics Important Questions Chapter 7 Alternating Current 7
Answer:
There will be no change in the current in figure (b) as the resistance of the resistor does not depend upon the frequency of the applied ac.

The reactance of the inductor in figure (a) is given by XL = 2πf L. An increase in frequency increases the value of inductive reactance. This decreases the current through the circuit.

The reactance of the capacitor in figure (c) is given by XC = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\). An increase in frequency decreases the value of capacitive reactance. This increases the current through the circuit.

Question 14.
An alternating voltage of frequency f is applied across a series LCR circuit. Let fr be the resonance frequency for the circuit. Will the current in the circuit lag, lead, or remain in phase with the applied voltage when (i) f > fr (ii) f < fr Explain your answer in each case.
Answer:
(i) When f > fr, then the circuit behaves as an inductive circuit. Thus emf leads current. This is because the inductive reactance is given by the expression XL = 2πfL. At high-frequency XL will be more.

(ii) When f < fr> then the circuit behaves as a capacitive circuit. Thus emf lags current. This is because the capacitive reactance is given by the expression XC = \(\frac{1}{2 \pi f C}\). This value is more at low 2πfC frequency.

Question 15.
For a series LCR circuit, connected to a sinusoidal ac voltage source, identify the graph that corresponds to ω > \(\frac{1}{\sqrt{L C}}\). Give reason.
Class 12 Physics Important Questions Chapter 7 Alternating Current 8
Answer:
When ω > \(\frac{1}{\sqrt{L C}}\), the circuit behaves as an inductive circuit. In an inductive circuit, emf leads current. This is depicted in the graph (a).

Question 16.
Draw the graphs showing the variations of (a) inductive reactance and (b) capacitive reactance, with the frequency of applied voltage in the ac circuit. How do the values of (a) inductive, and (b) capacitive reactance change, when the frequency of applied voltage is tripled?
Answer:
The graphs are as shown below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 9
(a) The inductive reactance is given by the expression XL = 2πfL. Therefore if the frequency is tripled then the value of XL also gets tripled.
(b) The capacitive reactance is given by the expression XC = \(\frac{1}{2 \pi f C}\). Therefore if the frequency is tripled then the value of XC becomes one-third of its previous value.

Question 17.
Can the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the ac source? Justify your answer.
Answer:
Yes, the voltage drop across the inductor or the capacitor in a series LCR circuit be greater than the applied voltage of the ac source. It is because the applied voltage is equal to the algebraic sum (as obtained by the use of a phasor diagram) of VR, VL, and VC, i.e.
V = l\(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

whereas VL = l XL and VC = l XC.
Thus, it is self-evident that VL or VC may be greater than V.

Question 18.
Mention the factors on which the resonant frequency of a series LCR circuit depends. Plot a graph showing the variation of the impedance of a series LCR circuit with the frequency of the applied ac source.
Answer:
In a series LCR circuit, the resonant frequency depends on the value of inductance L and capacitance C present in the circuit.
The graph showing the variation of impedance Z of a series LCR circuit with the frequency f of the applied ac source is shown below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 10
Question 19.
A lamp is connected in series with a capacitor. Predict your observations for dc and ac connections. What happens in each case if the capacitance of the capacitor is reduced? (NCERT)
Answer:
When a dc source is connected to a capacitor, the capacitor gets charged and after charging no current flows in the circuit and the lamp will not glow. There will be no change even if C is reduced. With ac source, the capacitor offers capacitive reactance (1/ωC), and the current flows in the circuit. Consequently, the lamp will shine. Reducing C will increase reactance and the lamp will shine less brightly than before.

Question 20.
A light bulb and an open coil inductor are connected to an ac source through a key as shown in the figure.
Class 12 Physics Important Questions Chapter 7 Alternating Current 11
The switch is closed and after some time, an iron rod is inserted into the interior of the inductor. The glow of the light bulb (a) increases; (b) decreases (c) is unchanged as the iron rod is inserted. Give your answer with reasons. (NCERT)
Answer:
As the iron rod is inserted, the magnetic field inside the coil magnetizes the iron rod thereby increasing the magnetic field inside it. Hence, the inductance of the coil increases. Consequently, the inductive reactance of the coil increases. As a result, a larger fraction of the applied ac voltage appears across the inductor, leaving less voltage across the bulb. Therefore, the glow of the light bulb decreases.

Question 21.
Draw the effective equivalent circuit of the circuit shown in the figure at very high frequencies and find the effective impedance. (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 7 Alternating Current 12
Answer:
At high frequency, the capacitive reactance is small while the inductive reactance is large. Therefore the capacitive reactance can be neglected while no current will flow through the inductors. Therefore the equivalent reactance of the circuit is Z = R1 + R3 and hence the circuit becomes
Class 12 Physics Important Questions Chapter 7 Alternating Current 13
Question 22.
Study the circuits (a) and (b) shown in the figure and answer the following questions.
Class 12 Physics Important Questions Chapter 7 Alternating Current 14
(a) Under which conditions would the rms currents in the two circuits be the same?
Answer:
This will happen when the impedance of both the circuits is the same, i.e. R. This is possible when circuit (b) is in resonance.

(b) Can the rms current in circuit (b) be larger than that in (a)? (NCERT Exemplar)
Answer:
No, because in circuit (b)
lrms = \(\frac{V_{\text {rms }}}{Z}=\frac{V_{\text {rms }}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\), Z cannot be less than R.

Question 23.
How does the sign of the phase angle Φ, by which the supply voltage leads the current in an LCR series circuit, change as the supply frequency is gradually increased from very low to very high values? (NCERT Exemplar)
Answer:
The phase angle for an LCR circuit is given by the expression
tan Φ = \(\frac{R}{Z}=\frac{X_{L}-X_{C}}{R}=\frac{2 \pi f L-1 / 2 \pi f C}{R}\)

At low frequencies XL < XC and at high frequencies XL > XC Therefore Φ changes from negative to zero and to positive; zero at the resonant frequency.

Question 24.
A device ‘X’ is connected to an ac source. The variation of voltage, current, and power in one complete cycle is shown in the figure.
(a) Which curve shows power consumption over a full cycle?
(b) What is the average power consumption over a cycle?
(c) Identify the device ‘X’. (NCERT Exemplar)
Class 12 Physics Important Questions Chapter 7 Alternating Current 15
Answer:
(a) A
(b) Zero
(c) L or C or LC

Question 25.
Both alternating current and direct current are measured in amperes. But how is the ampere defined for an alternating current? (NCERT Exemplar)
Answer:
An ac current changes direction with the source frequency and the charge flow would average to zero. Thus, the ac ampere must be defined in terms of some property that is independent of the direction of the current. Joule’s heating effect is such property and hence it is used to define misvalue of ac.

Question 26.
A sinusoidal voltage of peak value 10 V is applied to a series LCR circuit In which resistance, capacitance, and inductance have values of 10 0, 1 μF, and 1 H, respectively.
Find (i) the peak voltage across the inductor at resonance
(ii) the quality factor of the circuit. (CBSE Sample Paper 2018-19)
Answer:
Given Vm = 10V, R = 1o Ω, L = 1 H, C = 1 μF,
VL = ?, Q = ?
Class 12 Physics Important Questions Chapter 7 Alternating Current 16
Q= ωr L/R = (103 × 1)/10 = 100

Question 27.
Draw a labeled diagram of a step-down transformer. State its working principle Write one main cause of energy loss in this device and the method used for reducing it.
Answer:
The labeled diagram is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 17
It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil.

One main cause of loss of energy is heat produced due to the production of eddy currents. This can be reduced by laminating the iron core.

Alternating Current Important Extra Questions Long Answer Type

Question 1.
Prove mathematically that the average power over a complete cycle of alternating current through an Ideal inductor is zero.
Answer:
Let the instantaneous value of voltage and current in the ac circuit containing a pure inductor are
V = Vm sin ωt and
l = lm sin (ωt – π/2) = – lm cos ωt
where π/2 is the phase angle by which voltage Leads currently when ac flows through an inductor. Suppose the voltage and current remain constant for a small-time dt. Therefore, the electrical energy consumed in the small-time dt is
dW = V l dt

The total electrical energy consumed in one time period of ac is given by
Class 12 Physics Important Questions Chapter 7 Alternating Current 18
Therefore, the total electrical energy consumed in an ac circuit by a pure inductor is W = 0

Now average power is defined as the ratio of the total electrical energy consumed over the entire cycle to the time period of the cycle, therefore
Pav = \(\frac{W}{T}\) = 0

Hence, the average power consumed in an ac circuit by a pure inductor is Pav = 0
Thus a pure inductor does not consume any power when ac flows through it. Whatever energy is used in building up current is returned back during the decay of current.

Question 2.
Draw the phasor diagram of a series LCR connected across an ac source V= Vo sin ωt. Hence, derive the expression for the impedance of the circuit. Obtain the conditions for the phase angle under which the current is
(i) maximum and
(ii) minimum. (CBSE AI 2019)
Answer:
The voltages across the various elements are drawn as shown in the figure below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 19
From the diagram, we observe that the vector sum of the voltage amplitudes VR, VL, and VC equals a phasor whose length is the maximum applied voltage Vm, where the phasor Vm makes an angle φ with the current phasor lm. Since the voltage phasors, VL and VC are in opposite direction, therefore, a difference phasor (VL – VC) is drawn which is perpendicular to the phasor VR. Adding vectorially we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 20
Class 12 Physics Important Questions Chapter 7 Alternating Current 21
where XL = ω L and XC = 1 / ω C, therefore, we can express the maximum current as
lm = \(\frac{V_{\mathrm{m}}}{\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{\mathrm{C}}\right)^{2}}}\)

The impedance Z of the circuit is defined as Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{C}\right)^{2}}\)

For maximum lm, Z should be minimum (Z = R) or XC = XL = 0 and Φ = 0

For (lm)min Φ → 90° (|XC – XL| >> R) Z → ∞

Question 3.
(a) The graphs (i) and (ii) represent the variation of the opposition offered by the circuit element to the flow of alternating current with a frequency of the applied emf. identify the circuit element corresponding to each graph.
Class 12 Physics Important Questions Chapter 7 Alternating Current 22
Class 12 Physics Important Questions Chapter 7 Alternating Current 23
(b) Write the expression for the impedance offered by the series combinations of the above two elements connected across the ac source. Which will be ahead in phase in this circuit, voltage or current? (CBSE AI 2011C)
Answer:
(a) In figure (i) the opposition to the flow of current does not depend upon frequency, the circuit element is a resistor.

