Category: CBSE Class 12

NCERT Class 12 Chemistry Solutions for Chapter 11 provides an insight into the various concepts related to alcohols, phenols and ethers. This is an important chapter and hence requires an indepth knowledge of the topics. The subject experts have provided accurate explanations and step wise solutions for the questions provided in the textbook. This will help the students prepare well during the exams.

NCERT Solutions not only help the students appearing for UP board, MP board, CBSE, Maharashtra board, Gujarat board, etc. but also prepares them for competitive exams such as JEE and NEET. The students should refer to the NCERT Solutions to score well in the examinations.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Chemistry
Chapter	Chapter 11
Chapter Name	Alcohols, Phenols and Ehers
Number of Questions Solved	45
Category	NCERT Solutions

NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers

Alcohols, phenols, and ethers is an important chapter from the examination perspective. The chapter explains the structure, properties, and applications of alcohols, phenols, and ethers. It also explains the correlation and differences between the three.

NCERT Solutions for Class 12 Chemistry Chapter 11 gives a better idea of the related concepts. The students can refer to these for better practice.

NCERT INTEXT QUESTIONS

Question 1.
Classify the following into primary, secondary, and tertiary alcohols

Answer:
(i) Primary alcohol
(ii) Primary alcohol
(iii) Primary alcohol
(iv) Secondary alcohol
(v) Secondary alcohol
(vi) Tertiary alcohol

Question 2.
Identify allylic alcohol in the above examples.
Answer:
allylic alcohols:

Question 3.
Give the IUPAC names of the following compounds : (C.B.S.E 2008)

Answer:

Question 4.
Show how the following alcohols can be prepared by the action of suitable Grignard reagent on methanal ?

Answer:
(i) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is isopropyl magnesium halide.

(ii) The structure of alcohol suggests that the Grignard reagent that reacts with methanal is cyclohexyl magnesium halide.

Question 5.
Write the structures of the products of the following reactions :

Answer:
(i) The acidic hydration of propene gives propan-2-ol (isopropyl alcohol)

(ii) NaBH₄ is a weak reducing agent. It brings about the reduction of the ketonic group present in cyclohexane ring to secondary alcoholic group. However, it does not affect ester group.

(iii) NaBH₄ reduces aldehydic group to a primary alcoholic group.

Question 6.
Give structures of the products you would expect when each of the following alcohols reacts with (a) HCl/ZnCl₂ (b) HBr (c) SOCl₂ :
(i) Butan-l-ol
(ii) 2-Methylbutan-2-ol
Answer:
(i) Reactions of Butan-1-ol

The reaction of butan-l-ol (primary alcohol) can take place only upon heating. At room temperature, the reaction does not occur.

(ii) Reactions of 2-Methylbutan-2-ol

The alcohol being tertiary in nature reacts immediately with Lucas Reagent at room temperature.

Question 7.
Predict the major product of add catalysed dehydration of :
(i) 1-Methylcyclohexanol
(ii) Butan-1-ol
Answer:
(i) Acid catalysed dehydration of 1-methylcyclohexanol can give two products. However, 1-methylcyclohexene will be  preferably formed according to Satyzeff s rule since it is more substituted.

(ii) Butan-1-o1 upon acid dehydration will give but-2-ene as the major product along with but-l-ene as the minor product. Actually, primary carbocation formed can either lose a H⁺ to form but-1-ene or may undergo rearrangement and shift to secondary carbocation, which is more stable. The latter then forms but-2-ene by losing a H⁺.

Question 8.
Ortho and para nitrophenols are more acidic than phenol. Draw the resonating structures of the corresponding phenoxide ions.
Answer:
We know that the nitrophenols are more acidic than phenol. Their acidic strength can be compared in terms of the relative stabilities of the corresponding phenoxide ions based on resonance. For example,
(i) Phenoxide ion :


(ii) p-Nitrophenoxide ion :


(iii) p-Nitrophenoxide ion :

In case of nitrosubstituted phenoxides, the contributing structures that are enclosed in boxes have negative charge on the carbon atom to which die electron withdrawing nitro group is attached. They therefore, contribute more towards the acidic character than the rest of the contributing structures. Consequently, both ortho and para nitrophenol are stronger acids than phenol.

Question 9.
Write the equations involved in the following reactions :
(i) Reimer Tiemann Reaction
(ii) Kolbe’s Reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachioride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCI when salicylic acid is formed. This method is commonly used for the commercial preparation of saucy lic acid.

Question 10.
Write the reactions of Williamson’s synthesis of 2-ethoxy-3-methoxypentane starting from ethanol and 3-methylpentan- 2-ol.
Answer:
In the Williamson’s synthesis, the reactants are alkyl halide and sodium salt of an alcohol. In order to avoid the formation of alkene during the reaction, the alkyl halide should be primary while sodium salt must be of branched chain alcohol. In the present case, alkyl halide must be derived from ethanol upon heating with halogen acid (e.g HBr).

Question 11.
Which of the following is an appropriate set of reactants for the preparation of l-methoxy-4-nitrobenzene and why ?

Answer:
The second set of reactants is more appropriate to give the products i.e., l-methoxy-4-nitrobenzene. In the first set, cleavage of C – Br bond is involved. It is rather difficult since the carbon atom is sp² hybridised and the bond has partial double bond character as well. The product is formed as a result of Williamson’s synthesis.

Question 12.
Predict the products of the following reactions :

Answer:
(i) The reaction involves the cleavage of C – 0 bond. The Br atom of HBr is to combine with the smaller alkyl group to give the following products.

For more details and explanation, consult section 12.24.
(ii) This reaction also proceeds in the same manner. The Br atom of HBr is expected to combine with ethyl group (smaller in size) and not with phenyl group (bigger in size).

(iii) Nitrating mixture brings about the nitration of benzene ring. The ethoxy group (OC₂H₅) is an activating group and increases the electron density at the ortho and para positions due to +M effect. As a result, a mixture of o-nitro and p-nitro derivatives is formed. Out of these, the p-isomer is in excess since there is less steric hindrance due to OC₂H₅ group at the para position than at the ortho position in the ring.

(iv) In this reaction, the ether is initially protonated by H⁺ ion of the acid HI. to accomodate I^– ion (nucleophile). The reaction follows S_n1 mechanism.

NCERT EXERCISE

Question 1.
Write IUPAC names of the following compounds :

Answer:
(i) 2, 2, 4 – Trimethylpentan-3-ol
(ii) 5 – Ethylheptan-2, 4-diol
(iii) Butane – 2, 3, diol
(iv) Propane – 1, 2, 3-triol
(v) 2 – Methylphenol
(vi) 4 – Methylphenol
(vii) 2, 5 – Dimethylphenol
(viii) 2, 6 – Dimethyiphenol
(ix) 1 -Methoxy – 2 – methy lpropane
(x) Ethoxybenzene
(xi) 1 – Phenoxyheptane
(xii) 2 – Ethoxybutane.

Question 2.
Write the structure of the compounds, whose IUPAC names are as follows
(i) 2-Methylbutan – 2 – ol
(ii) I-Phenylpropan – 2 – ol
(iii) 3-Phenylhexane – l, 3, 5 – triol
(iv) 2, 3 – Diethylphenol
(v) 1 – Ethoxypropane
(vi) 2- Ethoxy – 3- methylpentane
(vii) Cyclohexylmethanol
(viii) 3-Cyclohexylpentan – 3 – ol
(ix) Cyclopent – 3 – en –  – ol
(x) 3 – Chloromethvlpentan – l – ol
Answer:

Question 3.
(i) Draw the structures of all isomeric alcohols of molecular formula C₅HI₂0 and give their IUPAC names.
(ii) Classify the isomers of alcohols in question 11.3 (i)as primary, secondary, and tertiary alcohols.
Answer:
Eight isomers are possible. These are:

Question 4.
Explain why is propanol higher boiling than butane?
Answer:
Propanol (Propan-l-ol) and butane are of comparable molecular masses 60m and 58u respectively but the boiling point of propanol is higher (391 K) because of the presence of intermolecular hydrogen bonding in the molecules. However, it is not present in butane due to the absence of polar OH group. The only attractive forces are weak van der Waals forces. Therefore, the boiling point of propanol is more than that of butane (309 K).

Question 5:
Explain why are alcohols more soluble in water than the corresponding hydrocarbons?
Answer:
The solubility of alcohols in water may be attributed to two factors :
(i) Both of them are of polar nature.
(ii) Molecules of both of them are involved in the intermolecular hydrogen bonding.
However, hydrocarbons are non-polar and are also not involved in any hydrogen bonding with alcohols. Alcohols readily dissolve in a water while the hydrocarbons are almost insoluble.

Question 6.
What is meant by hydroboration oxidation reaction ? Illustrate with an example.
Answer:
By hydroboration- oxidation of alkenes. Indirect hydration of alkenes can also bedone by hvdroboration-oxidation which is completed in two steps. In the first step. alkene reacts ith diborane (B’1{6) as boron hydride (BH3) dissolved in tetrahydrofuran (THF) to form an alkyl horane. In fact. the boron atom along i th the hydrogen gets attached to the double
bonded carbon atom with more number of hydrogen atoms less sterically hindered side). One hydrogen is then transferred to the other carbon atom. In this manner, all the three hydrogen atoms of boron are transferred to alkene molecule to form
trialkyl borane as the product. In the next step. the alkyl borane is oxidised by alkaline 11202 to form an alcohol. The indirect hydration proceeds according to Antimarkownikov s rule. For example.

Question 7.
Give the structures and IUPAC names of the phenols of molecular formula C₇H₈O.
Answer:

Question 8.
In separating a mixture of ortho and para nitrophenols by steam distillation, name the isomer which is steam volatile? Give reason.
Answer:
o-nitrophenol is steam volatile while p-nitrophenol is not. This is on account of intramolecular hydrogen bonding in the molecules of o-nitrophenol. As a result, its boiling point is less than that of p-nitrophenol in which the molecules are linked by intermolecular hydrogen bonding.
It is interesting to note that in the substituted phenols, the nature and position of the substituent influences the boiling point of phenol.
For example .o-nitrophenol is steam volatile while p-nitrophenol is not. This is supported by the fact that the boiling point temperature of o-nitrophenol (100°C) is less than that of p-nitrophenol, (279°C). In o-nitrophenol, there is intramolecular hydrogen bonding in OH and NO₂ groups placed in a adjacent positions. However, these are linked by intermolecular hydrogen bonding in the p-isomers. It is quite obvious that extra energy is needed to cleave the hydrogen bonds in the p-isomer. Consequently, its boiling point is more.

o-nitrophenol with lower boiling point is steam volatile while p-nitrophenol is not. This helps in the separation of the two isomers present in the liquid mixture. On passing steam, o-nitrophenol volatilises, and its vapours rise alongwith steam and after condensation, collect in the receiver p-nitrophenol is left behind in the distillation flask. e-nkrophenol p-nnrophenol.

Question 9.
Give the equations of reactions for the preparation of phenol from cumene.
Answer:

Question 10.
Write chemical equations involved in the preparation of phenol from chlorobenzene.
Answer:
From chlorobenzene, phenol is prepared by Dow’s process.In this method, chlorobenzene is heated with 6 to 8% solution of NaOH to about 623 K under a pressure of 300 atmospheres to form sodium phenate which upon acidification with dilute HCI gives phenol as follows:

Question 11.
Write the mechanism of hydration of ethene to yield ethanol.
Answer:
Ethene does not react with water as such. Water being little polar, is not in a position to provide H⁺ ion for initial electrophilic attack on ethene. The reaction is carried in the presence of H₂S0₄. The acid provides proton (H⁺) for the initial electrophilic attack.

In the second step, H₂0 attacks the carbocation in preference to HS0₄ ion as a nucleophile

Question 12.
You are given benzene, cone. H₂S0₄ and NaOH. Write equations for the preparation of phenol using these reagents.
Answer:

Question 13.
Show how well you synthesise:
(i) 1-phenyl ethanol from a suitable alkene
(ii) Cyclohexylmethanol using an alkyl halide by S_N² reaction
(iii) Pentan-1-ol using a suitable alkyl halide.
Answer:

Question 14.
Give two reactions to show the acidic nature of phenol. Compare the acidity of phenol with that of ethanol.
Answer:
Acidic nature of phenols. Phenols are weakly acidic in nature. The liquid form of phenol containing about 5 percent water is known as carbolic acid. The dissociation constant (K_a) for phenol is 10^-10 at 298 K (room temperature). The corresponding pK_a* value is 10.0. The acidic character is exhibited by the following properties:
(i) Reaction with active metals. Phenols evolve hydrogen with active metals such as sodium and potassium.

(ii) Reaction with alkalies. Phenols neutralise caustic alkalies such as NaOH or KOH to form salt and water.

In addition to these, phenols turn blue litmus red which is the characteristic property or acids. However, phenols do not
react with either alkali metal carbonates or bicarbonates since these are quite weak acids.

Question 15.
Explain why is ortho-nitrophenol more acidic than ortho-methoxy phenol?
Answer:
Due to strong -R and – I-effect of the -NO₂ group, electron density of the O – H bond decreases and hence the loss of a proton becomes easy.

Question 16.
Explain how does -OH group attached to a carbon atom of benzene ring activates it towards electrophilic substitution ?
Answer:
The OH group exerts a +M (or + R) effect on the ring under the influence of attacking electrophile.

As a result, there is an increase in the electron density in the ring particularly at the ortho and para positions. The electrophilic substitution readily takes place at these positions when electrophile attacks.

Question 17.
Give equations for the following chemical reactions :
(i) Oxidation of propan-1-ol with alkaline KMnO₄
(ii) Reaction of bromine in CS₂ with phenol
(iii) Action of dilute HNO₃ on phenol
(iv) Treating phenol with chloroform in the presence of aqueous NaOH at 343 K.
Answer:

Question 18.
Write short notes on (i) Kolbe reaction (ii) Reimer-Tiemann reaction.
Answer:
(i) In this reaction phenol is heated with chloroform alongwith aqueous NaOH (or KOH) to about 340 K. This is followed by acidification with dilute HCl when 2 – hydroxyhenzaldehyde (salicylaldehyde) is formed as the major product.

A small amount of para isomer is also formed in the reaction. In case, chLoroform is replaced by carbon tetrachloride, then 2—hydroxybenzoic acid (salicylic acid) is formed as the main product.

(ii) In this reaction, CO₂ gas is passed through sodium phenate at 400 K under a pressure of 4 to 7 atmospheres. This is followed by acidification with dilute HCl when salicylic acid is formed. This method is commonly used for the commercial preparation of salicylic acid.

Question 19.
Write a mechanism for the acid dehydration of ethanol to ethene.
Answer:
Mechanism of dehydration. The mechanism is illustrated with ethanol which is a primary alcohol.

The relative ease of dehydration of different alcohols i.e. primary, secondary and tertiary can be further justified on the basis of the relative stabilities of the carbocations formed in the slow step. Since tertiary carbocation is maximum stable while primary is the least, the tertiary alcohols are maximum reactive while the primary are the least reactive in nature.
In other words, greater the stability of carbocation formed, more is the reactivity of the alcohol.