In figure (ii) the opposition increases with frequency, the current element is an inductor.

(b) When the resistor R and the inductance L are connected in series across an ac source, the impedance Z of the circuit is given by
Z = \(\sqrt{R^{2}+X_{L}^{2}}\), where XL is the inductive reactance.

In an L – R circuit, the voltage is ahead of the current.

Question 4.
An Inductor L of inductance XL is connected in series with bulb B and an ac source. How would the brightness of the bulb change when
(a) the number of turns In the Inductor Is reduced,
(b) an Iron rod Is Inserted Into the Inductor and
(c) a capacitor of reactance XC = XL
Is Inserted in series In the circuit. Justify your answer In each case. (CBSE Delhi 2015)
Answer:
(a) When the number of turns of the inductor reduced it decreases the inductance of the inductor as (L ∝ n2) where n is the number of turns. This in turn decreases the inductive reactance XL which increases the current in the circuit and hence the brightness of the bulb decreases.
(b) When an iron rod is inserted in the inductor, it increases the inductive reactance, which in turn decreases the current and hence the brightness of the bulb.
(c) When XL = XC, the circuit acts as a resistive circuit, i.e. the impedance becomes minimum and maximum current flows. This makes the bulb glow more brightly.

Question 5.
An inductor L of reactance XL is connected in series with a bulb B to an ac source as shown in the figure below. Briefly explain how the brightness of the bulb changes, when
(a) number of turns of the inductor Is reduced? and
(b) a capacitor of reactance XC = XL is included in series in the same circuit?
Class 12 Physics Important Questions Chapter 7 Alternating Current 24
Answer:
If R is the resistance of the bulb, then the total impedance of the circuit is Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and the corresponding current is l = V l Z.
(a) If the number of turns of the inductor is reduced then its inductance L and consequently there is a decrease in the reactance. This leads to a decrease in the impedance of the circuit. As a result the current flowing through the circuit increases. This increases the brightness of the bulb.

(b) When a capacitor of reactance XC = XL is included in the circuit, then the new impedance becomes Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\) = R. Thus the impedance has its minimum value. This increases the current through the circuit, which results in an increase in the brightness of the bulb.

Question 6.
A capacitor, ‘C’ a variable resistor ‘R’, and a bulb ‘B’ are connected in series to the ac mains in the circuit as shown. The bulb glows with some brightness. How will the glow of the bulb change if (a) a dielectric slab is introduced between the plates of the capacitor, keeping resistance R to be the same (b) the resistance R is increased keeping the same capacitance? (CBSE Delhi 2014)
Class 12 Physics Important Questions Chapter 7 Alternating Current 25
Answer:
(a) As the dielectric slab is introduced between the plates of the capacitor, its capacitance will increase. Hence, the potential drop across the capacitor will decrease (V = Q/C). As a result, the potential drop across the bulb will increase (since both are connected in series). So, its brightness will increase.
(b) As the resistance (R) is increased, the potential drop across the resistor will increase. As a result, the potential drop across the bulb will decrease (since both are connected in series). So, its brightness will decrease.

Question 7.
What is meant by impedance? Give its unit. Using a phasors diagram or otherwise derive the expression for the Impedance of an ac circuit containing L, C, and R in series. Find the expression for the resonant frequency.
Answer:
It is the opposition offered by LR or CR or LCR circuit to the flow of ac. It is measured in ohm.

For derivation of Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 46
The current through an LCR circuit is given by the expression lv = \(\frac{E_{v}}{Z}\)

where Z is the impedance. Substituting the value of Z in the above equation we have
lv = \(\frac{E_{v}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\)

Because the impedance depends on the frequency of the source, we see that the current in the LCR circuit will depend upon frequency. The current, therefore, reaches its maximum value when the impedance is minimum. This happens when XL = XC, corresponding to Z = R. The frequency ω at which this occurs is called resonant angular frequency. To find ω0, we use the condition XL = XC, from which we get
Class 12 Physics Important Questions Chapter 7 Alternating Current 26
But ωo = 2 πfo, where fo is called resonance frequency. Therefore, the above equation becomes fo = \(\frac{1}{2 \pi} \frac{1}{\sqrt{L C}}\)

Question 8.
Explain the term ‘inductive reactance’. Show graphically the variation of an inductive reactance with the frequency of the applied alternating voltage.
An ac voltage E = Eo sin ωt is applied across a pure inductor of inductance L. Show mathematically that the current flowing through it lags behind the applied voltage by a phase angle of π/2.
Answer:
It is the opposition offered to the flow of current by a pure inductor.

Consider an ac circuit consisting of a pure Inductor connected to the terminaLs of an ac source. Let the Instantaneous value of the ac source be
V = Vm sin ωt ….(1)
Class 12 Physics Important Questions Chapter 7 Alternating Current 39
Let VL be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + VL = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since VL = – L\(\frac{di}{dt}\)
Using equations (1) and (2) we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 40
Integrating the above equation we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 41
where \(\frac{V_{m}}{\omega L}\) = lm , ωL = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity cos ωt = – sin (ωt – π/2) in equation (6) we have
lL =lm sin (ωt – π/2) …(7)
Comparing equation (1) with equation (7) we see that the current lags voltage by π/2 radian or 90°.
Class 12 Physics Important Questions Chapter 7 Alternating Current 42
Question 9.
Explain the term ‘capacitive reactance’. Show graphically the variation of a capacitive reactance with the frequency of the applied alternating voltage.
An ac voltage E=Eo sin cot is applied across a pure capacitor of capacitance C. Show mathematically that the current flowing through it leads the applied voltage by a phase angle of π/2.
Answer:
It is the opposition offered to the flow of current by a pure capacitor.
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = Vm sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – VC = 0
V = VC = Vm Sin ωt …(2)

But from the definition of capacitance, VC = Q / C or Q= VCC. Substituting for VC we have
Q = C Vm sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CVm sin ωt = ω C Vm cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
lC = ω C Vm cos ωt = lm cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
lC = lm sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of XC with f is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 44
Question 10.
Derive an expression for the impedance of a series LCR circuit connected to an ac supply of variable frequency. Plot a graph showing a variation of current with the frequency of the applied voltage. Explain briefly how the phenomenon of resonance in the circuit can be used in the tuning mechanism of a radio or a TV set. (CBSE Delhi 2011)
Answer:
The voltages across the various elements are drawn as shown in the figure below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 45
From the diagram, we observe that the vector sum of the voltage amplitudes VR, VL, and VC equals a phasor whose length is the maximum applied voltage Vm, where the phasor Vm makes an angle Φ with the current phasor lm. Since the voltage phasors, VL and VC are in opposite directions, a different phasor (VL – VC) is drawn which is perpendicular to the phasor VR.

Adding vectorially, we have
Vm = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{c}\right)^{2}}\)
= \(\sqrt{\left(l_{m} R\right)^{2}+\left(l_{m} X_{L}-l_{m} X_{c}\right)^{2}}\) …(1)
or
Vm = lm\(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\) …(2)

where XL = ω L and XC = 1/ω C, therefore we can express the maximum current as
lm = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\) …(3)

∴ lm = \(\frac{V_{m}}{Z}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 46
The graph is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 27
The antenna of a radio accepts signals from many broadcasting stations. The signals picked up in the antenna act as a source in the tuning circuit of the radio, so the circuit can be driven at many frequencies. But to hear one particular radio station, we tune the radio. In tuning, we vary the capacitance of a capacitor in the tuning circuit such that the resonant frequency of the circuit becomes nearly equal to the frequency of the radio signal received. When this happens, the amplitude of the current with the frequency of the signal of the particular radio station in the circuit is maximum and that particular radio station is tuned in.

Question 11.
A device ‘X’ is connected to an ac source V = Vo sin ωt. The variation of voltage, current, and power in one cycle is shown in the following graph.
Class 12 Physics Important Questions Chapter 7 Alternating Current 28
(a) Identify the device ‘X’.
Answer:
Capacitor

(b) Which of the curves A, B, and C represent the voltage, current, and power consumed in the circuit? Justify your answer.
Answer:
A: Power, B: Voltage, and C: Current.

(c) How does its impedance vary with the frequency of the ac source? Show graphically.
Answer:
For a capacitor, Impedance is given by XC = 1 /ωC = 1 /2πfC. This is shown graphically as
Class 12 Physics Important Questions Chapter 7 Alternating Current 29
(d) Obtain an expression for the current in the circuit and its phase relation with ac voltage. (CBSE AI 2017)
Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = Vo sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – VC = 0
or
V = VC = Vo sin ωt …(2)

But from the definition of capacitance,
VC = Q / C
or
Q= VC C.

Substituting for Vc we have
Q = CVo sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\) CVo sin ω t = ω C Vo cos ω t …(4)

But dQ/dt = ic, therefore the above equation becomes,
ic = ω C Vo cos ω t = lo cos ω t …(5)

Here the term 1/ωc has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm. Using the trigonometric identity cos ω t = sin (ω t + π/2) equation (5) can be written as
ic = lo sin (ω t + π/2) …(6)

Comparing this expression with equation (1) we see that the current is 90° (π/2 ) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

Question 12.
A device X is connected across an ac source of voltage V = Vo sin ωt. The current through X is given as l = lo sin (ωt + π/2).
(a) Identify the device X and write the expression for Its reactance.
(b) Draw graphs showing the variation of voltage and current with time over one cycle of ac, for X.
(c) How does the reactance of device X vary with the frequency of the ac? Show this variation graphically.
(d) Draw the phasor diagram for the device X. (CBSE AI 2018, Delhi 2018)
Answer:
(a) X: capacitor
Reactance XC = 1 / ωC = 1 / 2πfC

(b) The graphs area as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 30
(c) Reactance of the capacitor varies in inverse proportion to the frequency. Therefore, the graph is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 31
(d) The phasor diagram is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 32
Question 13.
A resistor of 200 Ω and a capacitor of 15.0 μF is connected In series to a 220 V, 50 Hz ac source, (a) Calculate the current in the circuit; (b) Calculate the voltage (rms) across the resistor and the capacitor. Is the algebraic sum of these voltages more than the source voltage? If yes, resolve the paradox. (NCERT, CBSE Delhi 2004)
Answer:
Given R = 200 Ω, C= 15.0 μF = 15.0 × 10-6 F, V = 220 V, f = 50 Hz
(a) In order to calculate the current, we need the impedance of the
Class 12 Physics Important Questions Chapter 7 Alternating Current 33
Therefore the current in the circuit is
l = \(\frac{V}{Z}=\frac{220}{291.5}\) = 0.7555 A

(b) Since the current is the same throughout the circuit we have
VR = IR = 0.755 × 200 = 151 V
VC = lXC = 0.755 × 212.3 = 160.3 V

The algebraic sum of the two voltages VR and Vc is 311.3 V which is more than the source voltage of 220 V. How to resolve this paradox? As you have learned in the text, the two voltages are not in the same phase. Therefore, they cannot be added like ordinary numbers. The two voltages are out of phase by ninety degrees. Therefore, the total of these voltages must be obtained using the Pythagorean theorem;
V = \(\sqrt{V_{R}^{2}+V_{C}^{2}}\) = 220 V

Thus if the phase difference between two voltages is properly taken into account, the total voltage across the resistor and the capacitor is equal to the voltage of the source.

Question 14.
Explain why the reactance offered by an inductor increases with the increasing frequency of an alternating voltage.
(NCERT Exemplar)
Answer:
An inductor opposes the flow of current through it by developing a back emf according to Lenz’s law. The induced voltage has a polarity so as to maintain the current at its present value. If the current is decreasing, the polarity of the induced emf will be so as to increase the current and vice versa. Since the induced emf is proportional to the rate of change of current, it will provide greater reactance to the flow of current if the rate of change is faster, i.e. if the frequency is higher. The reactance of an Inductor, therefore, is proportional to the frequency, being given by ωL.

Question 15.
(a) Prove that an Ideal capacitor in an ac circuit does not dissipate power.
Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = Vm sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – VC = 0
V = VC = Vm Sin ωt …(2)

But from the definition of capacitance, VC = Q / C or Q= VCC. Substituting for Vc we have
Q = C Vm sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CV<sub>m</sub> sin ωt = ω C Vm cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
lC = ω C Vm cos ωt = lm cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
lC = lm sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of XC with f is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 44
Power in capacitive circuit P = Ev lv cos Φ. In pure captor 0 = 90° phase difference between voltage and current is π/2.
So power consumed in pure capdtor Pav = Ev lv cos 90° = 0.

(b) An ideal inductor of 200 mH, capacitor of 400 μF, and a resistor of 10 Ω are connected In series to an ac source of 50 V and variable frequency. Calculate
(i)the angular frequency at which maximum power dissipation occurs in the circuit and the corresponding value of effective current and
(ii) the value of Q factor in the circuit. (CBSE AI 2017C)
Answer:
Class 12 Physics Important Questions Chapter 7 Alternating Current 34
Question 16.
(a) What is the principle of transformer?
(b) Explain how laminating the core of a transformer helps to reduce eddy current losses in it.
(c) Why the primary and secondary coils of a transformer are preferably wound on the same core?
Or
Show that In the free oscillations of an LC circuit, the sum of energies stored in the capacitor and the Inductor Is constant In time. (CBSE Sample Paper 2018-19)
Answer:
(a) It works on the principle of mutual induction.
(b) In order to reduce the eddy current loss, the resistance of the core should be Increased. In a transformer, the core Is made up of thin sheets of steel, each lamination being insulated from others by a thin layer of varnish. As the laminations are thin, they will have relatively high resistance.
(c) This is done to maximize the sharing of magnetic flux and also so that magnetic flux per turn should become the same in both the primary and secondary coils.
Or
Let at any instant t, q be the charge on the capacitor and ‘l’ be the current through the inductor
q (t) = q0 cos ωt
i (t) = – q0 ω sin ωt
Energy stored in the capacitor at time t is
UE = \(\frac{1}{2} \frac{q^{2}}{c}=\frac{1}{2} \frac{q_{0}^{2}}{C}\) cos2 ωt

Energy stored in the inductor at time t is
Class 12 Physics Important Questions Chapter 7 Alternating Current 35
This sum is constant in time as q0 and C, both are time-independent.

Question 17.
Explain why the reactance provided by a capacitor to an alternating current decreases with increasing frequency. (NCERT Exemplar)
Answer:
A capacitor does not allow the flow of direct current through it as the resistance across the gap is infinite. When an alternating voltage is applied across the capacitor plates, the plates are alternately charged and discharged. The current through the capacitor is a result of this changing voltage (or charge). Thus, a capacitor will pass more current through it if the voltage is changing at a faster rate, i.e. if the frequency of the supply is higher. This implies that the reactance offered by a capacitor is less with increasing frequency; it is given by 1 /ωC.

Question 18.
(a) With the help of a labeled diagram, describe briefly the underlying principle and working of a step-up transformer.
(b) Write any two sources of energy loss in a transformer.
(c) A step-up transformer converts a low input voltage into a high output voltage. Does it violate the law of conservation of energy? Explain. (CBSE Delhi 2011)
Answer:
(a) The diagram is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 36
It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil. In a step-up transformer, the number of turns in the secondary is more than that in the primary.

The ac source causes an alternating current in the primary, which sets up an alternating flux in the core; this induces an emf in each winding of the secondary, in accordance with Faraday’s law. The induced emf in the secondary gives rise to an alternating current in the secondary, and this delivers energy to the device to which the secondary is connected.

For step up : Ns> > Np
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) and \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\)

(b) Two major sources of energy loss in this device are:

  1. Heat produced due to production of eddy currents.
  2. Copper loss: heat produced in the copper coils

(c) No, a transformer does not change the power. Thus if a voltage increases current gets decreased such that the energy and power remain the same.

Question 19.
(a) Explain with the help of a labeled diagram, principle, and working of a transformer. Deduce the expression for its working formula.
Answer:
For diagram as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 56

  • Principle: A transformer works on the basis of mutual induction.
  • Working: In a 100% efficient transformer

εs ls = εp lp where l and lp, are the secondary and primary currents, therefore we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) ….(1)

Now a 1oo % efficient transformer
we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} \frac{\frac{d \phi}{d t}}{\frac{d \phi}{d t}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}\) …(2)

Therefore form (1) and (2) we have
\(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) = k

Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\).

(b) Name any four causes of energy loss in an actual transformer. (CBSEAI 2013C)
Answer:

  1. Copper loss
  2. Iron loss
  3. Flux loss
  4. Hysteresis loss

Question 20.
(a) Draw a schematic sketch of an ac generator describing its basic elements. State briefly its working principle. Show a plot of the variation of (i) Magnetic flux and alternating emf versus time generated by a loop of wire rotating in a magnetic field,
Answer:
A schematic sketch is as shown.

Working principle: Electromagnetic induction.
Class 12 Physics Important Questions Chapter 7 Alternating Current 37
The graphs are as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 38

(b) Why is a choke coil needed in the use of fluorescent tubes with ac mains? (CBSE Delhi 2014)
Answer:
It is used to regulate current with minimum loss of energy.

Question 21.
(a) Draw a labeled diagram of the AC generator. Derive the expression for the instantaneous value of the emf induced in the coil.
Answer:
It is based on the principle of electromagnetic induction. When a coil is rotated about an axis perpendicular to the direction of the uniform magnetic field, an induced emf is produced across it.

Working: The working of the ac generator can be understood with the help of the various positions of the armature as shown in the figure below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 58
Class 12 Physics Important Questions Chapter 7 Alternating Current 59
Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero.

As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to the maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coil

(b) A circular coil of cross-sectional area 200 cm2 and 20 turns is rotated about the vertical diameter with an angular speed of 50 rad s-1 in a uniform magnetic field of magnitude 3.O × 10-2 T. Calculate the maximum value of the current in the coil. (CBSE Delhi 2017)
Answer:
Given A = 200 cm2, ω = 50 rad s-1, n = 20
S = 3 × 10-2 T, εo = ?,

Using the relation ε0 = nBAω
= 20 × 3 × 10-2 × 200 × 10-4 × 50 = 0.6 V

Question 22.
Draw a labeled diagram of a step-up transformer. Obtain the ratio of secondary to primary voltage in terms of the number of turns and currents in the two coils.
A power transmission line feeds input power at 2200 V to a step-down transformer with its primary windings having 3000 turns. Find the number of turns in the secondary to get the power output at 220 V. (CBSE Delhi 2017)
Answer:
(a) The diagram is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 36
It works on the principle of mutual induction i.e., whenever magnetic flux linked with a coil changes an emf is induced in the neighboring coil. In a step-up transformer, the number of turns in the secondary is more than that in the primary.