Question 20.
How are the following conversions carried out?
(i) Propene → Propan -2-ol
(ii) Benzyl chloride → Benzyl alcohol
(iii) Ethyl magnesium chloride → Propan-1-ol
(iv) Methyl magnesium bromide → 2-Methylpropan-2-ol
Answer:

Question 21.
Name the reagents used in the following reactions :
(i) Oxidation of primary alcohol to carboxylic acid
(ii) Oxidation of primary alcohol to an aldehyde
(iii) Bromination of phenol to 2, 4, 6-tribromophenol
(iv) Benzyl alcohol to benzoic acid
(v) Dehydration of propan-2-ol to propene
(vi) Butan-2-one to butan-2-ol.
Answer:
(i) Acidified K₂Cr₂O₇
(ii) Pyridine chlorochromate (PCC)
(iii) Bromine water (Br₂/H₂O)
(iv) Alkaline KMnO₄
(v) 60% H₂S0₄ at 373 K
(vi) LiAlH₄ or NaBH₄.

Question 22.
Give a reason for the higher boiling point of ethanol in comparison to methoxymethane.
Answer:
Ethers are isomeric with alcohols but their boiling points are comparatively low due to the lack of hydrogen bonding. For example, boiling points of isomeric n – butyl alcohol (nC₄H₉OH) and diethyl ether (C₂H₅ – O – C₂H₅) are, 390 K and 308 K respectively.

Question 23:
Give the IUPAC names of the following ethers :


Answer:
(i) 1-Methoxy-2-methylpropane
(ii) 1-Chloro-2-methoxy ethane
(iii) 4-Nitroanisole
(iv) 1-Methoxypropane
(v) 4-Ethoxy-1, 1-dimethyl cyclohexane
(vi) Ethoxybenzene

Question 24.
Write the names of reagents and equations for the preparation of the following ethers by Williamson’s synthesis:
(i) 1-Propoxypropane
(ii) Ethoxybenzene
(iii) 2-Methoxy-2-methylpropane
(iv) 1-Methoxyethane
Answer:

Question 25.
Illustrate with examples the limitations of Williamson’s synthesis for the preparation of certain types of ethers.
Answer:
Preparation from Alkyl Halides
From alkyl halides, ethers can be prepared by the following methods
By Williamson’s synthesis. It is the best method for the laboratory preparation of both simple and mixed ethers and involves the action of sodium alkoxide (formed by reaction between alcohol and sodium metal) on a suitable alkyl halide.

Limitations of the reaction. In the preparation of unsymmetrical ethers, a proper choice of the reactants is necessary.
Elimination leading to alkene can take place since alkoxide ion can also abstract one of the 3—hydrogen atom alongwith acting as a nucleophile. Thus, in order ro prepare ethyl tertiary butyl ether, we must use ethyl halide (primary) and sodium tertiary
butoxide.

In case, the alkyl halide is tertiary and sodium ethoxide is employed, then C2H5O ion will cause the elimination of alkyl halide to form an alkene as the main product.

Since secondary and tertiary alkyl halides prefer to undergo elimination rather substitution, symmetrical ethers containing secondary and tertiary alkyl groups are obtained only in poor yields by Willamson’s synthesis. For example,

This method is also not successful for preparing aryl alkyl ethers by reacting sod. alkoxide with aryl halides because the cleavage of C – X bond is not so easy due to partial double bond character. In such cases, we must react sodium phenoxide with
alkyl halide as follows:

However, diaryl ethers (both the groups are aryl or phenyl groups) cannot be prepared since aryl halides fail to participate in the nucleophilic substitution reactions.

Question 26.
How is 1-propoxypropane synthesised from propan-1-ol? Write the mechanism of this reaction.
Answer:
(a) Williamson’s synthesis

Question 27.
Preparation of ethers by acid dehydration of secondary or tertiary alcohols is not a suitable method. Give reason.
Answer:

Question 28.
Write the equation for the reaction of HI with :
(i) 1-Propoxypropane
(ii) Methoxybenzene
(iii) Benzyl ethyl ether.
Answer:

Question 29.
Explain the fact that in aryl alkyl ethers (i) the alkoxy group activates the benzene ring towards electrophilic substitution and (ii) it directs the incoming substituents to ortho and para positions in benzene ring.
Answer:
In aryl alkyl ethers, the +R-effect of the alkoxy (OR) group increases the electron density in the benzene ring, thereby activating the benzene ring towards electrophilic substitution reaction.

Since the electron density increases more at the two ortho and one para position as compared to meta position therefore, electrophilic substitution reactions mainly occur at o-and -positions.

Question 30.
Write the mechanism of the reaction of Hl with methoxymethane.
Answer:

Question 31.
Write equations for the following reactions :
(i) Friedel Crafts reaction-alkylation in anisole.
(ii) Nitration of anisole
(iii) Bromination of anisole in ethanolic medium
(iv) Friedel Crafts acetylation of anisole.
Answer:
(i) The halogenation of the benzene ring occurs at the ortho and para positions. However, the para isomer is
formed in excess. For example, the bromination of anisole in ethanoic acid gives nearly 90 percent p-bromoanisole.

(ii)  The nitration of anisole carried with a nitrating mixture of conc. UNO3 and conc. H2SO4 upon heating gives a
a mixture of ortho and para nitro derivatives.

(iii) Sulphonation: Anisole upon sulphonation gives a mixture of isomeric sulphonic acid derivatives.

(iv) Friedel Craft’s reaction: Both alkylation and acylation are carried in the presence of anhydrous Aid3 catalyst which
behaves as a Lewis acid.

Question 32.
Show how would you synthesise the following alcohols from appropriate alkanes?

Answer:

Question 33.
When 3-methylbutan-2-ol is treated with HBr, the following reaction takes place :

Give the mechanism of the reaction.
Answer:
The mechanism is explained as follows :

We hope the NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers help you. If you have any query regarding NCERT Solutions for Class 12 Chemistry Chapter 11 Alcohols, Phenols and Ehers, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 12
Chapter Name	Colonial Cities Urbanisation, Planning and Architecture
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 12 Colonial Cities Urbanisation, Planning and Architecture

Question 1.
To what extent are census data useful in reconstructing patterns of urbanisation in the colonial context ?
Solution :
The census data are useful in reconstructing patterns of urbanisation in the colonial context in the following ways :

1. The censuses reveal that after 1800 the urbanisation in India was slow. The proportion of the urban population to the total population in India was almost stagnant. Between 1900 and 1940 the urban population increased from about 10 per cent of the total population to about 13 per cent.
2. Smaller towns did not grow economically but Bombay, Madras and Calcutta grew rapidly.
3. The introduction of railways changed the centre of economic activity from traditional towns to towns connected with the railways.

The above facts provide us the patterns of urbanisation but the historians have found the census data misleading because the census operation was a means by which social data were converted into convenient statistics about the population. There were many shortcomings in it. For example, classification of different sections of population was arbitrary. There were overlapping identities of people. People w7ere too suspicious of census operations and did not cooperate with the officials. Thus, census data is invaluable but should be studied carefully in restructuring patterns of urbanisation in colonial India.

Question 2.
What do the terms “White” and “Black” Town signify?
Solution :
The British had white skin as they were often described ‘white’ and they considered themselves as superior to others. On the other hand, the blacks had brown or black skin. So they were known as the ‘black’. The White signified their superiority over the black due to the colour of their skin. The British symbolised the Black areas full of chaos and anarchy, filth and disease and on the other hand, the white areas stood for cleanliness and hygiene. In Black areas, epidemics like cholera and plague often broke out. So the British took stringent measures to ensure sanitation and public health to prevent diseases of the Black areas. They ensured underground piped water supply and introduced sewerage and drainage system in White areas. Thus, we can say, the White Towns were those parts of the colonial towns where the White people lived. These towns had wide roads, barracks, churches, paradeground, big bungalows and gardens, symbolised settled city life, whereas the Indian lived in Black Towns, were said to be unorganised and a source of filth and disease.

Question 3.
How did prominent Indian merchants establish themselves in the colonial city?
Solution :
The prominent Indian merchants established themselves in the colonial city in the following ways :

1. With the expansion of railways, the countryside from where raw materials and labour were obtained was linked to the cities like Bombay, Madras and Calcutta. This gave an opportunity to the Indian merchants to set up modern factories. Thus, after the 1850s, cotton mills were set up by Indian merchants and entrepreneurs in Bombay.
2. Kanpur specialised in leather, woollen, cotton textiles and Jamshedpur where steel factory was established by J. Tata, specialised in steel.
3. The American Civil War started in 1861 gave another opportunity to the Indian merchants for earning huge profits. Bombay was the most important city of India. By the late nineteenth century, Indian merchants in Bombay established cotton mills.

Question 4.
Examine how concerns of defence and health gave shape to Calcutta.
Solution :
Sirajudaula, the Nawab of Bengal in 1756, sacked the small fort from Britisher. In this fort the British traders had built to house their goods. Consequently, when Sirajudaula was defeated in the Battle of Plassey, the British built a new fort, Fort William which could not be easily attacked. Around this a vast open space was left. This open space ‘ was called the Maidan or garermath. This was done for security reasons, because there would be no obstructions to a straight time of fire from the Fort against an advancing enemy army. Soon the British began to move out of the Fort. They built residences along the periphery of the Maidan. This indicates that how the English Settlement in Calcutta began to take shape. The vast open space around the Fort William became the significant town planning measure in Calcutta (Now Kolkata).

Lord Wellesley was more concerned about the conditions that existed in the cities. Cities were overcrowded, and had no sanitation facilities. He issued an administrative order in 1803 on the need for town planning and set up various committees for this purpose open places in the city would make the city healthier. As a result of this, many bazaars, ghats, burial ground and tanneries were cleared or removed. After Wellesley’s departure, the Lottery Committee carried on with the work of town planning in Calcutta.

Question 5.
What are the different colonial architectural styles which can be seen in Bombay city ?
Solution :
The different colonial architectural styles which can be seen in Bombay city are as mentioned below :

1. European style : In mid nineteenth century, the buildings were constructed in the European style to create a familiar landscape in an alien country and to symbolise their superiority.
2. Indian style : As the Indians used European architecture, the British adopted Indian style that can be seen in the construction of bungalows in Bombay and all over India.
3. Neo-classical or new classical style : Its characteristics are construction of geometrical structures fronted with lofty pillars. Town Hall is an example of this style.
4. Neo-Gothic style : Its characteristics are high-pitched roofs, pointed arches and detailed decoration: Secretariat, University of Bombay and High Court were made in this style. The most spectacular example of the neo-Gothic style is the Victoria Terminus.
5. Indo-saracenic style : It was a hybrid architectural style that combined the Indian with the European. It was inspired by the medieval buildings in India with their domes, chhatris, jalis and arches. The Gateway of India and the Taj Mahal Hotel that was built by Jamsetji Tata belong to this style.

Question 6.
How were urban centres transformed during the eighteenth century?
Solution :
(i) The disintegration of the mughal empire after the death of Aurangzab paved the way of emergence of paverful regional powers. The capital cities of these regional kingdom likes Lucknow, Poona, Nagpur and Barda now become important. Taking the advantage of this opportunity many nobles and officials created new urban settlements such as the qasbah and ganj.

(ii) The European companies too had set up their bases in different parts of India during the sixteenth and seventeenth centuries. For example the Portugues (in Panaji in 1570) and the British in Madras in 1639. With the expansion in commercial activity, towns began to emerge as trading centres.

(iii) From the mid-eighteenth century trading. Centres like Surat and Dhakha which had grown in the seventeenth century now began to decline as trade shifted to other places. When the British acquired Bengal and the east Indian’s Company’s trade hereafter expanded the colonial port cities likes Madras and Calcutta. These new part cities began to emerge as the new economic capitals.

(iv) In these newly developed cities many new buildings were built and new occupations developed. People flocked to these cities in large numbers. By the nineteenth century, these newly developed cities become the biggest cities in India.

Question 7.
What were the new kinds of public places that emerged in the colonial city ? What functions did they serve ?
Solution :
New kinds of public places that emerged in the colonial city were as given below :

1. Fort St. George (Madras), Fort William (Calcutta), the Fort George (Bombay). These were the fortified areas of British settlement.
2. The Writers’ Building in Calcutta : It was the building where the servants of the East India Company in India stayed on arrival in the country. Later this building became a government office.
3. Clubs, racecourses and theatres were built for the ruling elites exclusively on racial grounds.
4. Cantonment places were developed. Here Indian troops under European command were stationed. These were considered safe enclaves for Europeans.
5. Simla, founded during the course of Gurkha war, and Darjeeling were hill stations that became strategic places for billeting troops, guarding frontiers and launching campaigns against enemy rulers.
6. Public places such as public parks, theatres, and cinema halls came into existence for providing new forms of entertainment and social interaction.
7. Government House Calcutta : It was built by Lord Wellesley for himself in Calcutta.

Question 8.
What were the concerns that influenced town planning in the nineteenth century?
Solution :

1. Two concerns which influenced the town planning in the nineteenth century were defence and health.
2. In many towns British built forts to protect their factories. Around the fort, a vast open space was left open. This vast space was known as the Maidan.
3. It was done so that there would be no obstructions to a straight line of fire from the Fort against an advancing enemy.
4. Attempts were also made to improve the sanitation and cleanliness by creating open spaces in the city.
5. For this purpose, in Calcutta many bazaars, ghats and burial grounds were cleared.

Question 9.
To what extent were social relations transformed in the new cities ?
Solution :
The social relations were transformed in the new cities in the following ways :

1. New transport facilities as horse-drawn carriages, trams and buses meant that people could live at a distance from the city centre. This led to separation of the place of work from the place of residence. People travelled from home to office or factory.
2. The sense of coherence and familiarity of the old towns disappeared. The public places such as parks, theatres and cinema halls provided new forms of entertainment and social interaction.
3. New social groups came into existence. The “middle classes” increased due to the coming of all types of people i.e., clerks, doctors, teachers, lawyers and others. With the spread of education, people could put forward their views in newspapers and journals. People started questioning old customs and practices.
4. Women entered new professions as factory workers, teachers, theatres and film actresses. However, their entry into public spaces remained the objects of social censure.
5. There was a dramatic contrast between extreme wealth and poverty. The new cities were bewildering places where life seemed always in a flux. Paupers from the villages came to cities in search of employment. The male migrants left their families in the villages because jobs were uncertain and food was expensive. But yet the villagers participated in religious festivals, tamashas and swangs which mocked the pretensions of their masters, Indian and European.

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Here we are providing NCERT Solutions for Class 12 English Vistas Chapter 7 Evans Tries an O-level. Students can get Class 12 English Evans Tries an O-level NCERT Solutions, Questions and Answers designed by subject expert teachers.