The ac source causes an alternating current in the primary, which sets up an alternating flux in the core; this induces an emf in each winding of the secondary, in accordance with Faraday’s law. The induced emf in the secondary gives rise to an alternating current in the secondary, and this delivers energy to the device to which the secondary is connected.

For step up : Ns > Np
\(\frac{\varepsilon_{s}}{\varepsilon_{p}}=\frac{N_{s}}{N_{p}}\) and \(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\).

Principle – It works on the principle of electromagnetic induction. When the current in one circuit changes, an induced current is set up in the neighboring circuit.

(b) Ns = ?, Np = 3000
Ep = 2200 V, Es = 220 V

Since, Es/Ep = Ns/Np
220/2200 = Ns/3000
Ns = 300

Question 23.
(a) State the principle of working of a transformer.
Answer:
It works on the principle of mutual inductance i.e. whenever the magnetic flux Linked in a coil changes an induced emf Is produced in the neighboring coil.

(b) Define the efficiency of a transformer.
Answer:
It is the ratio of the output power to the input power.

(c) State any two factors that reduce the efficiency of a transformer.
Answer:
Copper Loss, flux Loss

(d) Calculate the current drawn by the primary of a 90% efficient transformer which steps down 220 V to 22 V If the output resistance Is 440 Ω. (CBSFAI 2018 C)
Answer:
Given η = 90%, Vp = 22OV, Vs = 22V, Ro = 440 W
Now ls = Vs/R = 22/440 = 0.05 A

Also η = \(\frac{P_{0}}{P_{1}}=\frac{V_{s} l_{s}}{V_{p} l_{p}}\) , therefore we have

\(\frac{90}{100}=\frac{22 \times 0.05}{220 \times l_{\mathrm{p}}}\)
or
lp = 0.0056 A

Question 24.
Draw an arrangement for winding of primary and secondary coil In a transformer with two coils on a separate limb of the core.
State the underlying principle of a transformer. Deduce the expression for the ratio of secondary voltage to the primary voltage In terms of the ratio of the number of turns of the primary and secondary winding. For an ideal transformer, obtain the ratio of primary and secondary currents In terms of the ratio al the voltages in the secondary and primary voltages. Write any two reasons for the energy losses which occur In actual transformers, (CBSE Dei hi 2016C)
Answer:
Principle: When an alternating voltage is applied to the primary, the resulting current produces an alternating magnetic flux which Links the secondary and Induces an emf in it.

Induced emf, or voltage, in the secondary, with ‘Ns‘ number of turns,
εs = – Ns \(\frac{d \phi}{d t}\)

Back emf In the primary with Np turns,
εp = – Np \(\frac{d \phi}{d t}\)

\(\frac{\varepsilon_{p}}{\varepsilon_{s}}=\frac{N_{p}}{N_{s}}\)

Since εs = Vs and εp = Vp
Therefore we have
\(\frac{V_{p}}{V_{s}}=\frac{N_{p}}{N_{s}}\)

For an ideal transformer,
Input power = output power
Therefore
lpVp = lsVs
Or
\(\frac{V_{p}}{V_{s}}=\frac{l_{s}}{l_{p}}\)

(a) Heat energy Loss.
(b) Humming effect Loss.

Question 25.
A pure inductor is connected across an ac source. Show mathematically that the current in it lags behind the applied emf by a phase angle of π/2. What is its inductive reactance? Draw a graph showing the variation of an inductive reactance with the frequency of the ac source.
Answer:
Consider an ac circuit consisting of a pure Inductor connected to the terminaLs of an ac source. Let the Instantaneous value of the ac source be
V = Vm sin ωt ….(1)
Class 12 Physics Important Questions Chapter 7 Alternating Current 39
Let VL be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + VL = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since VL = – L\(\frac{di}{dt}\)
Using equations (1) and (2) we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 40
Integrating the above equation we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 41
where \(\frac{V_{m}}{\omega L}\) = lm , ωL = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity cos ωt = – sin (ωt – π/2) in equation (6) we have
lL =lm sin (ωt – π/2) …(7)
Comparing equation (1) with equation (7) we see that the current lags voltage by π/2 radian or 90°.
Class 12 Physics Important Questions Chapter 7 Alternating Current 42

Question 26.
An alternating emf Is applied across a capacitor. Show mathematically that the current leads the emf by a phase angle π/2, What Is its capacitive reactance? Draw a graph showing the variation of a capacitive reactance with the frequency of the ac source.
Class 12 Physics Important Questions Chapter 7 Alternating Current 43
Answer:
Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = Vm sin ωt …(1)

Let Vc be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – VC = 0
V = VC = Vm Sin ωt …(2)

But from the definition of capacitance, VC = Q / C or Q= VCC. Substituting for VC we have
Q = C Vm sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CVm sin ωt = ω C Vm cos ωt …(4)

But dQ/dt = lC, therefore the above equation becomes,
lC = ω C Vm cos ωt = lm cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
lC = lm sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, currently leads emf by 90°.

The graph of variation of XC with f is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 44
Question 27.
(a) In a series LCR circuit connected across an ac source of variable frequency, obtain the expression for its impedance and draw a plot showing its variation with the frequency of the ac source.
Answer:
The voltages across the various elements are drawn as shown in the figure below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 45
From the diagram, we observe that the vector sum of the voltage amplitudes VR, VL, and VC equals a phasor whose length is the maximum applied voltage Vm, where the phasor Vm makes an angle Φ with the current phasor lm. Since the voltage phasors, VLand VC is in opposite directions, a different phasor (VL – VC) is drawn which is perpendicular to the phasor VR.

Adding vectorially, we have
Vm = \(\sqrt{V_{R}^{2}+\left(V_{L}-V_{c}\right)^{2}}\)
= \(\sqrt{\left(l_{m} R\right)^{2}+\left(l_{m} X_{L}-l_{m} X_{c}\right)^{2}}\) …(1)
or
Vm = lm\(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\) …(2)

where XL = ω L and XC = 1/ω C, therefore we can express the maximum current as
lm = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}}\) …(3)

∴ lm = \(\frac{V_{m}}{Z}\)

The impedance Z of the circuit is defined as
Z = \(\sqrt{R^{2}+\left(X_{L}-X_{c}\right)^{2}}\)

The graph is shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 46
(b) What is the phase difference between the voltages across the inductor and the capacitor at resonance in the LCR circuit?
Answer:
The phase difference between the voltage across the inductor and the capacitor at resonance is 180°.

(c) When an inductor is connected to a 200V dc voltage, a current of 1A flows through it. When the same inductor is connected to a 200V, 50 Hz ac source, only 0.5A current flows. Explain why? Also, calculate the self-inductance of the inductor. (CBSE Delhi 2019)
Answer:
Inductor will offer an additional impedance to ac due to its self-inductance.

Now R = \(\frac{V_{\text {rms }}}{l_{\text {rms }}}=\frac{200}{1}\) = 200 Ω

The impedance of the inductor
Z = \(\frac{V_{\mathrm{rms}}}{l_{\mathrm{rms}}}=\frac{200}{0.5}\) = 400 Ω

Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) , therefore
XL2 = Z2 – R2
XL = \(\left(\sqrt{(400)^{2}-(200)^{2}}\right)\) = 346.4 Ω

Hence inductance (L) is
L = \(\frac{X_{L}}{\omega}=\frac{346.4}{2 \times 3.14 \times 50}\) = 6.2 H

Question 28.
(a) Obtain the expression for the average power dissipated in a series LCR circuit driven by an ac source of voltage V = Vm sin ωt supplying the current i = im sin (ωt + Φ).
Answer:
Average Power dissipated in a series LCR circuit
Given V = Vm sin tot
And l = im sin (ωt + Φ)
Instantaneous Power
P = Vi
Class 12 Physics Important Questions Chapter 7 Alternating Current 47
Class 12 Physics Important Questions Chapter 7 Alternating Current 48

(b) Define the terms:
(i) Wattless current, and
Answer:
Wattles’s current: If the average power consumed due to the flow of current in a circuit is zero, the current is said to be wattless.

(ii) Q – a factor of LCR circuit.
Answer:
Q-factor of an LCR Circuit: Q-factor of the LCR circuit is the ratio of the potential difference across inductance (or capacitance) at resonance to the applied voltage.

Question 29.
Distinguish between the terms reactance and impedance of an ac circuit. Prove that an ideal capacitor connected to an ac source does not dissipate power.

S.No. Reactance Impedance
1. It is the opposition offered by a pure inductor or a pure capacitor or both to the flow of ac. 1. It is the opposition offered by LR or CR or LCR circuit to the flow of ac
2. It depends upon the frequency of ac. 2. It depends upon the frequency of ac.
3. It can be inductive or capacitive. 3. It can be inductive, capacitive, or resistive.
4. It can be zero. 4. It can never be zero.

Consider an ac circuit consisting of a capacitor connected to an ac source. Let the instantaneous value of emf be
V = Vm sin ωt …(1)

Let VC be the instantaneous voltage drop across the capacitor, then by Kirchoff’s loop rule,
V – VC = 0
V = VC = Vm Sin ωt …(2)

But from the definition of capacitance, VC = Q / C or Q= VCC. Substituting for Vc we have
Q = C Vm sin ωt …(3)

Differentiating equation (3) with respect to time we have,
\(\frac{d Q}{d t}=\frac{d}{d t}\)CVm sin ωt = ω C Vm cos ωt …(4)

But dQ/dt = lc, therefore the above equation becomes,
lC = ω C Vm cos ωt = lm cos ωt …(5)
where
lm = \(\frac{v_{m}}{\frac{1}{\omega C}}\)

Here the term \({\frac{1}{\omega C}}\) has the dimensions of resistance and is called the capacitive reactance of the circuit and ωCVm = lm.

Using the trigonometric identity cos ωt = sin (ωt + π/2), equation (5) can be written as
lC = lm sin (ωt + π/2) …..(6)

Comparing this expression with equation (1)

we see that the current is 90° (π/2) out of phase with the voltage across the capacitor. In other words, current leads emf by 90°.

The graph of variation of Xc with f is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 44
Question 30.
Define the term root mean square (rms) value of ac. Derive the relation between rms value and the peak value of ac.
Answer:
It is that value of steady current (dc) which when passed through a resistor in a given time produces the same heat as is produced by the ac. when passed through the same resistor for the same time. It is abbreviated as rms value of current. It is denoted by lrms.

Consider an ac source. Let the instantaneous value of current be represented by the equation
i = lm sin ωt

Let this current pass through a resistor of resistance R. Therefore in a small-time dt, the amount of heat produced in the resistor is
dH = i2 R dt

The total amount of heat produced in one complete cycle of ac is given by
Class 12 Physics Important Questions Chapter 7 Alternating Current 49
Class 12 Physics Important Questions Chapter 7 Alternating Current 50

Let lrms be the virtual value of ac, then the amount of heat produced in the same resistor of resistance R, in the same time T is
H = lrms2RT …(2)

Therefore from equations (1) and (2) we have
lrms = \(\frac{l_{m}}{\sqrt{2}}\) = 0.707 lm

Question 31.
(a) Prove that the current flowing through an Ideal Inductor connected across a.c. source lags the voltage In phase by nil.
Answer:
(a) Consider an ac circuit consisting of a pure inductor connected to the terminals of an ac source. Let the instantaneous value of the ac source be
V = Vm sin ω t …(1)

Let VL be the instantaneous voltage drop across the inductor, then Kirchoff’s loop rule when applied to the circuit gives V + VL = 0
or
V – L\(\frac{di}{dt}\) = 0 …(2)

since VL = – L \(\frac{di}{dt}\)

Using equations (1) and (2) we have
L\(\frac{di}{dt}\) = Vm sin ω t …(3)
or
di = \(\frac{V_{m}}{L}\) sin ω t dt …(4)

Integrating the above equation we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 51
C = 0, over a period of cycle

\(\frac{V_{\mathrm{m}}}{\omega L}\) = lm, ωL = 2πfL has the dimensions of resistance and is called the inductive reactance of the circuit. Using the trigonometric identity
cos ω t = – sin (ωt – π/2) in equation (6) we have
lL = lm sin (ωt – π/2) …(7)

Comparing equation (1) with equation (7) clearly shows that the current lags voltage by π/2 radian or 90°.

(b) An inductor of self-inductance 100 mH and a bulb are connected in series with a.c. source of rms voltage 10 V, 50 Hz. It is found that the effective voltage of the circuit leads the current in phase by π/4. Calculate the resistance of the bulb used and average power dissipated in the circuit, if a current of 1 A flows in the circuit. (CBSE Delhi 2017C)
Answer:
(b) V= 10 V, f = 50 Hz, l = 1.0 A, Φ = π/4, R = ?, L = 100mH = 0.1 H
Now Z = V/l = 10/1.0 = 10 ohm

Also Z = \(\sqrt{R^{2}+X_{L}^{2}}\) = 10
Now cos Φ = \(\frac{R}{Z}\)
or
Therefore, R = cos Φ × Z
Or
R = cos π/4 × 10 = 7 ohm

Power dissipated
Pav = lrms Vrms cos Φ = \(\frac{V_{m}^{2}}{Z}\) × cos Φ
0r
Pav = \(\frac{(10)^{2}}{10}\) × 0.707 = 7.07 W

Question 32.
(a) Draw graphs showing the variations of inductive reactance and capacitive reactance with the frequency of the applied ac source.
(b) Draw the phasor diagram for a series RC circuit connected to an ac source.
(c) An alternating voltage of 220 V is applied across a device X, a current of 0.25 A flows, which lag behind the applied voltage In phase by \(\frac{π}{2}\) radian. If the same voltage is applied across another device Y, the same current flows but now it is in phase with the applied voltage.
(i) Name the devices X and Y.
(ii) Calculate the current flowing in the circuit when the same voltage is applied across the series combination of X and Y. (CBSEAI 2018C)
Or
(a) State the principle of working of a transformer.
(b) Define the efficiency of a transformer.
(c) State any two factors that reduce the efficiency of a transformer.
(d) Calculate the current drawn by the primary of a 90% efficient transformer which steps down 220 V to 22 V, if the output resistance is 440 Ω.
Answer:
(a) The two graphs are as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 52
Class 12 Physics Important Questions Chapter 7 Alternating Current 53

(b) (The current leads the voltage by an angle 0 where 0 < θ < \(\frac{π}{2}\).) The required phasor diagram is as shown.
Here θ = tan-1 \(\frac{1}{ωCR}\)
Class 12 Physics Important Questions Chapter 7 Alternating Current 54

(c) In device X:
Current lags behind the voltage by \(\frac{π}{2}\)
∴ X is an inductor.

In device Y:
Current is in phase with the applied voltage
∴ X is a resistor.

We are given that 0.25 = \(\frac{220}{X_{L}}\)
or
XL = \(\frac{220}{0.25}\) Ω = 880 Ω

Also 0.25 = 0.25 = \(\frac{220}{X_{R}}\)

∴ XR = \(\frac{220}{0.25}\) Ω = 880 Ω

For the series combination of X and Y, Equivalent impedance
= \(\sqrt{X_{L}^{2}+X_{R}^{2}}\) = (880\(\sqrt{2}\) Ω

∴ Current flowing = \(\frac{220}{880 \sqrt{2}}\) A = 0.177 A
Or
(a) A transformer works on the principle of mutual induction. (Alternatively, an emf is induced in the secondary coil when the magnetic flux linked with it changes with time due to changing magnetic flux linked with the primary coil).

(b) The efficiency of a transformer equals the ratio of the output power to the input power.
Efficiency = \(\frac{\text { output power }}{\text { input power }}\)
or
Efficiency = \(\frac{V_{S} I_{S}}{V_{P} I_{p}}\)
(c)

  • Eddy current Losses
  • route heat Losses
  • hysteresis Losses
  • magnetic flux leakage Losses

We have
Class 12 Physics Important Questions Chapter 7 Alternating Current 55

Question 33.
(a) Draw a schematic diagram of a step-up transformer. Explain its working principle. Assuming the transformer to be 100% efficient, obtain the relation for
(i) the current in the secondary in terms of the current in the primary, and
(ii) the number of turns in the primary and secondary windings.
(b) Mention two Important energy losses in actual transformers and state how these can be minimized. (CBSE Delhi 2011C)
Answer:
(a) For diagram as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 56

  • Principle: A transformer works on the basis of mutual induction.
  • Working: In a 100% efficient transformer

εs ls = εp lp where l and lp are the secondary and primary currents, therefore we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) ….(1)

Now a 1oo % efficient transformer
we have
\(\frac{\varepsilon_{\mathrm{s}}}{\varepsilon_{\mathrm{p}}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}} \frac{\frac{d \phi}{d t}}{\frac{d \phi}{d t}}=\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}\) …(2)

Therefore form (1) and (2) we have
\(\frac{N_{\mathrm{s}}}{N_{\mathrm{p}}}=\frac{l_{\mathrm{p}}}{l_{\mathrm{s}}}\) = k

Efficiency of transformer = \(\frac{V_{s} I_{s}}{V_{p} I_{p}}\)

(b) The two important energy losses in actual transformers are:

  1. Iron losses in the core of the transformer.
  2. Loss of energy in the primary and secondary due to Joule heating. The iron losses due to eddy currents are minimized by laminating the iron core.

Question 34.
(a) Draw the diagram of a device that is used to decrease high ac voltage into a low ac voltage and state its working principle. Write four sources of energy loss in this device.
Answer:
The labeled diagram is as shown.
Class 12 Physics Important Questions Chapter 7 Alternating Current 57
It works on the principle of mutual induction, i.e. whenever magnetic flux linked with a coil changes, an emf is induced in the neighboring coil.

The possible causes of energy losses in transformers are:

  • Flux leakage
  • Copper loss
  • Eddy currents
  • Hysteresis loss

(b) A small town with a demand of 1200kW of electric power at 220 V is situated 20 km away from an electric plant generating power at 440V. The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets the power from the line through a 4000 – 220V step-down transformer at a sub-station in the town. Estimate the line power loss in the form of heat. (CBSE Delhi 2019)
Answer:
Total resistance of the line = length × resistance per unit length = 40 km × 0.5 = 20 Ω

Current flowing in the line l = P/V
= 1200 × 103 /4000 = 300 A

Power loss = l2R = (300)2 × 20 = 1800 kW

Question 35.
(a) State the principle of an ac generator and explain its working with the help of a labeled diagram. Obtain the expression for the emf induced in a coil having N turns each of cross-sectional area A, rotating with a constant angular speed ‘co’ in a magnetic field B, directed perpendicular to the axis of rotation.
Answer:
For principle and diagram
Class 12 Physics Important Questions Chapter 7 Alternating Current 58
Class 12 Physics Important Questions Chapter 7 Alternating Current 59
Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to the maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coiL.