Evans Tries an O-level NCERT Solutions for Class 12 English Vistas Chapter 7

Evans Tries an O-level NCERT Text Book Questions and Answers

Evans Tries an O-level Reading with insight

Question 1.
Reflecting on the story, what did you feel about Evans having the last laugh?
Answer:
In this story, James Roderick Evans, a young criminal, manipulated a situation to outsmart a team of officers, headed by the Governor of HM Prison, Oxford. His ingenuity impresses the readers. One views him as a law breaker; he was a kleptomaniac, not a ferocious criminal.There was no record against him for any violence. On the contrary, he was praised by the Governor as a pleasant and amusing person. He was one of the stars at the Christmas concert and was good at imitation.

He was called “Evans the Break” by the prison officers as he had escaped three times from prison. His ability to outsmart the officers is viewed as more of a battle of wits, than a serious offence. One marvels at his cleverness—his plea to keep the hat on, impersonating the invigilator and having his friend pose as a tutor, and above all hoodwinking the police into escorting him out. One cannot but help admire him for his never-say- die attitude.

Question 2.
When Stephens comes back to the cell he jumps to a conclusion and the whole machinery blindly goes by his assumption, without even checking the identity of the injured ‘McLeery’. Does this show how hasty conjectures prevent one from seeing the obvious? How is the criminal able to predict such negligence?
Answer:
When Stephens walked beside ‘McLeery’, he noticed that his Scots accent was more pronounced and he looked slimmer. He was happily ruminating on the fact that the Governor had asked him, and not Jackson, to see McLeery off the premises. But much to his horror, when he decided to take just one last look at Evans, he saw the injured ‘McLeery’ instead and raised an alarm. McLeery, who was actually Evans in disguise, managed to have himself escorted to the hospital from where he escaped.

Though the police had seen Evans day in and day out, yet he had managed to give them the slip. Negligence to corroborate evidence and to fall for the general frenzy, outwitted the officers and the governor. The criminal, on the other hand, observed the people around him and plotted to take advantage of their complacence and weakness. Jackson and Stephens were so anxious to outdo one another that they were taken on a ride by the criminal.

Question 3.
What could the Governor have done to securely bring back Evans to the prison when he caught him at the Golden Lion? Does that final act of foolishness really prove that “he was just another good-for-a-giggle, gullible governor, that was all”?
Answer:
The Governor was not necessarily foolish. It was he who tracked down Evans at the Golden Lion. He was perceptive and was shown to unravel the mystery, almost single-handedly.Yet, when the prison officer handcuffed the recaptured Evans, the Governor put him in the van and easily let him escape. He committed the same mistake of assuming that Evans could no longer pull another trick. His self¬assured attitude caused him to lose to Evans who seemed alert as ever, and quick-witted.

Though the Governor was outsmarted by Evans once, he failed to anticipate that Evans could pull off a similar trick with the police officer. However, Evans comes across as a mastermind who outwitted one and all, it seemed impossible to keep up with his antics.

Question 4.
While we condemn the crime we are sympathetic to the criminal. Is this the reason why the prison guards develop a soft corner for those in custody?
Answer:
The reason why the prison guards develop a soft comer for those in custody is that constant interaction brings people closer. There is increased understanding between them. With the passage of time they grow to empathise with the criminal. As ordinary individuals, we too identify with the characters in a book or a film, we can identify with their pain and suffering. The prison guards tend to form their personal favourites and alleviate their suffering by the little compassion they can, within legal boundaries. In this story, Evans knew of Jackson’s compassionate nature which he took advantage of, to stage his escape. Jackson let Evans keep on his tattered hat and spared him the discovery of his new haircut, facilitating his escape.

Question 5.
Do you agree that between crime and punishment it is mainly a battle of wits?
Answer:
Between crime and punishment, it is mainly a battle of wits. Courts have become battlegrounds of sophistry, hyperbole, and obfuscation. This is the reason why seasoned criminals, taking advantage of the situation, go unpunished. There is a relationship between intelligence and crime. As in the story, Evans was a kleptomaniac but managed to outsmart the entire team of police officers. So much so that he was escorted out of the prison by the men in uniform, arrested and he managed to escape again in the wake of a confession.

His plan was brilliant and his tactics ingenious, such as getting his friends to masquerade as a German teacher and as the invigilator, concealing blood in a rubber ring, getting directions through the means of a question paper, and the correction of the paper. He manages to outwit the entire police force.

Evans Tries an O-level Extra Questions and Answers

Evans Tries an O-level Short Answer Questions

Question 1.
Who was Evans? Why were precautions taken for the smooth conduct of the examination?
Answer:
Evans was a young prisoner in the Oxford Prison who wished to appear for his O-level German exam. He was notorious for breaking out of prison, having tried it thrice. The prison officers called him “Evans the Break”. Hence, precautions were taken to guard against any of his ploys to escape.

Question 2.
Why had Evans been imprisoned?
Answer:
James Roderick Evans had no record of violence. In fact, he was reputed to be a pleasant fellow, and was well known for his performances at the Christmas concert. However he was a habitual kleptomaniac—he stole things.

Question 3.
What reason did Evans give for wearing a hat? Why did he do so?
Answer:
When the senior prison guard, Mr Jackson, asked Evan to take his filthy hat off, Evans begged to keep it on as it was his lucky charm. In reality, the hat was a device to hide his closely cropped hair that he would consequently use to pull off a disguise.

Question 4.
Who was Stuart McLeery? What did he look like?
Answer:
Reverend Stuart McLeery was a priest who was called in to invigilate the exam Evans was to take. He wore a long black overcoat and a shallow-crowned clerical hat to protect himself from the steady drizzle. He wore spectacles that had thick lenses, in his right hand he carried a small brown suitcase, which contained all that he would need for his morning duties, including a sealed question paper envelope, a yellow invigilation form, a special “authentication” card from the Examinations Board, a paper knife, a Bible and a current copy of The Church Times. However, the author only provides the readers with a description of the impostor dressed as McLeery.

Question 5.
What were the Governor’s fears?
Answer:
The Governor had made secure arrangements to guard against any possible plans for Evans to escape. He was suddenly apprehensive because Evans, he thought, could take advantage of McLeeiy. He could get him to smuggle in a chisel or a rope ladder. He was worried if McLeery had, unwittingly, brought in something— like a jack-knife, Evans could hold him hostage with such a weapon.

Question 6.
What was the one thing about McLeery that puzzled Jackson? How did McLeery explain this?
Answer:
Jackson lightly frisked McLeery’s clothes, and came upon his reading glasses, in the spectacle case. One of the objects puzzled him sorely. It was a smallish semi-inflated rubber ring. McLeery explained that he suffered from haemorrhoids, and it served him to sit comfortably.

Question 7.
What was the phone call for during the examination?
Answer:
The phone call, was supposedly, from the Examinations Board that instructed candidates offering German, 021-1 to note a correction. McLeery was instructed to call out the correction carried in the examination paper.

Question 8.
What did Evans ask for during the course of the examination? How did Stephens react to this?
Answer:
Evans wanted to put a blanket around his shoulders as he claimed to be feeling cold. Stephens was sceptical of allowing Evans the benefit of the blanket to keep him from trying to suffocate McLeery. But he changed his mind, considering the prison cell was indeed cold, away from the sun’s warmth.

Question 9.
How was the invigilator escorted out of the prison? What was Stephens’ reaction?
Answer:
Stephens escorted McLeery to the main gates. He noticed that the invigilator’s Scottish accent seemed more pronounced than ever, and his long black overcoat created the illusion that he had suddenly grown slimmer. Nevertheless, Stephens felt pleased that the Governor had asked him, and not Jackson, to see McLeery off the premises.

Question 10.
What shock awaited Stephens when he looked into Evan’s cell?
Answer:
Stephen decided to take one last look at Evans after escorting the invigilator out. Much to his horror, he discovered the injured parson in place of Evans. He found the parson with a blanket on his shoulders, and his irregularly cropped hair covered with blood, dripping onto his priestly white collar. He shouted frantically for Jackson.

Question 11.
Who was the bleeding man in the cell? What was he doing?
Answer:
The bleeding man in the cell was none other than Evans himself who the officials presumed to be the invigilator. McLeery, as he was presumed to be, seemed to hear the officers at hand, as his hand felt feebly for a handkerchief from his pocket, and he held it to his bleeding head. He tried to speak but could give only a long low moan.

Question 12.
Where did Carter take McLeery?
Answer:
Carter was alarmed to see McLeery hurt. He immediately called for an ambulance, but McLeery insisted on leading them to Elsfield Way, where he claimed Evans was. The Governor told Carter to take McLeery for guidance as he was the only one who seemed to know what was happening. Carter helped McLeery into the car and McLeery told them to drive from Elsfield Way to the Headington roundabout. They headed for the Examinations Board, suspecting to find there an accomplice to Evans.

Question 13.
How did the Governor realize that they had mistaken Evans to be McLeery?
Answer:
On being told by Carter, that the parson had been taken to the hospital from the Examination Board, he called to confirm at the Radcliffe. To his shock, the hospital staff said that McLeery had not been admitted there. The hospital claimed to have sent an ambulance which had failed to trace McLeery. It was only then that the realization dawned on the Governor how Evans had managed to pose as the injured McLeery.

Question 14.
Where was Evans after his escape from the prison?
Answer:
Evans was at a hotel, Golden Lion, dressed smartly in a smart new hat to conceal his hair that he had cropped so closely. When the narrative returns to him, he was seen directing the receptionist to send him the morning papers.

Question 15.
How was Evans apprehended the last time? How did he outsmart the police again?
Answer:
When Evans entered his hotel room to find the Governor, he made no effort to avoid being caught, and unravelled his plan, good-naturedly to the Governor. Following which, a prison officer handcuffed Evans and took him to the prison van. As the prison van turned right from Chipping Norton on to the Oxford Road, the ‘prison officer’ who had been silent till then unlocked the handcuffs and revealed himself to be yet another accomplice of Evans. Evans managed to stage his last escape by having an impostor stage the act of a police officer.

Evans Tries an O-level Long Answer Questions

Question 1.
What did the police in Oxford Prison think about James Evans?
Answer:
The Governor of Oxford Prison felt that Evans was not a violent criminal. He found him to be a pleasant and an amusing character. Evans was not a violent sort of person and there was no record of his violence. In fact, he was considered quite a pleasant sort of chap. He was also an entertaining person and was one of the stars at the Christmas concert where he did imitations like that of the comedian Mike Yarwood. He was an incorrigible kleptomaniac and had an obsessive impulse to steal regardless of economic need.

Evans was called “Evans the Break” by the prison officers. He had thrice escaped from prison, and would not have been in Oxford Prison but for the recent wave of unrest in the maximum-security establishments in the north. Tight security kept Evans behind the bars. The Governor, personally, made sure that he did no such thing that would shame them. But, Evans was viewed as real trouble. His presence was persistently irritating to the jail authorities. The Governor doubted Evans’ interest in O-level German and thought it was a ruse to break out of prison again.

Question 2.
Write in detail about the preparations that were made for Evans James to write his examination.
Answer:
When Evans applied for permission to appear for the O-Level Examination in German, the Governor decided to look at the security measures personally. Evans was to write the examination in his own cell. To invigilate the exam, they had decided to call one of the parsons from St Mary Mags. The authorities assumed that they would not have much trouble preventing his communication with others. The examination was to be taken in D Wing, which had a heavy outer door and a heavy inner door that led to the cell.

The two-hour examination was scheduled to start at 9.15 a.m. So the next morning, at 8.30 a.m., two prison officers, Mr Jackson and Mr Stephens, went to Evans’ cell. Mr Jackson asked Mr Stephens to take away his razor after he had finished shaving. Even his nail-scissors were taken away. Evans was allowed to retain his hat as he begged to keep with him his lucky charm. Two small square tables were brought into his cell and placed opposite each other.

They had a listening device installed in his cell, above the door, so that the Governor could listen in. Stephens had to dutifully peer through the peephole at regular intervals to keep a close watch on Evans.

Question 3.
How did Evans plan his escape amidst tight security?
Answer:
Evans was known as “Evans the Break” by the prison officers as he had escaped three times from prison, previously. At Oxford Prison, he managed to outsmart everybody once again. He arranged for a man who came disguised as Reverend Stuart McLeery, an accomplice of Evans, while the actual parson had been locked in his apartment.

He came carrying a ring to sit on, filled with pig’s blood which helped Evans mislead the police. While he pretended to give instructions to Evans to fill in the examination sheet, McLeery actually told him of the escape plan. His corrections were clues of the escape plan. The blanket given to Evans at 10.50 a.m. helped Evans with the disguise. The cap, that was his lucky mascot, was an aid to hide the closely cropped hair which helped him look like the parson. Stuart McLeery, wore two black fronts and two collars to provide one set to Evans. After McLeery was escorted out, Mr Stephens was aghast to see McLeery sprawled in Evans’ chair with blood oozing out. He refused to go to the hospital but offered his help to find Evans. He also showed the Governor a photocopied sheet that had been carefully and cleverly superimposed over the blank page of the question paper.

McLeery led the police to Elsfield Way and from there to Headington and finally to Newbury in order to put the police off Evans’ track. When Superintendent Carter informed the Governor that McLeery was in the Radcliffe Hospital, and the Governor went to check on him, he found McLeery missing and discovered the truth about Evans’ plan. All along, it was not McLeery, but Evans who had been in the cell, pretending to be injured. He had executed the plan through his so-called invigilator, who was in reality his friend.

Question 4.
How did the Governor track down Evans? How did Evans outsmart him again?
Answer:
Once the Governor found out that McLeery was not in the hospital, he realized the truth. They found the real McLeery bound and gagged in his study in the Broad Street. They also realized it had not been Evans impersonating McLeery who had walked out; it had been Evans impersonating McLeery who had stayed in. On the other hand, Evans, after a delightful meal, returned to the hotel Golden Lion wearing a hat to conceal his closely cropped hair. He noticed the pretty receptionist.

As he went up to his room, to his surprise, he saw the Governor sitting in his bedroom and realized that the blonde girl at the reception was an informer. The Governor, who knew a little German, had deduced “zum goldenen Lowen” meant the Golden Lion and using Evans’ Index number 313 and Centre number 271 as coordinates on an Ordnance Survey Map, he figured out the location of the Hotel. When Evans was handed to the police once again, the Governor heaved a sigh of relief. But much to everyone’s surprise, the prison van driver and the officer were impersonators who helped Evans flee right under their nose.

 

Here we are providing NCERT Solutions for Class 12 English Flamingo Chapter 6 Poets and Pancakes. Students can get Class 12 English Poets and Pancakes NCERT Solutions, Questions and Answers designed by subject expert teachers.

Poets and Pancakes NCERT Solutions for Class 12 English Flamingo Chapter 6

Poets and Pancakes NCERT Text Book Questions and Answers

Poets and Pancakes Think as you read 

Question 1.
What does the writer mean by ‘the fiery misery’ of those subjected to make-up?
Answer:
The writer means the misery caused by the incandescent lights that poured out intense heat. The make-up room of the Gemini Studios had bright bulbs in the room full of large mirrors that reflected the glowing lights. Under such blazing heat make-up was done.

Question 2.
What is the example of national integration that the author refers to?
Answer:
The make-up team and also those who came and went were from different states. It was headed by a Bengali and next in hierarchy was a Maharashtrian, assisted by an Andhraite, a Madras Indian Christian, an Anglo-Burmese and other local Tamils. It was truly a gang of nationally integrated make-up men.