(b) An airplane is flying horizontally from west to east with a velocity of 900 km h-1. Calculate the potential difference developed between the ends of its wings having a span of 20 m. The horizontal component of the Earth’s magnetic field is 5 × 10-4 T and the angle of dip is 30°. (CBSEAI 2018, Delhi 2018)
Answer:
Potential difference developed between the ends of the wings
ε = BLv

Given Velocity V = 900 km h-1 = 250 m s-1
Wing span (L1) = 20 m

Vertical component of Earth’s magnetic field
Bv = BH tan δ = 5 × 10-4 (tan 30°) T

∴ Potential difference
= 5 × 10-4 (tan 30°) × 20 × 250 = 1.44 V

Question 36.
Explain, with the help of a diagram, the principle and working of an ac generator. Write the expression for the emf generated in the coil in terms of its speed of rotation.
Answer:
It is based on the principle of electromagnetic induction. When a coil is rotated about an axis perpendicular to the direction of a uniform magnetic field, an induced emf is produced across it.

Working: The working of the ac generator can be understood with the help of the various positions of the armature as shown in figure below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 58
Class 12 Physics Important Questions Chapter 7 Alternating Current 59
Suppose at time t = 0, the plane of the loop is perpendicular to B. As the loop rotates from position t = 0 to position t T/2, the Induced emf changes from zero to maximum value and then becomes zero again as shown in the diagram For angles 0° and 1800 the instantaneous rate of magnetic flux is zero, hence induced emf Is zero. As the Loop moves from position t = T/2 to position t = T, the emf again changes from zero to a maximum value and then again becomes zero. But this time it reverses Its direction. For angles 90° and 270° maximum magnetic flux are linked with the coil hence the emf is a maximum. Thus the output of the ac generator varies sinusoidally with time. The Induced emf does not depend upon the shape of the Loop but depends only upon the area of the loop.

The emf generated Is given by the expression ε = nBAω Sin ωt, where ω is the speed of rotation of the coiL.

Numerical Problems:
Formulae for solving numerical problems

  • Capacitive reactance XC = \(\frac{1}{\omega C}=\frac{1}{2 \pi f C}\)
  • Inductive reactance XL = ωL = 2πfL
  • When ac flows through an LR circuit then,
    Z = \(\sqrt{R^{2}+X_{L}^{2}}\) and tan Φ = \(\frac{X_{L}}{R}=\frac{\omega L}{R}\)
  • When ac flows through a CR circuit then,
    Z = \(\sqrt{R^{2}+X_{C}^{2}}\) and tan Φ = \(\frac{X_{c}}{R}=\frac{1}{\omega C R}\)
  • For a senes LCR circuit driven by voltage V = Vm sin ωt, the current is given by
    l = lm sin (ωt + Φ) where
    lm = \(\frac{V_{m}}{\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}}\)
  • The impedance of this circuit is given by
    Z = \(\sqrt{R^{2}+\left(X_{L}-X_{C}\right)^{2}}\)
    \(\frac{V_{s}}{V_{p}}=\frac{N_{s}}{N_{p}}=\frac{I_{p}}{I_{s}}\) = k.

Question 1.
The figure shows how the reactance of an Inductor varies with frequency. (a) Calculate the value of the Inductance of the Inductor using Information given In the graph. (b) If this Inductor is connected In senes to a resistor of 8 ohms, find what would be the impedance at 300 Hz.
Class 12 Physics Important Questions Chapter 7 Alternating Current 60
Answer:
(a) We know that XL = 2πfL or L = \(\frac{X_{L}}{2 \pi f}\).
Now slope of the graph is
\(\frac{X_{L}}{f}=\frac{8-6}{400-300}=\frac{2}{100}\) = 0.02

Therefore L is L = \(\frac{X_{L}}{2 \pi f}=\frac{0.02}{2 \times 3.14}\) = 0.0032 H

(b) NowR = 8 ohm, f = 300 Hz, Z = ?
Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\) . Therefore we have
Z = \(\sqrt{R^{2}+X_{L}^{2}}\) = \(\sqrt{(8)^{2}+(6)^{2}}\) = 10 Ω

Question 2.
A 25.0 μF capacitor, a 0.10-henry Inductor, and a 25.0-ohm resistor are connected in series with an ac source whose emf is E= 310 s. In 314 t (i) what is the frequency of the .mf? (ii) Calculate (a) the reactance of the circuit (b) the Impedance of the circuit and (C) the current in the circuit.
Answer:
Given C = 25.0 μF, L = 0.10 henry, R = 25.0 ohm, Eo = 310V,
Comparing with the equation E = Eo sin ωt,
we have
(i) ω = 314 or f = 50 Hz
Class 12 Physics Important Questions Chapter 7 Alternating Current 61

Question 3.
A sinusoidal voltage V = 200 sin 314 t Is applied to a resistor of 10 ohms. Calculate (i) rms value of current (ii) rms value of voltage and (iii) power dissipated as heat in watt.
Answer:
Vo = 200 V, ω = 314 rads-1 , Vrms = ?, lrms = ?, P = ?
Class 12 Physics Important Questions Chapter 7 Alternating Current 62

Question 4.
Find the inductance of the inductor used in series with a bulb of resistance 10 ohms connected to an ac source of 80 V, 50 Hz. The power factor of the circuit is 0.5. Also, calculate the power dissipation in the circuit.
Answer:
Given R = 10 ohm, V = 80 Hz, f = 50 Hz, cos Φ = 0.5 , L = ?, P = ?
Using the formula
cos Φ = \(\frac{R}{\sqrt{R^{2}+X_{L}^{2}}}\)
or
0.25(R2 + X2) = R2
Or
O.25XL2 = O.75R2
or
XL = 17.32 ohm

Now using XL = 2πfL we have
L = \(\frac{X_{L}}{2 \pi f}=\frac{17.32}{2 \times 3.14 \times 50}\) = 0.055 H

Now Z = \(\sqrt{R^{2}+X_{L}^{2}}\)
= \(\sqrt{(10)^{2}+(17.32)^{2}}\) = 20 Ω

Now Pav = lrms Vrms cos Φ = \(\frac{V_{m}^{2}}{Z}\) × cos Φ
or
Pav = \(\frac{(80)^{2}}{2 \times 20}\) × 0.5 = 160 W

Question 5.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and Is in phase with the applied voltage. When the same voltage is applied across a device Y, the same current again flows through it, but it leads the voltage by π/2. If element ‘X’ is a pure resistor of 100 ohms,
(a) name the circuit element ‘Y’ and
Answer:
The element Y is a capacitor.

(b) calculate the rms value of current, if rms value of voltage is 141 V.
Answer:
The value of Xc is obtained as below

XC = \(\frac{V}{I}=\frac{220}{0.5}\) = 440 ohm

Therefore impedance of the circuit
Z = \(\sqrt{R^{2}+X_{C}^{2}}=\sqrt{(100)^{2}+(440)^{2}}\) = 451.2 ohm

Therefore rms value of current V 141
l = \(\frac{V_{r m s}}{Z}=\frac{141}{451.2}\) = 0.3125 A

Question 6.
When an alternating voltage of 220 V is applied across a device X, a current of 0. 5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across a device V, the same current again flows through it, but it lags the voltage by π/2.
(a) Name the devices X and Y.
Answer:
The element X is a resistor and Y is an inductor.

(b) Calculate the current flowing through the circuit when the same voltage is applied across the series combination of the two devices X and Y.
Answer:
Now both R and XL are the same and are given by
R = XL = \(\frac{220}{0.5}\) = 440 ohm

Hence impedance of the circuit
Z = \(\sqrt{R^{2}+X_{L}^{2}}\)
= \(\sqrt{(440)^{2}+(440)^{2}}\)
= 622.2 ohm

Therefore current flowing through the circuit is
l = \(\frac{V}{Z}\) = \(\frac{220}{622.2}\) = 0.353 A

Question 7.
A 15.0 μF capacitor Is connected to a 220 V, 50 Hz source. Find the capacitive reactance and the current (rms and peak) In the circuit. If the frequency Is doubled, what happens to the capacitive reactance and the current?
Answer:
Given C= 15.0 μF= 15 × 10-6 F, V= 220 V,
f = 50 HZ, XC = ? lm = ?

The capacitive reactance is
Class 12 Physics Important Questions Chapter 7 Alternating Current 63
Now lrms = \(\frac{V_{r m s}}{X_{c}}=\frac{220}{212}\) = 1.04 A

Peak value of current
lm = \(\sqrt{2}\) × lrms = 1.4.1 × 1.04 = 1.47 A

This current oscillates between + 1.47A and – 1.47 A, and is ahead of the voltage by π/2.

If the frequency is doubled, the capacitive reactance is halved and consequently, the current is doubled.

Question 8.
A sinusoidal voltage of peak value 283 V and frequency 50 Hz is applied to a series LCR circuit in which R = 3 Ω, L = 25.48 mH and C = 796 μF. Find
(a) the impedance of the circuit;
(b) the phase difference between the voltage across the source and the current;
(c) the power dissipated in the circuit; and
(d) the power factor.
Answer:
Given Vm = 283 V, f = 50 Hz, R = 3 Ω
L = 25.48 mH, and C = 796 μF.
(a) To find the impedance of the circuit, we first calculate XL and Xc
XL = 2πfL = 2 × 3.14 × 50 × 25.48 × 10-3 = 8 Ω
XC = \(\frac{1}{2πfC}\)
= \(\frac{1}{2 \times 3.14 \times 50 \times 796 \times 10^{-6}}\) = 4 Ω

Therefore impedance of the circuit is
Z = \(\sqrt{R^{2}+\left(X_{\mathrm{L}}-X_{C}\right)^{2}}\)
= \(\sqrt{3^{2}+(8-4)^{2}}\)
= 5 Ω

(b) Phase difference
Φ = tan-1\(\frac{X_{L}-X_{c}}{R}\) = tan-1 \(\frac{8-4}{3}\) = 53.1°
Since Φ is positive therefore voltage leads current by the above phase.

(c) The power dissipated in the circuit is
P = \(\frac{V_{\mathrm{rms}}^{2}}{Z}=\frac{V_{\mathrm{m}}^{2}}{2 \mathrm{Z}}=\frac{(283)^{2}}{2 \times 5}\) = 8008.9 W
(d) power factor = cos Φ = cos 53.1° = 0.6

Question 9.
A capacitor and a resistor are connected in series with an ac source. If the potential difference across the C, R is 120 V and 90 V respectively, and if rms current of the circuit is 3 A, calculate the (i) impedance and (ii) power factor of the circuit.
Answer:
Given VC = 120 V, VR = 90 V, f = 3 A. and R = 90/3 = 30 ohm ,

Effective voltage in the circuit
V = \(\sqrt{V_{C}^{2}+V_{R}^{2}}=\sqrt{(120)^{2}+(90)^{2}}\)= 150 V .
(i) Therefore impedance of the circuit
Z = \(\frac{V}{l}=\frac{150}{3}\) = 50 Ω.

(ii) Now power factor of the circuit is
cos Φ = \(\frac{R}{Z}=\frac{30}{50}\) = 0.6

Question 10.
An inductor 200 mH, a capacitor C, and a resistor 10 ohm are connected in series with 100 V, 50 Hz ac source. If the current and the voltage are in phase with each other, calculate the capacitance of the capacitor.
Answer:
When current and voltage are in phase then XL = XC
Class 12 Physics Important Questions Chapter 7 Alternating Current 64

Question 11.
An ac voltage of 100V, 50 Hz is connected across a 20-ohm resistor and 2 mH inductors in series.
Calculate (i) impedance of the circuit, and
(ii) rms current in the circuit. (NCERT)
Answer:
Given, V = 100 V, f = 50 Hz, R = 20 ohm, L = 2 mH = 2 × 10-3 H, Z = ?, lrms = ?
Using the relation
(a) Z = \(\sqrt{(R)^{2}+\left(X_{L}\right)^{2}}=\sqrt{(R)^{2}+(2 \pi f L)^{2}}\)
Z = \(\sqrt{(20)^{2}+\left(2 \times 3.14 \times 50 \times 2 \times 10^{-3}\right)^{2}}\)
Z = 20 ohm
(b) lrms= V/Z= 100/20 = 5 A

Question 12.
A light bulb is rated at 100 W for a 220 V supply. Find (a) the resistance of the. bulb; (b) the peak voltage of the source; and (c) the rms current through the bulb. (NCERT)
Answer:
(a) We are given P = 100 W and V = 220 V.
The resistance of the bulb is
R = \(\frac{V^{2}}{P}=\frac{(220)^{2}}{100}\) = 484 Ω

(b) The peak voltage of the source is
Vm = \(\sqrt{2}\)V = \(\sqrt{2}\) × 220 = 311V

(c) Since P = I V, therefore l = P/V= 100/220 = 0.45 A

Question 13.
A pure Inductor of 25.0 mH is connected to a source of 220 V. Find the inductive reactance and rms current in the circuit if the frequency of the source is 50 Hz. (NCERT)
Answer:
The inductive reactance,
XL = 2 πf L = 2 × 3.14 × 50 × 25 × 10-3
= 7.85 ohm

The rms current in the circuit is
l = \(\frac{V}{X_{L}}=\frac{220}{7.85}\) = 28 A

Question 14.
A 100 Ω resistor is connected to a 220 V, 50 Hz ac supply.
(a) What is the rms value of current in the circuit?
(b) What is the net power consumed over a full cycle? (NCERT)
Answer :
Given urms = 220 V, f = 50 Hz, lrms = ?, P = ?
(a) lrms = \(\frac{V_{\mathrm{rms}}}{R}=\frac{220}{100}\) = 2.2 A

(b) Power consumed over one cycle of ac P = Vrms × lrms = 220 × 2.2 = 484 W

Question 15.
(a) The peak voltage of an ac supply is 300 V. What is the rms voltage?
(b) The rms value of current in an ac circuit is 10 A. What is the peak current? (NCERT)
Answer:
(a) Vm = 300V, Vrms =?

Using Vrms = \(\frac{V_{\mathrm{m}}}{\sqrt{2}}=\frac{300}{\sqrt{2}}\) = 212.16 V

(b) Using
lm = lrms × V2 = 10 × \(\sqrt{2}\) = 14.14 A

Question 16.
A 44 mH inductor is connected to 220 V, 50 Hz ac supply. Determine the rms value of the current in the circuit. (NCERT)
Answer:
Given L = 44 mH = 44 × 10-3 H, Vrms = 220 V, f = 50Hz,lrms = ?
Using the relation
lrms = \(\frac{V_{\text {rms }}}{x_{L}}=\frac{V_{\text {rms }}}{2 \pi f L}=\frac{220}{2 \times 3.14 \times 50 \times 44 \times 10^{-3}}\)
= 15.9 A

Question 17.
A coil of inductance 0.50 H and resistance 100 Ω is connected to a 240 V, 50 Hz ac supply, (a) What is the maximum current in the coil? (b) What Is the time lag between the voltage maximum and the current maximum? (NCERT)
Answer:
Given L = 0.5 H, R = 100 Ω, Vrms = 240 V,
f= 50 Hz, ω = 2πf= 2 × 3.14 × 50 = 314 s-1 lm = ?
(a) Now
Z = \(\sqrt{R^{2}+X_{L}^{2}}=\sqrt{(100)^{2}+(0.5 \times 314)^{2}}\)
Z = 186.1 Ω

Therefore
l<sub>m</sub> = \(\frac{V_{m}}{Z}=\frac{\sqrt{2} \times V_{\mathrm{rms}}}{Z}=\frac{\sqrt{2} \times 240}{186.1}\)
lm = 1.82 A

(b) Let Φ be the phase angle by which current lags emf, then
tan Φ = \(\frac{X_{L}}{R}=\frac{0.5 \times 314}{100}\) = 1.571
Φ = 57.5° or Φ = 57.5 × π/180 rad

Therefore timw lag
t = \(\frac{\phi T}{2 \pi}=\frac{57.5 \times \pi}{180 \times 2 \pi \times 50}\) = 3.194 × 10-3 s

Question 18.
A coil of 0.01-henry Inductance and 1-ohm resistance is connected to 200 volts, 50 Hz ac supply. Find the Impedance of the circuit and time lag between max alternating voltage and current. (NCERT Exemplar)
Answer:
Given L = 0.01 henry, R = 1 ohm,
Vrms = 200V, f = 50Hz, Z = ?, t = ?
We know that
XL = 2πfL = 2 × 3.14 × 50 × 0.01 = 3.14 ohm
Now
Z = \(\sqrt{R^{2}+X_{\mathrm{L}}^{2}}=\sqrt{(3.14)^{2}+(1)^{2}}\) = 3.3 Ω
Also tan Φ = \(\frac{X_{L}}{R}=\frac{3.14}{1}\)= 3.14
Or
Φ = tan-1 (3.14)
Or
Φ = 72° or 72 × π/180 rad
Now time lag Δt is
Δt = \(\frac{\phi}{\omega}=\frac{72 \times \pi}{180 \times 2 \pi \times 50}=\frac{1}{250}\) s

Question 19.
An electrical device draws 2 kW power from AC mains (voltage 223 V (rms) = V50,000 V). The current differs (lags) in phase by Φ (tan Φ = – 3/4) as compared to voltage.
Find (i)R,
(ii) XC – XL
(iii) lm
Another device has twice the values for R, XC, XL. How are the answers affected? (NCERT Exemplar)
Answer:
Given P = 2 kW = 2000 W, Vrms = 223
= \(\sqrt{50,000}\) V, tan Φ = – 3/4, R = ?,XC – XL = ?, lm = ?

(i)Class 12 Physics Important Questions Chapter 7 Alternating Current 65
Solving for R we have
R = 20 ohm

(ii) Also XC – XL = (-3/4) R = – 15 Ω
Hence lrms =\(\frac{V}{Z}=\frac{223}{25}\) = 9 A
(iii) lm = \(\sqrt{2}\) × lrms = 1.414 × 9 = 12.6 A

If R, XC, XL, are all doubled, tan Φ does not change. Therefore as Z is doubled, the current becomes halved.

Question 20.
An alternating voltage given by V = 140 sin 314 t is connected across a pure resistor of 50 Ω. Find
(a) the frequency of the source.
(b) the rms current through the resistor. (CBSEAI 2012)
Answer:
R = 50 Ω, f = ?, lrms = ?, Vm = 140 V
Comparing with V = Vm sin 2πft,
we have
2πft = 314 t
or
f = 50 Hz
lrms = \(\frac{V_{r m s}}{R}=\frac{140}{50 \sqrt{2}}\) = 1.98 A

Question 21.
The figure shows a series LCR circuit with L = 5.0 H, C = 80 μF, R = 40 Ω connected to a variable frequency 240 V source. Calculate
Class 12 Physics Important Questions Chapter 7 Alternating Current 66
(a) The angular frequency of the source drives the circuit at resonance.
(b) The current at the resonating frequency.
(C) The rms potential drops across the capacitor at resonance. (CBSE Delhi 2012)
Answer:
Given L= 5.0 H, C= 80 μF , R = 40 Ω, Vrms = 240 V ωo = ?, lrms = ?, VC = ?
(a)Angutar frequency at resonance
ω0 = \(\sqrt{\frac{1}{L C}}=\sqrt{\frac{1}{5 \times 80 \times 10^{-6}}}\) = 50 s-1

(b)At resonance Z = R = 4O Ω
Now lrms = \(\frac{V_{\mathrm{rms}}}{\mathrm{Z}}=\frac{240}{40}\) = 6 A

(C) VC = lrms × XC = 6 (50 × 80 × 10-6) = 1500 V

Question 22.
The figure shows a series LCR circuit connected to a variable frequency 200 V source with L = 50 mH, C = 80 μF and R = 40 Ω.
Determine
(a) the source frequency which derives the circuit in resonance;
(b) the quality factor (Q) of the circuit. (CBSEAI 2014C)
Class 12 Physics Important Questions Chapter 7 Alternating Current 67
Answer:
(a) Given V = 200 V, L = 50 mH, C = 80 μF, R = 40 Ω
Source frequency at resonance
Class 12 Physics Important Questions Chapter 7 Alternating Current 68

Question 23.
When an alternating voltage of 220 V is applied across a device X, a current of 0.5 A flows through the circuit and is in phase with the applied voltage. When the same voltage is applied across another device Y, the same current again flows through it, but it leads the voltage by π/2 rattan.
(a) Name the devices X and Y.
(b) Calculate the current flowing In the circuit when the same voltage Is applied across the series combination of the two devices X and Y.
Answer:
(a) The two devices are resistor and capacitor, respectively.