Question 3.
What work did the ‘office boy’ do in the Gemini Studios? Why did he join the studios? Why was he disappointed?
Answer:
The office boy applied make-up to the crowds, mixing his paint in a giant vessel and slapping it on the crowd players. He had joined the studios in the hope of becoming a star actor or a top screen writer, director or lyrics writer. He was a bit of a poet. He was disappointed as he was placed low even in the hierarchy of make-up men.

Question 4.
Why did the author appear to be doing nothing at the studios?
Answer:
The author’s job was to cut out newspaper clippings on a wide variety of subjects and store them in files. Many of these had to be written out in hand. Seeing him sitting at his desk and tearing up newspapers most people thought he had nothing to do at the studios.

Question 5.
Why was the office boy frustrated? Who did he show his anger on?
Answer:
The office boy was frustrated because his hopes of making big in the movie world failed. He vent his anger and frustration on Kothamangalam Subbu, the No. 2 in the studios, whom he held responsible for his dishonour and neglect.

Question 6.
Who was Subbu’s principal?
Answer:
S.S. Vasan, the founder of Gemini Studios, was the boss and Subbu’s Principal in the studios. Subbu had a great loyalty to him. This made him identify himself with his principal completely. He turned his entire creativity to his principal’s advantage.

Question 7.
Subbu is described as a many-sided genius. List four of his special abilities.
Answer:
Subbu was a many-sided genius. He was born a Brahmin. It is a virtue in itself as it exposed him to more affluent situations and people. Second, he had the ability to look cheerful at all times. Third, he always had work for somebody. Fourth, he had great loyalty to his principal, S.S. Vasan, the Boss.

Question 8.
Why was the legal adviser referred to as the opposite by others?
Answer:
The lawyer was the only one at the studios who wore pants, tie and sometimes a coat, unlike others who wore khadi dhoti and shirt. His job was to give support and advise on problems, but in fact he created problems. He brought the career of a brilliant actress to an end by terrorizing her. He was rightly called an illegal adviser.

Question 9.
What made the lawyer stand out from the others at Gemini Studios?
Answer:
The lawyer wore pants, a tie and sometimes a coat, while all wore khadi dhoti and white khadi shirt. He looked alone and helpless. He was a man of cold logic in a crowd of dreamers. He was a neutral man among Gandhiites and Khadiites.

Question 10.
Did the people at Gemini Studios have any particular political affiliations?
Answer:
The people at Gemini Studios wore Khadi and worshipped Gandhi, but beyond that they had no particular political interests or understanding. They only had opinions on communism, which they loathed and looked down on communists. They considered communists as heartless atheists who are devoid of emotions. They went about letting loose anarchy in the society.

Question 11.
Why was the Moral Re-Armament Army welcomed at Gemini Studios?
Answer:
The Moral Re-Armament Army was invited to stage two plays, which were more like plain homilies ‘ (sermons/lectures) for the Gemini family. It was discovered only later that the group was part of the movement countering international communism and Vasan had invited them under the influence of his political interests.

Question 12.
Name one example to show that Gemini Studios was influenced by the plays staged by MRA?
Answer:
MRA staged two plays ‘Jotham Valley’ and ‘The Forgotten Factor’. Their high quality costumes and
well made sets earned a lot of admiration. Their sunrise and sunset scene impressed them so much that all Tamil plays started reproducing the scene with a bare stage, a white background curtain and a tune playing on the flute.

Question 13.
Who was the Boss of Gemini Studios?
Answer:
Mr. S.S. Vasan, the founder of Gemini Studios was the Boss. Apart from producing films, he was an editor of a popular Tamil weekly ‘Ananda Vikatan’. He was a great admirer of scholarly people. Subbu seemed to enjoy an intimate relationship with him. Mr. Vasan is projected as a bit of showman here.

Question 14.
What caused the lack of communication between the Englishman and the people at Gemini Studios?
Answer:
The Englishman’s speech was peppered with words like ‘freedom’ and ‘democracy’ and the Gemini
family had no political interests, so they were dazed and a silent audience. Also, the Englishman’s accent was difficult to understand, because of which all communication had failed. He was basically a poet and that made no sense to the people whose life centered around a film studio.

Question 15.
Why was the Englishman’s visit referred to as unexplained mystery?
Answer:
The Englishman was a poet whose name was not familiar. In his speech he talked about the thrills and travails of an English poet, which made no sense for the simple people at Gemini Studios who had had no exposure other than films and so they were not interested. These simple people had neither taste for English poetry nor political interests. Hence, his visit is referred to as an unexplained mystery.

Question 16.
Who was the English visitor to the studios?
Answer:
The English visitor to the studios was poet Stephen Spender, editor of British periodical ‘The Encounter’.

Question 17.
How did the author discover who the English visitor to the studio was?
Answer:
The author discovered his identity by reading his name on the pages of ‘The Encounter’ in the British Council Library. He also knew about him from the paperback edition of the book The God That Failed.

Question 18.
What does The God That Failed refer to?
Answer:
The God That Failed refers to a book that was a compilation of six essays by six eminent men. It was a low priced student edition released to commemorate the 50th anniversary of the Russian Revolution. It dealt with the author’s disillusionment with communism.

Poets And Pancakes Understanding the text

Question 1.
The author has used gentle humour to point out human foibles. Pick out instances of this to show how this serves to make the piece interesting?
Answer:
‘Poets and Pancakes’ has an underlying tone of humour which is satirical and has been deployed by the author to point out human foibles. It is mainly manifested in his description of the make-up room people.

The make-up room, he says, was in a building that had once been the stables of Robert Clive. He further makes fun of the make-up team that slapped make-up. Ironically, the make-up turned any normal man into a hideous monster, far from being presentable. He also refers to the fiery misery of the actors when their make-up was done under the bright bulbs, large mirrors reflecting blazing heat. His description of Subbu’s No. 2 position in Gemini Studios, the frustration of the office boy and the opposite role played by the legal adviser in the acting career of a countryside girl are humorously dealt with but effectively bring out the flaws in the set-up.

The showmanship of the boss and what influences his guest list point out human weaknesses in a light-hearted manner. The humour is at its peak in the description of the visit of Stephen Spender. S.S. Vasan’s reading a long speech in his honour but he too knew precious little about him. Spender’s accent is highly unintelligible. Then the author’s establishing long lost brother’s relationship with the English visitor is also funny and humorous. All these slight digs at human foibles tickle in us humour.

Question 2.
Why was Kothamangaiam Subbu considered No. 2 in Gemini Studios?
Answer:
Kothamangaiam Subbu was on the attendance roll with the story department and was No. 2 at Gemini Studios not by virtue of any merit, but because he was a Brahmin with affluent exposure. He was cheerful and had a sense of loyalty that placed him close to the Boss. He was quick to delegate work to others. As if tailor-made for films, sparks of his creativity showed in his suggestions on how to create shots. He composed poetry, scripted a story and a novel. He gave direction and definition to Gemini Studios during its golden years. He performed in a subsidiary role better than the main players. He had a genuine love for his relatives and near and dear ones. His extravagant hospitality was popular among his relatives and acquaintances, probably that is why he had enemies.

Question 3.
How does the author describe the incongruity of an English poet addressing the audience at Gemini Studios?
Answer:
The Gemini Studios witnessed a surprising visit by a tall Englishman who was proclaimed to be a poet. The welcome speech by the Boss was delivered in the most general terms, which only showed that even the Boss did not know much about him. The poet talked about the thrill and travails of an English poet which made no sense to the simple people at Gemini Studios. They had no exposure other than films and so, they were not interested. Also, words like democracy and freedom that featured in his speech held no interest for them as they had no political thought or interests. Moreover, the Englishman’s accent was difficult to understand, because of which all communications failed. He was basically a poet and that made no sense to the people whose life centred round a film studio. Therefore, his visit remained an unexplained mystery for much time.

Question 4.
What do you understand about the author’s literary inclinations from the account?
Answer:
The author, Asokamitran, was entrusted with the job of maintaining the newspaper clippings of movies and other articles. Though to others, who just saw him tearing papers, he appeared to be doing nothing, the job kept the author well informed. Also, there prevailed an intellectual environment to some extent because the poets and script writers used to hang out there in the mess that served coffee any time of the day. The author would pick up fifty paisa copies of journals from the footpath and took part in the poetry writing competition. He actually read essays ‘The God Who Failed’ to know more about the poet Stephen Spender. All these are evidence that he had some literary taste.

Poets And Pancakes Extra Questions and Answers

Poets And Pancakes Short Answer Questions

Question 1.
How does the writer describe the make-up room of the Gemini Studios?
Answer:
The make-up room of the Gemini Studios had incandescent lights. It also had lights at all angles around large mirrors. Those subjected to make-up had to face bright light and a lot of heat there. It was on the upper floor of the building that was believed to have been Robert Clive’s stables.

Question 2.
Bring out the humour in the job of the make-up men.
Answer:
The make-up men came from all corners of the country and could transform any decent-looking person into a repulsive crimson coloured fiend and made people look uglier than they were in real life. They used truck loads of pancakes and locally manufactured potions and lotions to transform the looks of the actors.

Question 3.
How was the make-up room a fine example of national integration?
Answer:
Transcending all the barriers of regions, religions and castes, people from all over India came to Gemini Studios for jobs. The make-up department was headed by a Bengali, succeeded by a Maharashtrian, assisted by a Dharwar Kannadiga, an Andhra, a Madrasi, Christian and an Anglo Burmese and the usual local Tamils. Hence, the writer finds in the make-up department a perfect example of national integration.

Question 4.
Why did the author appear to do nothing in the studio?
Answer:
The author’s job in the studio was to cut newspaper dippings of all the relevant news items and articles that appeared in different newspapers and maintain a record of the same. This tearing of newspaper gave an impression that he was free and simply whiling away his time. People used to barge in his cubicle and lectured him.

Question 5.
Why was the office boy frustrated? Who did he show his anger on and how?
Answer:
The office boy had joined the studio years back. He aspired to be a top film star, or top screen writer, lyricist or director. He felt frustrated on not being able to realise his dreams and had been given a job much below his calibre and dignity. He blamed Kothamangalam Subbu for all his woes, ignominy and neglect. He often gave vent to his frustrations in the narrator’s cubicle. The narrator yearned for relief from the never-ending babble of the office boy.

Question 6.
Subbu is described as a many-sided genius. Justify.
Answer:
Kothamangalam Subbu may not have had much formal education but, by virtue of his being born as a Brahmin, he had had exposure to many affluent situations and people. He had the ability to look cheerful at all times, even after a setback. He was always full of creative ideas. Above all, he was a charitable and extravagant man and hospitable to his relations. His loyalty had put him close to his boss. But he seemed to others a sycophant and a flatterer and, probably, that was the reason he had enemies.

Question 7.
How did the lawyer unwittingly bring an end to a brief and brilliant career of a young actress?
Answer:
A talented but very temperamental actress lost her cool on the sets. The lawyer recorded her outburst and played it back, much to her embarrassment. The actress from the countryside was so terror- struck that she retreated and never got back to films. In this way, his mischief making brought an abrupt end to the brilliant actress’ career.

Question 8.
Why did the magazine, ‘The Encounter’, ring a bell in the writer’s mind?
Answer:
The writer wanted to participate in a short story writing contest organized by ‘The Encounter’, a British publication. Before sending his entry, he waited, confirm the authenticity of the periodical, so he visited the British Council Library. When the author read the editor’s name, a bell rang in his mind. It was Stephen Spender, the poet who had visited the Gemini Studios.

Question 9.
What was significant about the book which the author took from roadside?
Answer:
‘The God That failed’ was the name of the book which caught the attention of the author. It contained the essays of six eminent men, who described their journey into communism and their return from it after being disillusioned. It suddenly assumed great significance for the author as he discovered that one of the essays had been written by Stephen Spender, the poet, who had visited the studio. He now understood the reason for his having been invited.

Question 10.
What do you understand about the author’s literary inclinations from the account?
Answer:
The author was very knowledgeable young man whose job required him to pour over the newspaper all day long. His interest in creative writing and participating in story writing contests indicates his interest in literature. This interest was so keen that he read books on varied subjects and went about buying them even when he was short of money.

Question 11.
What kind of people, according to the author, are meant for prose writing?
Answer:
According to the author, prose writing is not the pursuit of a genius. It is for the patient, persistent and persevering drudge whose heart can take rejections and whose spirit to keep trying does not get killed so easily.

Question 12.
Why was Gemini Studios a favourite haunt of poets?
Answer:
Gemini Studios was a favourite haunt of poets as it had an excellent mess which supplied good coffee at all times of the day and for most part of the night. Meeting there was a satisfying entertainment. Moreover, Mr. Vasan was a great admirer of scholarly people.

Question 13.
‘Prose writing is not and cannot be the true pursuit of a genius’, says the author. Explain the statement.
Answer:
In this statement, the author says that prose writing can’t be the true pursuit of a genius because it is always rejected. A genius is not that is accepted everywhere. The author states all this with criticism that prose writing is actually meant for rejection. Prose writers are patient, persistent and persevering drudges. They can’t be down played by rejection slips. Everytime he gets a rejection slip for his manuscript, he starts making a fresh copy and sends it to another publisher with return postage.

Question 14.
‘Suddenly the book assumed tremendous significance.’ Explain the statement.
Answer:
The author bought one copy of the book ‘The God That Failed’ from the footpath. Six eminent men of letters in six separate essays describe their journey into communism and their disillusioned return. Among them one was Stephen Spender. The author at once recollected that Stephen Spender had visited Gemini Studios. He knew about the mystery of his visit now. So, the book assumed tremendous significance for him.

Question 15.
Explain the appropriateness of the title ‘Poets and Pancakes’?
Answer:
The chapter describes Gemini Studios and its functioning very clearly. Its employees are little unrecognized poets. Though they work in a film studio, the focus is on the author’s station in the Studios as a make-up boy using pancakes on crowd players, and how he failed as a poet. So, the title is appropriate.

Question 16.
How humorously does the author describe Frank Buchman’s Moral Re-Armament Army?
Answer:
The author humorously calls the Moral Re-Armament Army after someone as ‘an international circus’. Then he states that they were not very good on the trapeze. Their acquaintance with animals should have been much as animals play tricks in a circus. “But the group ate animals”, says the author their acquaintance with animals was only at the dining table.

Question 17.
What was thought of a communist by the studios people?
Answer:
According to these people, a communist was a godless man. He had no filial or conjugal love. He had compunction about killing his own parents and children. He was always out to cause and spread unrest and violence among the innocent and ignorant people.

Poets And Pancakes Long Questions and Answers

Question 1.
How does the author come to know about the periodical ‘The Encounter’?
Answer:
The Hindu published an advertisement about a short story contest organized by ‘The Encounter’. The author wanted to know about it before he spent a lot of money on postage sending his manuscript to England. So, he visited the British Council Library. There he saw many untouched copies of ‘The Encounter’. When he read the editor’s name, he felt like he had found a long lost brother. He sang as he sealed the envelope and felt that the editor would also be singing the same song when he would open the envelope. Actually the editor was the same Englishman, Stephen Spender who visited Gemini Studios long time ago.

Question 2.
‘Poets and Pancakes’ is a beautiful example of humour in its chatty and rambling style’. Comment.
Answer:
‘Poets and Pancakes’ is an account of Asokamitran’s experiences at the Gemini Studios. Asokamitran
deals with a wide variety of ideas where despite one thought leading to another, the thematic coherence is never lost. The author has adopted a chatty and rambling style. The style has a flow and the reader glides smoothly with the flow of the narrative. Asokamitran highlights human foibles and unusual behaviour with the help of subtle humour. All the characters are so life like that the reader seems to come across in real life. The subtle humour is neither superimposed nor superfluous.

Throughout the chapter humour seems to be spontaneous and interwoven. Even the choice of the title shows that Asokamitran has a flair for natural humour. He ridicules without hurting and the entire account becomes very interesting. Throughout the chapter the reader doesn’t find even the slightest trace of malice. But every now and then the author speaks with his tongue-in-cheek way and pays left-handed compliments to different characters to the great amusement of the readers.

Question 3.
What was the opinion of most of the people at the studios regarding communism?
Answer:
The people at the Gemini Studios wore Khadi dhoti and a clumsily tailored white khadi shirt. It was a crowd of dreamers and an assembly of Gandhiites and khadietes. The Congress rule meant prohibition and most employees worshipped Gandhiji but beyond that they had no admiration for political thought of any kind. They hated the term ‘communism’ as to them a communist was a godless man, incapable of love and always out to spread unrest and violence among the innocent and ignorant people.

When Frank Buchman’s Moral Re-Armament Army visited the Gemini Studios in 1952, they gave them a warm reception. Later they learnt that MRA was a counter movement to international communism and the big bosses of Madras like Mr. Vasan simply played into their hands. Later, the mystery of the visit of Stephen Spender was resolved and the writer came to know the reason of his visit and that the Boss of Gemini Studios had not been interested in Spender’s poetry but in communism. But the khadi-clad poets of Gemini Studios fell the same aversion for communism.

Question 4.
The English poet who visited Gemini Studios was as baffled as the audience. Why was that?
Answer:
When Gemini Studios prepared to welcome Stephen Spender, they did not know the identity of the
visitor. All they knew was that he was a poet from England. However, he was not one of the poets from England that they had heard of. Later they heard that he was an editor. But again he wasn’t the editor of any of the known British publications. When the gentleman arrived, the mystery’ of his identity deepened.

The Englishman left the audience dazed and silent as no one knew what he was talking about. The people of Gemini Studios led lives that least offered them the possibilities of cultivating a taste for English poetry. The English man talked of the thrills and problems of an English poet. His British accent defeated any attempt to understand what he was saying. His lecture lasted an hour but it left the people confused. The poet himself was equally baffled. He too must have felt the strangeness of his talk. His visit remained an unexplained mystery for a long time.

Question 5.
Explain the statement ‘Subbu is tailor-made for films.’
Answer:
Kothamangalam Subbu has many qualities of head and heart with a great fund of humaneness. He has the ability’ to look cheerful at all times. His loyalty to his principal is complete. He is always ready to say nice things about everything. He is resourceful. He has an inventive brain as he can create new ways and means to dramatise some difficult propositions in films. Film-making goes very easy with him. He is a poet also and has written a novel titled Thillana Mchanambal He successfully recreated the mood and manner of the Devadasis of the early 20thcentury.

He is an amazing actor and has performed better than the supposed main players. Above all. he has a heart of gold for his relatives and near and dear ones. His house is a permanent residence for them. In fact, he is tailor-made for films. He gave direction and definition to Gemini Studios during its golden years. His success in films overshadowed and dwarfed his literary achievements. He is a poet, an actor and a creative assistant roled into one.

Question 6.
What idea do you get about the narrator from the chapter ‘Poets and Pancakes’?
Answer:
Asokamitran used to work in a cubicle tearing up newspapers all the time. Most of the people thought that he did nothing. So anyone who felt so would enter his cubicle and deliver an extended lecture. He never tried to say anything in his defence. He kept himself busy with his work, without poking his nose into anyone’s affairs but he was wide awake and highly observant. He very well knew what was going on around him. He had a deep psychological insight into the human mind.

He could see through Subbu’s sycophancy, the lawyer’s smartness and the office boy’s frustration. He was very fond of books and bought books all the time even when there was paucity of money. His description of S.S. Vasan’s battling with half a dozen pedestal fans while reading his welcome address and an account of Stephen Spender’s accent are examples of his brilliant sense of humour. He was a man of cool temperament who never jumped to conclusions or pounced upon opportunities rashly. On the whole, he emerges as a responsible and conscientious man.

Question 7.
Describe Stephen Spender’s visit to Gemini Studios.
Answer:
There was a lot of speculation about Spender’s visit. Initially, everyone thought he was a poet but later they heard that he was an editor. The author describes him as a tall, very English and a serious person. It was evident from Boss’ speech that he knew very little about the visitor. When Spender addressed the gathering they were all dazed as they couldn’t understand his accent and didn’t understand what he was talking about.

Even Spender must have sensed the incongruity of being called to talk about the travails of an English poet in a film studio making simple Tamil films. His visit remained an unexpected mystery. It was only much later that Asokamitran learnt that the reason why the boss had invited Spender. Spender was disillusioned with communism and had contributed an essay in a book on communism called ‘The God That Failed’.

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 10 Wave Optics

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 10
Chapter Name	Wave Optics
Number of Questions Solved	21
Category	NCERT Solutions

Question 1.
Monochromatic light of wavelength 589 nm is incident from air on a water surface. What are the wavelength, frequency and speed of
(a) reflected, and
(b) refracted light ? Refractive index of water is 1.33. (C.B.S.E. Sample Paper 1991)
Answer:
(a) For reflected light wavelength is unchanged i. e.
X = 589 x 10^-9 m = 589 nm
Also, speed of light in air c = 3 x 10⁸ m s ^-1

Question 2.
What is the shape of the wavefront in each of the following cases:
(a) Light diverging from a point source.
(b) Light emerging out of a convex lens when a point source is placed at its focus.
(c) The portion of the wavefront of light from a distant star intercepted by the Earth.
Answer:
(a) Spherical shape
(b) Plane wavefront
(c) Plane wavefront

Question 3.
(a) The refractive index of glass is 1.5. What is the speed of light in glass? (Speed of light in a vacuum is 3.0 x 10⁸ m s^-1)
(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) Here ,n=105,c=3.0 x 10⁸ ms^-1

speed of light when passing through glass depends on colour of light. λ_r > λ_υ , therefore the speed of violet light is less than the red light.

Question 4.
In a Young’s double-slit experiment, the slits are separated by 0.28 mm and the screen is placed 1.4 m away. The distance between the central bright fringe and the fourth bright fringe is measured to be 1.2 Determine the wavelength of light used in the experiment.
Answer:

Question 5.
In Young’s double-slit experiment using monochromatic light of wavelength λ, the intensity of light at a point on the screen where path difference is λ is K units. What is the intensity of light at a point where path difference is λ/2 ?
Answer:

Question 6.
A beam of light consisting of two wavelengths 650 nm and 520 nm is used to obtain interference fringes in a Young’s double-slit experiment.
(a) Find the distance of the third bright fringe on the screen from the central maximum for wavelength 650 nm.
(b) What is the least distance from the central maximum where the bright fringes due to both the wavelengths coincide ?
The distance between two slits is 2 mm and distance between the plane of the slits and the screen is 1.2 m.
Answer:
(a) λ = 650 nm = 650 x 10^-9 m,
d = 2 mm = 2 x 10^-3 m,
D = 1.2 m
Distance of mth bright fringe from the central maximum is given by

Question 7.
In a double-slit experiment, the angular width of a fringe is found to be 0.2° on a screen placed 1 m away. The wavelength of light used is 600 nm. What will be the angular width of the fringe if the entire experiment apparatus is immersed in water ? Take
refractive index of water to be \(\frac { 4 }{ 3 } \)
Answer:


Question 8.
What is the Brewster angle for air to glass transition? (Refractive index of glass = 1.5.)
Answer:
µ = 1.5
According to Brewster’s law,
µ = tan p
tan p = 1.5
⇒ p = 56.31º

Question 9.
Light of wavelength 5000 A falls on a plane reflecting surface. What are the wavelength and frequency of the reflected light ? For what angle of incidence is the reflected ray normal to the incident ray ?
Answer:
Here X = 5000 A = 5000 X 10^-10 m,
c =3 x 10⁸ m s^-1
Wave length of reflected light
= Wavelength of incident light = 5000 Å


Question 10.
Estimate the distance for which ray optics is good approximation  for an aperture  of
4 mm and wavelength 400 nm.
Answer:
Here X = 400 nm = 400 x 10^-9 m, Aperture, a = 4 mm = 4 x 10^-3 m
.’. Distance for which  ray optics is a good approximation is Fresnel’s distance


Question 11.
The 6563 Å Hα line emitted by hydrogen in a star is found to be red-shifted by 15Å. Estimate the speed with which the star is receding from the Earth.
Answer:

Negative sign shows that the star is receding away from the earth.

Question 12.
Explain how Corpuscular theory predicts the speed of light in a medium, say, water, to be greater than the speed of light in vacuum. Is the prediction confirmed by experimental determination of the speed of light in water ? If not, which alternative picture of light is consistent with experiment ?
Answer:
According to Corpuscular theory, when light in the form of particles enters into denser medium from a rarer medium, a force of attraction comes into play on the particles normal to the surface. Thus, the component of velocity normal to the surface of water increases whereas the component of velocity parallel to surface does not change. Therefore,

velocity in water is greater than velocity of light in air. However, in actual case, c > υ. Huygen’s wave theory of light is consistent with the experiment.

Question 13.
You have learnt in the text how Huygens’ principle leads to the laws of reflection and refraction. Use the same principle to deduce directly that a point object placed in front of a plane mirror produces a virtual image whose distance from the mirror is equal to the object distance from the mirror.
Answer:
Let there be a point object A at a distance y from a plane mirror. Treating this point to be a point source of light, we can assume spherical wavefronts progressing from A of radius y. Let there be no mirror then after time t, the wavefront will reach A’ as wavefront I. If a mirror is placed as shown in the figure then image will be formed at A’ represented by II.

It is seen that OA’ = OA i. e. virtual image is formed at a distance equal to the distance of object from the mirror.

Question 14.
Let us list some of the factors, which could possibly influence the speed of wave propagation:
(1) nature of the source,
(2) direction of propagation.
(3) the motion of the source and/or observer.
(4) wavelength
(5) intensity of the wave.
On which of these factors, if any, does
(a) the speed of light in vacuum,
(b) the speed of light in a medium (say, glass or water), depend ?
Answer:
(a) Speed of light in a vacuum is an absolute constant (universal constant). It is independent of any factor. It is independent of the relative motion between source and observer even.
(b)

1. Speed of light in a medium depends upon wavelength.
2. It is independent of the nature of the source and motion of the source relative to the medium.
3. It depends upon the properties of the medium of propagation and motion of the observer relative to the medium,
4. It is independent of the direction of propagation for isotropic medium,
5. It is independent of the intensity of the wave.

Question 15.
For sound waves, the Doppler formula for frequency shift differs slightly between the two situations :
(1) a source at rest; observer moving, and
(2) source moving ; observer at rest. The exact Doppler formulae for the case of light waves in vacuum are, however, strictly identical for these situations. Explain why this should be so. Would you expect the formulae to be strictly identical for the two situations in case of light travelling in a medium ?
Answer:
Sound requires material medium for propagation. Though situations
(1) and (2) may correspond to the same relative motion, yet they are not identical physically as the motion of observer relative to medium may be different in both situations. Hence, Doppler effect for sound cannot be same in both situations. Light when passing through material medium is also governed by different Doppler formulae for
(1) a source at rest; observer moving and
(2) source moving; observer at rest.
But when light passes through vacuum the formulae become exactly same for the two different situations because speed of light and frequency/wavelength of light remain unchanged in vacuum.

Question 16.
In a double-slit experiment using light of wavelength 600 nm, the angular width of a fringe formed on a distant screen is 0.1°. What is the spacing between two slits?
Answer:

Question 17.
Answer the following questions :
(a) In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction
band ?
(b) In what way is diffraction from each slit related to the interference pattern in a double-slit experiment? (C.B.S.E. 2013, 2013 )
(c) When a tiny circular obstacle is placed in the path of light from a distant source, a bright spot is seen at the center of the shadow of the obstacle.
Explain why? (C. B. S. E. 2013 )
(d) Two students are separated by a 7 m partition wall in a room 10 m high. If both light and sound waves can bend around obstacles, how is it that the students are unable to see each other even though they can converse easily ?   (C.B.S.E. 1990 )
(e) Ray optics is based on the assumption that light travels in a straight line. Diffraction effects (observed when light propagates through small apertures/slits or around small obstacles) disprove this assumption. Yet the ray optics assumption is so commonly used in understanding location and several other properties of images in optical instruments. What is the justification?   (C.B.S.E. 1990)
Answer:
(a) The width of central maxima = 2λD/d.
When the width (d) of slit is doubled, then the width of central diffraction maxima reduces to half and the intensity of the central band increases four times as amplitude of light wave is doubled.
(b) Intensity of fringes produced in the double-slit experiment is changed due to diffraction pattern superposing due to each slit.
(c) Light waves diffract at the edges of the circular obstacle. These diffracted waves interfere constructively and give rise to the bright spot at the center of the geometrical shadow.
(d) Diffraction is observed when the wavelength of the wave is of the order of the size of the obstacle. The wavelength of sound wave (≈ 0.33 m) is larger than the light wave (≈10^-7 m) and is also comparable to wall, so diffraction of sound waves takes place and hence the students can converse easily. On the other hand, the wavelength of light is very small as compared to the obstacle e. 1 m high wall so the diffraction of light waves does not take place.
(e) In optical instruments, size of apertures are much larger than the wavelength of light. So diffraction of light is negligible. Hence, the assumption that light can travel in straight line is used in optical instruments.

Question 18.
Two towers on top of two hills are 40 km apart. The line joining them passes 50 m above a hill halfway between the towers. What is the longest wavelength of radio waves, which can be sent between the towers without appreciable diffraction effects ?
Answer:
If A and B are two hills and C is the hill peak midway

Question 19.
A parallel beam of light of wavelength 500 nm falls on a narrow slit and the resulting diffraction pattern is observed on a screen 1 m away. It is observed that the first minimum is at a distance of 2-5 mm from the center of the screen. Find the width of the slit.( C.B.S.E. 2013)
Answer:

Question 20.
Answer the following questions :
(a) When a low flying aircraft passes overhead, we sometimes notice a slight shaking of the picture on our TY screen. Suggest a possible explanation.
(b) As you have learned in the text, the principle of linear superposition of wave displacement is basic to understanding distributions in diffraction and interference patterns. What is the justification of this principle?
Answer:
(a) When a low flying aircraft passes overhead, the metallic body of the aircraft reflects the TV signal. A slight shaking of the picture on the TV screen takes place due to the interference of the reflected signal from the aircraft and the direct signal received by the antenna.
(b) The linear combination of wave equations is also a wave equation. This is the very basis of the superposition principle.