(b) Given l = 0.5 A, V = 220 V, XL = ?, R = ? , l = ?
Now XC and R are equal and are given by
XC = R = \(\frac{220}{0.5}\) = 440 Ω

Hence impedance of the circuit
Z = \(\sqrt{R^{2}+X_{C}^{2}}=\sqrt{2(440)^{2}}\) = 622.25 Ω

Therefore current through the combination is
l = \(\frac{V}{Z}=\frac{220}{622.25}\) = 0.354 A

Question 24.
In the following circuit, calculate (a) the capacitance of the capacitor, if the power factor of the circuit is unity, (b) the Q-factor of this circuit. What is the significance of the Q-factor in a.c. circuit? Given the angular frequency of the a.c. source to be 100 s-1. Calculate the average power dissipated in the circuit. (CBSE Delhi 2017C)
Class 12 Physics Important Questions Chapter 7 Alternating Current 69
Answer:
(a) when the power factor is unity the circuit is in resonance, therefore, we have
Class 12 Physics Important Questions Chapter 7 Alternating Current 70
Quality factor determines the sharpness of tuning at resonance. More the Q factor more the sharpness.
Class 12 Physics Important Questions Chapter 7 Alternating Current 71

Question 25.
The figure below shows how the reactance of a capacitor varies with frequency.
Class 12 Physics Important Questions Chapter 7 Alternating Current 72
(a) Use the Information on the graph to calculate the value of capacity of the capacitor.
(b) An inductor of inductance L has the same reactance as the capacitor at 100 Hz. Find the value of L.
(c) Using the same axes, draw a graph of reactance against frequency for the inductor.
(d) If this capacitor and inductor were connected in series to a resistor of 10 Ω, what would be the impedance of the combination at 300 Hz?
Answer:
(a) From the graph we find that for a frequency of 100 Hz the capacitive reactance is 6 ohm, therefore
Class 12 Physics Important Questions Chapter 7 Alternating Current 73
(b) Now given XL = XC at 100 Hz, therefore XL = 6 ohm
Class 12 Physics Important Questions Chapter 7 Alternating Current 74
(c) The XL – f graph is as shown below.
Class 12 Physics Important Questions Chapter 7 Alternating Current 75
(d) For a frequency of 300 Hz, XL = 18 ohm and XC = 2 ohm. If resistance is 10 ohm then the impedance of the combination is
Class 12 Physics Important Questions Chapter 7 Alternating Current 76

Question 26.
The given graphs (i) and (ii) represent the variation of the opposition offered by the circuit element to the flow of alternating current, with the frequency of the applied emf. Identify the circuit element corresponding to each graph.
Class 12 Physics Important Questions Chapter 7 Alternating Current 77
A circuit is set up by connecting L= 100 mH, C = 5 μf and R =100 Ω in series. An alternating emf of (150 \(\sqrt{2}\)) volt, (500/π) Hz is applied across this series combination. Calculate the impedance of the circuit. What is the average power dissipated in (a) the resistor, (b) the capacitor, (c) the inductor, and (d) the complete circuit?
Answer:
(a) Resistor and (b) Inductor
Given L = 100 mH = 0.1 H, C = 5μF = 5 × 10-6F, R=100 Ω,
Class 12 Physics Important Questions Chapter 7 Alternating Current 78
Average power dissipated in
(a) the resistor P = l2R = (1.5)2 × 100 = 225 W
(b) power in inductor is zero.
(c) power in capacitor is zero.
(d) the complete circuit is same as that in a resistor.

Question 27.
The output voltage of an ideal transformer, connected to a 240 V ac mains, is 24 V. When this transformer is used to light a bulb with rating 24 V, 24 W, calculate the current in the primary coil of the circuit.
Answer:
Class 12 Physics Important Questions Chapter 7 Alternating Current 79

Question 28.
An ac generator consists of a coil of 50 turns and an area of 2.5 m2 rotating at an angular speed of 60 rads-1 in a uniform magnetic field B = 0.2 tesla, between the two fixed pole pieces. The resistance of the circuit including that of the coil is 500 ohm (a) Calculate the maximum current drawn from the generator. (b) What is the flux when the current is zero and (c) Would the generator work if the coil were stationary and instead the pole pieces rotated together with the same speed as above? Give a reason for your answer.
Answer:
Given n = 50, A = 2.5 m2, ω = 60 rads-1, B = 0.2 T, R = 500 ohm, l0 = ?
(a) The maximum induced emf produced in the coil is
ε0 = nBAω = 50 × 0.2 × 2.5 × 60 = 1500 V

Therefore maximum current in the circuit is
l0 = \(\frac{\varepsilon_{0}}{R}=\frac{1500}{500}\) = 3 A

(b) The flux in this case is maximum and is given by Φ = nBA = 50 × 0.2 × 2.5 = 25 Wb

(c) Yes, for a generator to work there should be relative motion between the magnet and the coil.

Question 29.
The primary coil of an ideal step-up transformer has 100 turns and the transformation ratio is also 100. The input voltage and the powers are 220 V and 1100 W respectively. Calculate:
(a) number of turns in the secondary
(b) the current In the primary
(C) the voltage across the secondary
(d) the current in the secondary
(e) power in the secondary
Answer:
Here, N = 100, Transformation ratio = 100
Vp = 220 V, Pp = 1100W
(a) Number of turns in the secondary
Ns = Np × Transformation ratio
= 100 × 100 = 10,000

(b) Current in the primary
lp = \(\frac{P_{p}}{V_{p}}=\frac{1100}{220}\) = 5 A

(C) Voltage across the secondary Vs = Vp × transformation ratio
= 220 × 1oo = 22000 V

(d) Current in the secondary
ls = \(\frac{V_{p} l_{p}}{V_{s}}=\frac{220 \times 5}{22000}\) = 0.05 A

(e) Power in the secondary = power in the primary (ideal transformer) = 1100 W

Question 30.
Calculate the current drawn by the primary of a transformer which steps down 200 V to 20 V to operate a device of resistance 20 Q. Assume the efficiency of the transformer to be 80 %.
Answer:
Class 12 Physics Important Questions Chapter 7 Alternating Current 80

Question 31.
The primary of a transformer has 200 turns and the secondary has 1000 turns. If the power output from the transformer at 1000 V is 9 kW, calculate
(a) The primary voltage and
(b) The heat loss in the primary coil if the resistance of the primary is 0.2 Ω and the efficiency of the transformer is 90%.
Answer:
Class 12 Physics Important Questions Chapter 7 Alternating Current 81

Question 32.
A power transmission line feeds input power at 2300 V to a step-down transformer with its primary windings having 4000 turns. What should be the number of turns in the secondary In order to get output power at 230 V? (NCERT)
Answer:
Class 12 Physics Important Questions Chapter 7 Alternating Current 82

Question 33.
A small town with a demand of 800 kW of electric power at 220 V situated 15 km away from an electric plant generating power at 440 V The resistance of the two wirelines carrying power is 0.5 Ω per km. The town gets power from the line through a 4000 – 220 V step-down transformer at a sub-station in the town.
(a) Estimate the line power loss in the form of heat.
(b) How much power must the plant supply, assuming their negligible power loss due to leakage?
(c) Characterise the step-up transformer at the plant. (NCERT)
Answer:
Electric power required P = 800 kW = 800 × 103 W
Voltage required V0 = 220 V

Distance of the power station
d = 15km = 15 × 103m

Resistance per kilometre = 0.5 ohm per km
Total resistance R = 0.5 × 2 × 15 = 15 ohm

Input voltage Vi = 440 V

Primary voltage of transformer εp = 4000 V
Secondary voltage of transformer εs = 220 V
(a) Rms value of current in the two wire line
Class 12 Physics Important Questions Chapter 7 Alternating Current 83

Therefore power loss along the line
= l2rmsR = (200)2 × 15 = 600 kW

(b) Assuming negligible loss due to leakage Total power supply = power demand of town + power loss along the line
P = 800 kW + 600 kW = 1400 kW

(c) Voltage drop on the line = 200 × 15 = 3000 V
Hence voltage at the generation station
V= 4000 + 3000 = 7000 V

Therefore the step-up transformer at the plant is 440 V – 7000 V.

Question 34.
A 60 W load is connected to the secondary of a transformer whose primary draws line voltage. If a current of 0.54 A flows in the load, what is the current in the primary coil? Comment on the type of transformer being used. (NCERT Exemplar)
Answer:
Given Ps = 60 W, ls = 0.54 A
We know that P=V I, therefore we have Vs = Ps/ls = 60 /0.54 = 110 V

Therefore the transformer is a step-down transformer, whose transformation ratio is 1/2
Therefore
lp = 1/2 × ls = 1/2 × 0.54 = 0.27 A