Question 21.
In deriving the single slit diffraction pattern, it was stated that the intensity is zero at angle of nλ/α. Justify this by suitably dividing the slit to bring out the cancellation.
Answer:
Let us suppose that we have n slits each of width

Therefore, each of the n slits of width d’ each sends zero intensity in the direction 9. As a result, the net resultant of intensity due to n such slits is zero.

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NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments are part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 9
Chapter Name	Ray Optics and Optical Instruments
Number of Questions Solved	37
Category	NCERT Solutions

Question 1.
A small candle 2.5 cm in size is placed 27 cm in front of a concave mirror of a radius of curvature of 36 cm. At what distance from the mirror should a screen be placed in order to receive a sharp image? Describe the nature and size of the image. If the candle is moved closer to the mirror, how would the screen have to be moved?
Answer:
Here, h = 2.5 cm, u = – 27 cm, R = – 36 cm.

Nature of image : Real, inverted and magnified. When the position of the object (i.e. candle) is moved closer to the concave mirror, the distance of the image moves away from the screen till the distance of the candle from the concave mirror is less than 18 cm. Hence, the screen has to be moved away from the concave mirror. When the distance of the candle is less than 18 cm from the concave mirror, a virtual and magnified image of the candle is formed behind the mirror. This image is not obtained on the screen.

Question 2.
A 4.5 cm needle is placed 12 cm away from a convex mirror of focal length 15 cm. Give the location of the image and magnification. Describe what happens as the needle is moved farther from the mirror.
Answer:
Here, h₁ = 4.5 cm, μ = – 12 cm, f= 15 cm

Thus, the image is virtual, erect and diminished. As we move the needle away from the mirror, the image goes on decreasing in size and moves towards the principal focus on the other side.

Question 3.
A tank is filled with water to a height of 12.5 cm. The apparent depth of the needle lying at the bottom of the tank is measured by a microscope to be 9.4 cm. What is the refractive index of water ? If water is replaced by a liquid of refractive index 1.63 up to the same height, by what distance would the microscope have to be moved to focus on the needle again ? (C.B.S.E. 2009 )
Answer:

Question 4.
The following figures (a) and (b) show refraction of an incident ray in air at 60° with the normal to glass-air and water-air interface, respectively. Predict the angle of refraction of an incident ray in water at 45° with the normal to a water-glass interface [Fig. (c)]

Answer:

Question 5.
A small bulb is placed at the bottom of a tank containing water to a depth of 80 cm. What is the area of the surface of water through which light from the bulb can emerge out ? Refractive index of water is 1.33 (consider the bulb to be a point source).
Answer:

The light rays from the bulb B, which fall on the surface of water at an angle equal to critical angle (θ_C), grazes on the surface of water and the rays of light which fall on the surface of water at an angle greater than θ_C are totally internally reflected back into the water. The rays of light images emerges out of water through a circular patch of radius r.

Question 6.
A prism is made of glass of unknown refractive index. A parallel beam of light is incident on a 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°.
Answer:

Question 7.
Double-convex lenses are to be manufactured from a glass of refractive index 1.55, with both faces of the same radius of curvature. What is the radius of curvature required if the focal length is to be 20 cm ?
Answer:
Here, n = 1.55, R₁ = R and R₂ = – R, f= 20 cm Using lens maker formula, we get

Question 8.
A beam of light converges to a point P. A lens is placed in the path of the convergent beam 12 cm from P. At what point does the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm?
Answer:
(a) When a convex lens is placed in the path of light converging at P, the beam converges at P_t. Thus, point P acts as virtual object for the convex lens.
Now, u = 12 cm, f= 20 cm.

Question 9.
An object of size 3.0 cm is placed 14 cm in front of a concave lens of focal length 21 cm. Describe the image produced by the lens. What happens if the object is moved farther from the lens?
Answer:

Thus, the image is virtual, erect, diminished and is formed on the same side of the lens at a distance of 8.4 cm from the lens. If the object is moved away from the lens, the image moves towards the principal focus and goes on decreasing in size.

Question 10.
What is the focal length of a convex lens of focal length 30 cm in contact with a concave lens of focal length 20 cm? Is the system a converging or a diverging lens? Ignore thickness of the lenses.
Answer:
Here, f₁ = 30 cm and f₂ = -20 cm
For the combination of two thin lenses, the focal length of the combination is given by

Since the focal length of the system of lens is negative, therefore, the combination behaves as a diverging lens.

Question 11.
A compound microscope consists of an objective lens of focal length 2-0 cm and an eye-piece 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 (a) the least distance of distinct   vision (25 cm), (b) infinity ? What is the magnifying power of the microscope in each case? (C.B.S.E. 2008)
Answer:
Here, f_n =2.0 cm, f = 6.25 cm,
Distance between object lens and eye piece = 15 cm (a) For the formation of image at the least distance of distinct vision,

Question 12.
A person with a normal near point (25 cm) using a compound microscope with objective of focal length 8.0 mm and an eye-piece of focal length 2.5 cm can bring an object placed 9.0 mm from the objective in sharp focus. What is the separation between the two lenses ? How much is the magnifying power of the microscope?
Answer:

Question 13.
A small telescope has an objective lens of focal length 144 cm and an eye-piece of focal length 6.0 cm. What is the magnifying power of the telescope? What is the separation between the objective and the eye-piece?
Answer:
Here, focal length of objective lens, f₀ = 144 cm Focal

Question 14.
(a) A giant refracting telescope at an observatory has an objective lens of focal length 15m. If an eye-piece 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 x 10⁶m and the radius of the lunar orbit is 3.8 x 10⁸m. (C.B.S.E. 2008, 2011)
Answer:

Question 15.
Use the mirror equation to deduce that
(a) an object placed between f and 2f of a concave mirror produces a real image beyond 2f.
(b) a convex mirror always produces a virtual image independent of the location of the object.
(c) the virtual image produced by a convex mirror is always diminished in size and is located between the focus and the pole.
(d) an object placed between the pole and focus of a concave mirror produces a virtual and enlarged image. (C.B.S.E. 2011)
Answer:

Question 16.
A small pin fixed on the table top is viewed from above from a distance of 50 cm. By what distance would the pin appear to be raised if it is viewed from the same point through a 15 cm thick glass slab held parallel to the table?
Refractive index of glass = 1.5. Does the answer depend upon the location of the slab?
Answer:
Here, t = 15 cm, n = 1.5
The lateral displacement

For small angles of incidence, the answer does not depend upon the location of the slab.

Question 17.
(a) Following figure shows a cross-section of a ‘light pipe’ made of glass fibre of refractive index 1.68. The outer covering of the pipe is made of material of refractive index 1.44. What is the range of the angles of incident rays with the axes of the pipe for which the total internal reflection inside the pipe take place as shown in the figure ?
(b) What is the answer if there is no outer covering of pipe ?

Answer:


Therefore all incident rays in the range 0 to 90°suffer total internal refletion.

Question 18.
Answer the following questions:
(a) You have learned that plane and convex mirrors produce virtual images of objects. Can they produce real images under some circumstances ? Explain.
(b) A virtual image, we always say, cannot be caught on a screen. Yet when we ‘see’ a virtual image, we are obviously bringing it on to the ‘screen’ (i.e., the retina) of our eye. Is there a contradiction?
(c) A diver under water, looks obliquely at a fisherman standing on the bank of a lake, would the fisherman look taller or shorter to the diver than what he actually is ?
(d) Does the apparent depth of a tank of water change if viewed obliquely ? If so, does the apparent depth increase or decreases
(e) The refractive index of diamond is much greater than that of ordinary glass. Is this fact of some use to a diamond cutter ?
Answer:
(a) Yes. They can produce real images if the object is a virtual object as shown in figure.

(b) here is no contradiction in this case. The virtual image of the object acts as an object for the convex lens of our eye and the lens of our eye make a real image of this object on the ratina.
(c) Let AB be the fisherman standing on the bank of the lake. The rays of light from the head of the fisherman bends towards the normal on refraction at the interface separating water and air. The refracted rays appear to come from point B’ instead of point B for the fish. Thus, for a diver the height of the fisherman is AB’ which is greater than his actual height AB.

(d) The apparent depth of a pond of water decreases when viewed obliquely. This is due to the refraction of light from the surface of water.

Question 19.
The image of a small electric bulb fixed on the wall of a room is to be obtained on the opposite wall 3 m away by means of a large convex lens. What is the maximum possible focal length of the lens required for the purpose ?
Answer:
The minimum distance between a real object and its real image formed by a convex lens of focal length/is given by L = 4f

Question 20.
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 locations separated by 20 cm. Determine the focal length of the lens.
Answer:
Let O be the position of object and I is the position of image when lens is at L₁ and then at L₂


Question 21.
(a) Determine the effective focal length of the combination of the two lenses in the question 910 if they are placed 8.0 cm apart with their principal axes coincident. Does the answer depend on which side a beam of parallel light is incident ? Is the notion of effective focal length of this system useful at all ?
(b) An object 1.5 cm in size is placed on the side of the convex lens in the above arrangements. The distance between the object and the convex lens is 40 cm. Determine the magnification produced by the two-lens system, and the size of the image.
Answer:

Question 22.
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.
Answer:
Let the ray of light be incident on the face AB at angle i so that it is totally internally reflected at face AC.

Question 23.
You are given prism made of crown glass and flint glass with a wide variety of angles. Suggest a combination of prisms which will (a) deviate a pencil of white light without much dispersion (b) disperse (and displace) a pencil of white light without much deviation.
Answer:
(a) Angular dispersion produced by two prisms i.e. crown glass and flint glass should be zero in this case


In the combination of prisms, flint glass prism of greater angle may be tried but in any case still this angle will be smaller than the angle of the crown glass prism in opposite order as shown in figure.

Question 24.
For a normal eye, the far point is at infinity and the near point of distinct vision is about 2.5 cm in front of the eye. The cornea of the eye provides a converging power of about 40 dioptres, and the least converging power of the eye lens behind the cornea is about 20 dioptres. From this rough data estimate the range of accommodation {i.e., the range of converging power of the eye-lens) of a normal eye.
Answer:
When the object is placed at infinity, the eye makes use of the least converging power, Therefore, total converging power of cornea and the eye lens = 40 + 20 = 60 dioptre.


Question 25.
Does short-sightedness (myopia) or long-sightedness (hypermetropia) imply necessarily that the eye has partially lost its ability of accommodation ? If not, what might cause these defects of vision ?
Answer:
No. Myopia may arise due to the elongation of the eye ball and hypermetropia may arise due to the decrease in the size of the eye ball even when the eye has the normal ability of accommodation. There is another defect in the eye called presbyopia similar to hypermetropia. However, the causes of presbyopia and hypermetropia are different. Presbyopia arises in elderly persons and is corrected by using a bi-focal lens.

Question 26.
A myopic person has been using spectacles of power – 1.0 dioptres for distant vision. During old age he also needs to use separate reading glass of power +2 dioptres. Explain what may have happened.
Answer:
For -1.0 dioptre, the far point for eyes is 1 m i.e. 100 cm. The near point is 25 cm. The objects lying at infinity are brought at 100 cm from his eyes using the concave lens and the objects lying in between 25 cm and 100 cm are brought to focus using the ability of accommodation of the eye lens. In the old age, this ability of accommodation is reduced and the near point reaches 50 cm from his eyes.


Question 27.
A person looking at a mesh of crossed wires is able to see the vertical wires more distinctly than the horizontal wires. What is this defect due to ? How is such a defect of vision corrected ?
Answer:
This is due to the defect of lenses called astigmatism. The defect arises because of the fact that curvature of the eye-lens and the cornea is not same in different planes. This defect is removed by using cylindrical lens with vertical axis.

Question 28.
A man with normal near point (25 cm) reads a book with small print using a magnifing glass: a thin convex lens of focal length 5 cm.
(a) What is the closest and the farthest distance at which he can read the book when viewing through the magnifying glass ?
(b) What is the maximum and minimum angular magnification (magnifying power) possible using the above simple microscope ?
Answer:
(a) To see the object at a closest distance, the image of object should be formed at the least distance of distinct vision.


Question 29.
A card sheet divided into squares each of size 1 mm² is being viewed at a distance of 9 cm through a magnifying glass (a converging lens of focal length 10 cm) held close to eye.
(a) What is the magnification (image size/object size) produced by the lens ? How much is the area of each square in the virtual object ?
(b) What is the angular magnification (magnifying powers) of the lens ?
(c) Is the magnification in
(1) equal to magnifying power in
(2) ? Explain.
Answer:

(c) Clearly magnification and power magnification are not equal to each other unless the image is located near the least distance of distinct vision, e. v = D.

Question 30.
(a) At what distance should the lens be held from the figure in the above exercise in order to view the squares distinctly with maximum possible magnifying power ?
(b) What is the magnification (image size/object size) in this case ?
(c) Is the magnification equal to magnifying power in this case ? Explain.
Answer:
(a) The magnifying power is maximum if the image is formed at the least distance of distinct point from the eye, i.e., if υ = -25 cm ; Also, f = 10 cm

=> the linear magnification and magnifying power is equal in this case.

Question 31.
What should be the distance between the object in the previous exercise and the magnifying glass if the virtual image of each square in the figure is to have an area 6.25 mm²? Would you be able to see the squares distinctly with your eyes very close to the magnifier?
Answer:


Question 32.
Answer the following questions :
(а)  The angle subtended at the eye by an object is equal to the angle subtended at the eye by the virutal image produced by a magnifying glass. In what sense then does a magnifying glass provide angular magnification?
(b) In viewing through a magnifying glass, one usually positions one’s eyes very close to the lens. Does angular magnification change if the eye is moved back?
(c) Magnifying power of a simple microscope is inversely proportional to the focal length of the lens. What then stops us from using a convex lens of smaller and smaller focal length and achieving greater and greater magnifying power?
(d) Why must both the objective and the eye-piece of a compound microscope have short focal lengths?
(e) When viewing through a compound microscope, our eyes should be positioned not on the eye-piece but a short distance away from it for best viewing. Why? How much should be that short distance between the eye and eyepiece?
Answer:
(a) When magnifying glass is not used, object to be seen clearly is to be placed at 25 cm. However, while using magnifying glass, object can be placed closer to eye than at 25 cm. The closer object has large angular size than the same object placed at 25 cm. It is in this sense that magnifying glass provides angular magnification.
(b) Yes. The angular magnification decreases slightly because angle subtended at eye is somewhat less than the angle subtended at the lens.
(c) The aberrations like spherical and chromatic aberrations start croping up if the convex lens of smaller and smaller focal length is made.
(d) Angular magnification of eye piece is given by (1+D/f_e)  and angular magnification of objective is approximately given by υ/f₀. Clearly for better magnification focal length of eye piece f_e and focal length of objective f_e should be small.
(e) If we position our eye very close to the eyepiece, the whole light will not fall on our eye and the field of view will decrease. So we place our eye a short distance away from the eye-piece to collect the large amount of light refracted through the eyepiece to increase the field of view.

Question 33.
An angular magnification (magnifying power) of 30 X is desired using an objective of focal length 1.25 cm and eye piece of focal length 5 cm. How will you set up the compound microscope ?
Answer:
For the image formed at the least distance of distinct vision, the magnifying power is given by



Question 34.
A small telescope has an objective lens of focal length 140 cm and an eye piece of focal length 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 formed at infinity) ?
(b) the final image is formed at the least distance of distinct vision (25 cm) ?
Answer:

Question 35.
For the telescope described in the last exercise, in 9.34
(a) what is the separation between the objective lens and the eye-piece ?
(b) If this telescope is used to view a 100 m tall tower 3 km away, what is the height of image of the tower formed by objective lens ?
(c) What is the height of final image of the tower if it is formed at 25 cm ?
Answer:
(a) Since the final image is formed at infinity, the distance between the object lens and the eye-piece is f₀ + f_e = 140 + 5 = 145 cm


Question 36.
A cassegrain telescope uses two mirrors as shown in figure. Such a telescope is built with mirrors 20 mm apart. If the radius of curvature of large mirror is 220 mm and the small mirror is 140 mm, where will be the final image of an object at infinity be ?

Answer:

Question 37.
The adjoining figure shows an equiconvex lens (of refractive index 1.50) in contact with a liquid layer on top of a plane mirror. A small needle with its tip on the principal axis is moved along the axis until its inverted image is found at the position of the needle. The distance of the needle from the lens is measured to be 45.0 cm. The liquid is removed and the experiment is repeated. The new distance is measured to be 30.0 cm. What is the refractive index of liquid ?

Answer:
In the presence of the liquid, the distance of the needle from the lens is equal to the focal length f of the combination of the convex lens and the piano concave lens formed by the liquid below it i.e. f = 45 cm. Also n = 1.5
In the absence of the liquid, the distance of the needle and the lens is equal to the focal length of the convex lens only i.e. f = 30 cm
.’.  If f₂ is the focal length of plane concave lens formed

We hope the NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments, help you. If you have any query regarding NCERT Solutions for Class 12 Physics Chapter 9 Ray Optics and Optical Instruments, drop a comment below and we will get back to you at the earliest.

Here we are providing NCERT Solutions for Class 12 English Flamingo Poem 4 A Thing of Beauty. Students can get Class 12 English A Thing of Beauty NCERT Solutions, Questions and Answers designed by subject expert teachers.

A Thing of Beauty NCERT Solutions for Class 12 English Flamingo Poem 4

A Thing of Beauty NCERT Text Book Questions and Answers

A Thing of Beauty Think it out 

Question 1.
List the things of beauty mentioned in the poem.
Answer:
Keats begins the poem by saying that a beautiful thing is a source of everlasting pleasure. It keeps people from being overwhelmed by worldly concerns. Pleasure is the escape from reality. The things of beauty that provide happiness to the soul are the sun, the moon, a bower of trees, daffodils, and clear streams. He also talks about musk-roses that flower in the forest. The magnificence of the heroic tales also inspires us. Keats primarily talks about the beauty of nature in the poem.

Question 2.
List the things that cause suffering and pain.
Answer:
Keats lists the things that cause suffering and pain to human soul. It includes the scarcity of noble people, overcast days, and the unhealthy and miserable ways in which humanity searches for meaning in life.

Question 3.
What does the line, “Therefore are we wreathing a flowery band to bind us to earth” suggest to you?
Answer:
For Keats, each “flowery band” serves as a reminder of the fragility of whatever people have achieved. Beauty signifies a constant battle to guard one’s well-being against sadness and defeat. The poet praises nature and believes that man’s connection with nature was essential in contributing to his happiness.

Question 4.
What makes human beings love life in spite of troubles and sufferings?
Answer:
Keats elaborates through the poem that beautiful experiences helps in overcoming one’s problems. It is because of the eternal pleasures that despite much unhappiness and misery, one continues to love and value life. He exults over how he considers beauty to be immortal. He affirms that it always lifts the pall—the funeral cover—from the coffin of confused misery to provide some light for humanity’s anxious quest for meaning. The pleasure one derives from beautiful objects and experiences dispels any misery.

Question 5.
Why is “grandeur” associated with the “mighty dead”?
Answer:
The word “grandeur” is associated with the “mighty dead” because the splendour of such deeds is inspirational. The legends and stories of martyrs inspire people. Through their legends, the mighty dead continue to live and act as spiritual guides. This enhances the quality of one’s life — divine influence, like an eternal fountain, showers its grace on earthly existence.

Question 6.
Do we experience things of beauty only for short moments or do they make a lasting impression on us?
Answer:
Beauty is a quality in a person or an object that gives intense pleasure or deep satisfaction to the mind. Beauty has inspired many,throughout history. Beauty of nature has been celebrated by humans since time immemorial. Those who cherish the beauty of the earth find reserves of strength that would endure for life. According to Keats, things of beauty give eternal pleasure and erase the misery in our soul. They do not fade away into nothingness, their beauty increases with time.

Question 7.
What image does the poet use to describe the beautiful bounty of the earth?
Answer:
Romantics have praised the majesty of nature in many of the literary works, Keats’ poetry in Endymion, is dominated by a description of the natural world. He uses the images of nature to describe the beautiful bounty of the earth. He specifically refers to the sun, moon, daffodils, and trees, stating that their beauty gives life true meaning and significance. Keats also states how beauty continue to inspire ceaselessly. The poem also illustrates a connection with nature—“flowery band to bind us to the earth”.

A Thing of Beauty Extra Questions and Answers

A Thing of Beauty Short Answer Questions

Question 1.
Explain: “A thing of beauty is a joy forever”.
Answer:
This is the first line of Book 1 of Endymion and one of the most quoted lines in English poetry. It is multi¬layered in meaning. It implies that beauty can create joy in the soul that will last forever. The sights of beauty also decrease the misery of dark and gloomy days. It underlines the power of nature to heal and give happiness.

Question 2.
What are the things that cause suffering and pain to human beings on earth?
Answer:
Keats feels that beauty makes life worth living despite the unhappiness and misery that one goes through on earth. He feels that one encounters days that are sad and depressing; causing gloom, natural sorrow, sadness or dejection.

A Thing of Beauty Value Based Questions

Question 1.
Keats feels that nature binds us to the earth. How does he justify this?
Answer:
According to Keats, man has a strong relationship with nature. Man is bound to the earth despite the gloom and misery that abounds his existence. Beauty relieves man of gloom and misery. The beauty that man sees around him, brings him joy and respite; it “is a joy forever” because of the bond he forges with nature. Like other Romantics, Keats is a believer in the healing powers of nature. He talks about the different elements of nature that surround man and overwhelm him with their magnificence.

Question 2.
Bring out the theme of love and beauty in the poem.
Answer:
Keats, an advocate of beauty, begins the verse by celebrating a thing of value which is always considered beautiful. Hence, despite human misery on earth, man is inexplicably tied to this world. The poet discusses how the beauty of nature brings about joy and drives away sorrow. Keats also discusses how the legends and stories of martyrs instil inspiration in people. The various things of beauty, love and inspiration that the poet celebrates through his poem contribute to the pleasure; he calls these divine influence that, like an eternal fountain, is bestowed upon man.

Give examples from the poem of the following poetic devices.

Rhyme scheme
Rhyming couplets
Metaphor

(a) “will keep
A bower quiet for us”
“Flowery band”

(c) “Some shape of beauty moves away the pall
From our dark spirits.”

(d) “An endless fountain of immortal drink,
Pouring unto us from the heaven’s brink”

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields part of NCERT Solutions for Class 12 Physics. Here we have given. NCERT Solutions for Class 12 Physics Chapter 1 Electric Charges and Fields

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	Physics
Chapter	Chapter 1
Chapter Name	Electric Charges and Fields
Number of Questions Solved	34
Category	NCERT Solutions

Question 1.
What is the force between two small charged spheres having charges of 2 x 10^-7 C and 3 x 10^-7 C placed 30 cm apart in the air?
Answer:

Question 2.
The electrostatic force on a small sphere of charge 0.4 μC due to another small sphere of charge -0.8 μC in air is 0.2 N.
(a) What is the distance between the two spheres?
(b) What is the force on the second sphere due to the first?
Answer:
(a) Force on charge 1 due to charge 2 is given by the relation

Question 3.
Check that the ratio ke²/G m_em_p. is dimensionless. Lookup a Table of Physical Constants and determine the value of this ratio. What does the ratio signify?
Answer:

Thus, the given ratio is a number and is dimensionless. This ratio signifies that electrostatic force between electron and proton is very-very large as compared to the gravitational force between them.

Question 4.
(a) Explain the meaning of the statement ‘electric charge of a body is quantised.’
(b) Why can one ignore quantisation of electric charge when dealing with macroscopic i.e., large-scale charges?
Answer:
(a) Charge occurs in nature as a discrete entity. One packet of charge (least quantity) is called quantum charge. It is represented by ‘e‘. Generally, charge of an electron is represented by ‘e’ and charge of a proton is represented by ‘+ e’. Therefore, the charge possessed by any charged body will be an integral multiple of ± e, ie., ne where n = 1, 2, 3,
∴ q = ± ne
The fact that electric charge collections are integral multiples of the fundamental electronic charge was proved experimentally by Millikan.

(b) While dealing with the large-scale electrical phenomenon, we ignore the quantization of charge because the magnitude of charge of proton and electron is so small. For continuous charge distribution, charge can be accounted in terms of charge density such as linear charge density λ etc. We need not go for individual charges.

Question 5.
When a glass rod is rubbed with a silk cloth, charges appear on both. A similar phenomenon is observed with many other pairs of bodies. Explain how this observation is consistent with the law of conservation of charge.
Answer:
Before rubbing, both the glass rod and the silk cloth are electrically neutral. In other words, the net charge on the glass rod and the silk cloth is zero. When the glass rod is rubbed with silk cloth, a few electrons from the glass rod get transferred to the silk cloth. As a result, the glass rod becomes positively charged and the silk cloth negatively charged. Since the magnitude of positive charge on the glass rod is the same as that of negative charge on the silk, the net charge on the system is zero. Thus the appearances of charge on the glass rod and the silk cloth is in accordance with the law of conservation of charges.

Question 6.
Four-point charges q_A = 2 μC, q_B = -5 μC, or q_c = -2 μC and q_D = -5 μC are located at the corners of a square ABCD of side 10 cm. What is the force on a charge of 1μC placed at the center of the Square?
Answer:
The symmetry of the figure clearly indicates that 1μC charge will experience equal and opposite forces due to equal charges of 2μC placed at A and C. Similarly, 1μC charge will experience equal and opposite forces due to -5 μC charges placed at D and B.

Thus, net force = zero.

Question 7.
(a) An electrostatic field line is a continuous curve.
That is, a field line cannot have sudden breaks. Why not?
(b) Explain why two field lines never cross each other at any point?
Answer:
(a) Electric lines of force exist throughout the region of an electric field. The electric field of a charge decreases gradually with increasing distance from it and becomes zero at infinity (i)e., electric field can’t vanish abruptly. So a line of force can’t have sudden breaks, it must be a continuous curve.

(b) If two lines of force intersect, then there would be two tangents and hence two directions of electric fields at the point of intersection, which is not possible.

Question 8.
Two-point charges q_A = 3μC and q_B = -3 μC are located 20 cm apart in vacuum.
(a) What is the electric field at the midpoint O of the line AB joining the two charges?
(b) If a negative test charge of magnitude 1.5 x 10^-9 C is placed at this point, what is the force experienced by the test charge? (C.B.S.E. 2003)
Answer:
(a) Electric field at the midpoint of the separation between two equal and opposite charges is given by

(b) Force experienced by test charge = q₀^E
= (1.5 x 10^-9) (5.4 x 10⁶)
= 8.1 x 10^-3 N along BA

Question 9.
A system has two charges q_A = 2.5 x 10^-7 C and q_B = -2.5 x 10^-7 C located at points
A : (0,0, -15 cm) and B : (0, 0, + 15 cm), respectively. What are the total charge and electric dipole moments of the system?
Answer:
Clearly, the given points are lying on the z-axis.
Distance between charges, 21

= 15 + 15 = 30 cm = 0.3 m
Total charge = (2.5 x 10-⁷) – (2.5 x 10-⁷) = 0
Dipole moment
= q x 21 = 2.5 x 10-⁷ x 0.3
= 7.5 x 10^-8 C m along negative z-axis.

Question 10.
An electric dipole with dipole moment 4 x 10^-9 Cm is aligned at 30° with the direction of a uniform electric field of magnitude 5 x 10⁴ NC^-1. Calculate the magnitude of the torque acting on the dipole.
Answer:
Using τ = pE sin 0, we get
= (4 x 10-⁹) (5x 10-⁴ sin 30°)
= 2 x 10-⁴ x \(\frac { 1 }{ 2 } \) = 10-⁴ N m 2

Question 11.
A polythene piece rubbed with wool is found to have a negative charge of 3.2 x 10-⁴ C.
(a) (Estimate the number of electrons transferred (from which to which ?)
(b) Is there a transfer of mass from wool to polythene?
Answer:
(a) Using q = ne, we get

(b) Yes, but of negligible amount because mass of an electron is very-very small (mass transferred = m_e x n = 91 x 10-³¹ x 2 x 10¹² = 1.82 x 10^-18 kg).

Question 12.
(a) Two insulated charged copper spheres A and B have their centers separated by a distance of 50 cm. What is the mutual force of electrostatic repulsion if the charge on each is 6.5 x 10–⁷ C ? The radii of A and B are negligible compared to the distance of separation.
(b) What is the force of repulsion if each sphere is charged double the above amount, and the distance between them is halved?
Answer:

Question 13.
Suppose the spheres A and B in Q 1.12 have identical sizes. A third sphere of the same size but uncharged is brought in contact with the first, then brought in contact with the second, and finally removed from both. What is the new force of repulsion between A and B?
Answer:
Charge on sphere A on contact with the third sphere (say C) having no charge is given by

When third sphere, how having charge 3.25 10^-7 C is brought in contact with sphere B, the charge left on sphere B is given by,

Question 14.
Figure shows tracks of three charged particles in a uniform electrostatic field. Give the signs of the three charges. Which particle has the highest charge to mass ratio?

Answer:
Unlike charges attract each other, therefore particle 1 and 2 are negatively charged whereas particle 3 has a positive charge. Particle 3 gets maximum deflection so it has the highest charge (e) to mass (m) ratio because deflection, y α e/m

Question 15.
Consider a uniform electric field \(\overrightarrow { E } \)
= 3 x 10³ \(\hat { i } \)N/C.
(a) What is the flux of this field through a square of 10 cm on a side whose plane is parallel to the yz plane?
(b) What is the flux through the same square if the normal to its plane makes a 60° angle with the x-axis?
Answer:
(a) Electric flux through the square,

Question 16.
What is the net flux of the uniform electric field of Problem 1.15 through a cube of the side 20 cm oriented so that its faces are parallel to the co­ordinate planes?
Answer:
Zero, because the number of field lines entering the cube is equal to the number of field lines coming out of the cube.

Question 17.
Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 x 10³ Nm2/C.
Answer:
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the box? Why or why not?
Ans.
(a) Using φ = φ /ε₀ we get q =φ ε₀
= (8 x 10³) (8.854 x 10^-12)
= 70.8 x 10-⁹ C = 0.07 μC
(b) No, it cannot be said so because there may be an equal number of positive and negative elementary changes inside the box. It can only be said that the net charge inside the box is zero.

Question 18.
A point charge + 10 μC is a distance 5 cm directly above the center of a square of side 10 cm, as shown in the given figure. What is the magnitude of the electric flux through the square? [Hint. Think of the square as one fact of a cube with edge 10 cm.]

Answer:
The charge can be assumed to be placed as shown in the figure.

Question 19.
A point charge of 2.0 μC is at the center of a cubic Gaussian surface 9.0 cm on edge. What is the net electric flux through the surface?
Answer:

Question 20.
A point charge causes an electric flux of -1.0 x 10³ Nm²/C to pass through a spherical Gaussian surface of a 10.0 cm radius centered on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge?
Answer:
(a) The electric flux depends only on the charge enclosed by the Gaussian surface and independent of the size of the Gaussian surface. The electric flux through new Gaussian surface remains same
i.e. -1 x 10³ Nm² C-¹ because the charge enclosed remains same in this case also.
(b) Using φ = q/ε₀, we get q = ε₀ φ = (8.85 x 10-¹²) (-1 x 10³)
i.e. q = -8.85 x 10^-9 C.

Question 21.
a conducting sphere of radius 10 cm has an unknown charge. If the electric field 20 cm from the center of the sphere is 1.5 x 10³ N/C and points radially inward, what is the net charge on the sphere?
Answer:

Question 22.
A uniformly charged conducting sphere of 2.4 m diameter has a surface charge density of 80.4 μC /m².
(a) Find the charge on the sphere.
(b) What is the total electric flux leaving the surface of the sphere? (B.S.E. 2009 C)
Answer:
(a) Charge on the sphere is given by

Question 23.
An infinite line charge produces a field of 9 X 10⁴ N/C at a distance of 2 cm. Calculate the linear charge density
Answer:

Question 24.
Two large, thin metal plates are parallel and close to each other. On their inner faces, the plates have surface charge densities of opposite signs and of magnitude 17.0 x 10^-22 C/m². What is E?

(a) in the outer region of the first plate,
(b) in the outer region of the second plate, and
(c) between the plates?
Answer:

Question 25.
An oil drop of 12 excess electrons is held stationary under a constant electric field of 2.55 x 10⁴ NC^-1 in Millikan’s oil drop experiment. The density of the oil is 1.26 g cm^-3. Estimate the radius of the drop, (g = 9.81 ms^-2 ; e = 1.60 x 10^-19 C.)
Answer:
Charge on the drop,

Question 26.
Which among the curves shown in the figure cannot possibly represent electrostatic field lines?



Answer:
(a) Incorrect, because the field lines should be normal to the surface of a conductor.
(b) Incorrect, because the field lines cannot start from a negative charge.
(c) Correct
(c) Incorrect, because electric field lines cannot intersect with each other.
(d) Incorrect, because electrostatic field lines cannot form a closed loop.

Question 27.
In a certain region of space, the electric field is along the z-direction throughout. The magnitude of the electric field is, however, not constant but increases uniformly along the positive z-direction, at the rate of 10⁵ N C^-1 per meter. What are the force and torque experienced by a system having a total dipole moment equal to 10^-7 C m in the negative z-direction?
Answer:
Suppose the dipole is along the z-axis.


Question 28.
(a) A conductor A with a cavity as shown in figure (a) is given a charge Q. Show that the entire charge must appear on the outer surface of the conductor.
(b) Another conductor B with charge q is inserted into the cavity keeping B insulated from A. Show that the total charge on the outside surface of A is Q + q [Figure (b)].
(c) A sensitive instrument is to be shielded from the strong electrostatic fields in its environment. Suggest a possible way.

Answer:
Select a Gaussian surface lying wholly inside the conductor but very near to the surface of the conductor.
(a) There is no electric field inside the conductor so electric flux through Gaussian surface is zero or in other words, net charge inside the Gaussian surface is zero. Then it can be said that the charge lies outside the Gaussian surface e. on the outer surface of the conductor.
(b) Charge q inside the cavity will induce a charge -q on the inner side of the cavity and thus +q will appear on the outer surface. Thus total charge will be (q + Q).
(c) The instrument should be enclosed in a metallic shell so that the effect of the electrostatic field is cancelled out.

Question 29.
A hollow charged conductor has a tiny hole cut into its surface. Show that the electric field in the hole is (σ/2ε₀) \(\hat { n } \), where \(\hat { n } \) is the unit vector in the outward normal direction, and a is the surface charge density near the hole.
Answer:
Let the tiny hole of the conductor be considered as filled up. The field inside the conductor is zero, whereas outside it is given by

This field is infact due to
(1) field (E₁) due to plugged hole and
(2) field E₂ due to rest of the charged conductor. Inside the conductor, these fields are equal but opposite, whereas outside they are exactly same. i.e.

Question 30.
Obtain the formula for the electric field due to a long thin wire of uniform linear charge density X without using Gauss’s law.
[Hint. Use Coulomb’s law directly and evaluate the necessary integral.]
Answer:
Consider a long thin wire of uniform linear charge density X placed along the X-axis. Let P be a point lying on the y-axis






Question 31.
It is now believed that protons and neutrons (which constitute nuclei of ordinary matter) are themselves build-out of more elementary units called quarks. A proton and a neutron consist of three quarks each. Two types of quarks, the so called ‘up’ quark (denoted by u) of charge + (2/3)e, and the ‘down’ quark (denoted by d) of charge
(-l/3)e, together with electrons build up ordinary matter. (Quarks of other types have also been found which give rise to different unusual varieties of matter.) Suggest a possible quark composition of a proton and neutron.)
Answer:

Question 32.
(a) Consider an arbitrary electrostatic field configuration. A small test charge is placed at a null point (i.e., where E = 0) of the configuration. Show that the equilibrium of the test charge is necessarily unstable.
(b) Verify this result for the simple configuration of two charges of the same magnitude and sign placed a certain distance apart.
Answer:
(a) To prove the result, let us assume that the test charge placed at the null point is in stable equilibrium. If it is so, then on being displaced slightly away from the null point, the test charge should return to its position., It implies that if a closed surface is drawn around the test charge, there will be a net inward flux of the electric field through its surface. According to Gauss law, there cannot be any electric flux through its surface as it does not enclose any charge. Hence our assumption is wrong and the test

(b) For the configuration of the two charges of the same magnitude and sign, the null point is the midpoint of the line joining the two charges. If the test charge is displaced slightly from the null point along the line, it will return back due to the restoring force that comes into the day. But if the charge is displaced slightly from the null – point along normal to the line it will not return. This is because the resultant force due to the configuration of two charges will take it away from the null point. For the test charge to be in stable equilibrium restoring force must come into play, when it is displaced in any direction. Hence the test charge cannot be in stable equilibrium.

Question 33.
A particle of mass m and charge (-q) enters the region between the two charged plates initially moving along x-axis with speed σ_x, (like particle 1 in figure.) The length of plate is L and a uniform electric field E is maintained between the plates. Show that the vertical deflection of the particle at the far edge of the plate is qEL²/(2m υ_x²).
Compare this motion with motion of a projectile in gravitational field discussed in Section 4.10. of Class XI Textbook of Physics.
Answer:
Consider a uniform electric field \(\overrightarrow { E } \) set up between two oppositely charged parallel plates (Figure). Let a positively charged particle having charge +q and mass m enters the region of electric field E at O with velocity E along X-direction.
Step 1.

Force acting on the charge +q due to electric field E is given by
\(\overrightarrow { F } \) =Q \(\overrightarrow { E } \)
The direction of the force is along the direction of
\(\overrightarrow { E } \)and hence the charged particle is deflected accordingly.
Acceleration produced in the charged particle is given by

Step 2.
The charged particle will accelerate in the direction of E . As soon as the particle leaves the region of electric field, it travels due to inertia of motion and hits the screen at point P. Let t be the time taken by the charged particle to traverse the region of electric field of length L. Let y be the distance travelled by the particle along y-direction (i.e. direction of electric field). Using a standard equation of motion,
S = ut + \(\frac { 1 }{ 2 } \) at².
For horizontal motion. S = L, u = υ_x and a = 0.
(∴ no force acts on the particle along x-direction)
From equation (ii), we have

Equation (i) is the equation of a parabola.
Hence a charged particle moving in a uniform electric field follows a parabolic path.

Question 34.
Suppose that the particle in Q 1.33 is an electron projected with velocity
υ_x = 2.0 x 10⁶ ms^-1. If E between the plates separated by 0.5 cm is 9.1 x 10² N/C, where
will the electron strike the upper plate ? ( |e| = 1.6 x 10^-19 C, m_e = 9.1 x 10-³¹ kg.)
Answer:

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NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 6
Chapter Name	Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts

Question 1.
Explain with examples what historians mean by the integration of cults.
Solution :
The integration of cults means that there was composition, compilation and preservation of Puranic texts in simple Sanskrit verse, explicitly meant to be accessible to women and Shudras, who were generally excluded from Vedic learning. The Brahmanas also accepted and reworked the beliefs and practices of these and other social categories. The example of this integration of cults is at Puri, Orissa where the principal deity was identified as Jagannatha (literally, the lord of the world), a form of Vishnu. Here the deity is represented in a very different way. In this case, a local deity, whose image was and continues to be made of wood by local tribal specialists, was recognised as a form of Vishnu. At the same time, Vishnu was visualised in a way, that was very different from that in other parts of the country. Such instances of integration were evident amongst goddess cults as well where the local deities were provided an identity as the wife of the principal male deities – Lakshmi, the wife of Vishnu or Parvati, the wife of Shiva.

Question 2.
To what extent do you think the architecture of mosques in the subcontinent reflects a combination of universal ideals and local traditions?
Solution :
With the arrival of Islam in the Medieval ages, the architecture of Islam also came to India. However, the Arab-cum-Islamic architecture got impacted by the local traditions and rites too. Hence, we see a fusion of the two. This can be further elaborated by the examples of architecture mainly the constructions of the mosques of those days.

Some features of the architecture of mosques are universal. All mosques have orientation towards Mecca. This is manifested in the placement of Mehrab and Minar within a mosque. But at the same time we have influences that can be described only as local influences. A 13th Century mosque in Kerala has a shikhar like roof unlike a normal mosque where it is dome. The Shah Hamdan Mosque in Kashmir is made of Kashmiri woods and its facade is like that of a temple. The Atia Mosque in Bangladesh is made of bricks, though its roof is round. Thus, we can see that the architecture of Mosques is that of fusion.

Question 3.
What were the similarities and differences between the be-shari‘a and ba- shari‘a sufi traditions ?
Solution :
(a) Similarities :

• Both be-sharia and ba-shari‘a sufis turned to asceticism and mysticism and were against the growing materialism of the Caliphate as a religious and political institution.
• They are critical of the dogmatic definitions and scholastic methods of interpreting the Quran and sunna adopted by the theologians.
• Both laid emphasis on seeking salvation through intense devotion and love for God.
• Both sought an interpretation of the Quran on the basis of their personal experience.

(b) Differences :

Be-Shari’s	Ba-Shari’s
(i)   They scorned the Khanqah and took to mendicancy and observed Celibacy.	(i) They organised communities around the khanqah controlled by a teaching master known as Shaikh, Pir or Murshid.
(ii)   They ignored rituals and observed extreme forms of asceticism.	(ii) They observed special rituals of initiation. The initiates took an oath of allegiance, wore a patched garment and shaved their hair. Khanqah was the centre of social life.
(iii)   They were known by different names as Qalandars, Madaris, Malangs, and Haidaris.	(iii) The Chishti silsila is the most important and influential tradition.
(iv)   They deliberately defied the shari‘a i.e., law governing the Muslim community. So they were called as be-shari‘a.	(iv)     They complied with the sharia.

 Question 4.
Discuss the ways in which the Alvars, Nayanars and Virashaivas expressed critiques of the caste system.
Solution :
The Alvars, Nayanars and Virashaivas expressed critiques of the caste system in the following ways :
(a) Alvars and Nayanars :

1. The Alvars and Nayanars protested against the caste system or at least attempted to reform the system. Their bhaktas hailed from diverse social backgrounds ranging from Brahmanas to artisans and cultivators and even from castes considered “untouchable”. Thus, people from all walks of society were welcomed by them.
2. They stated that their compositions were as important as the Vedas. For example, one of the major anthologies of compositions by the Alvars, the Nalayira Divyaprabandham, was described as the Tamil Veda.
3. Tondaradippodi, a Brahmana Alvar, opposed the cast system in the following way :
   “You (Vishnu) manifestly like those “servants” who express their love for your feet, though
   they may be bom outcastes, more than the Chaturvedins who are strangers and without allegiance to your service.”
4. Another saint, a Nayanar, composed the following verse in protest of the caste system : “O rogues who quote the law books, Of what use are your gotra and kula, Just bow to Marperu’s lord (Shiva who resides in Marperu, in Thanjavur, Tamil Nadu) as your sole refuge.”

(b) Virashaiva:

1. Virashaiva or Lingayats challenged the idea of caste and the “pollution” attributed to certain groups by Brahmanas. As a result of it, marginalised people within the Brahmanical order joined this tradition.
2. They questioned the theory of rebirth.
3. The Lingayats encouraged certain practices disapproved in the Dharmashastras, such as post-puberty marriage and remarriage of widows.

Question 5.
Describe the major teachings of either Kabir or Baba Guru Nanak, and the ways in which these have been transmitted.
Solution :

1. The major teachings of Kabir and Baba Guru Nanak are as given below :
   • Teachings of Kabir :
     
     • He described the Ultimate Reality as Allah, Khuda, Hazrat and Pir. He also used terms drawn on Vedantic traditions, alakh (the unseen), nirakar (formless), Brahman and Atman. Terms such as shabda (sound) or shunya (emptiness) were drawn from yogic traditions.
     • There is only one God. He was against the distinction made between gods of different communities.
     • Kabir was against idol worship and Hindu polytheism.
     • He was in favour of the Hindu practice of nam-simaran (remembrance of God’s name).
   • Teachings of Guru Nanak : The message of Baba Guru Nanak is spelt out in his hymns and teachings which are as given below :
     • Baba Guru Nanak believed in nirguna bhakti.
     • He was against the external practices of religions.
     • He rejected sacrifices, ritual baths, image worship, austerities and the scriptures of both Hindus and Muslims.
     • For him, the Absolute or “arab” had no gender or form.
     • He favoured a simple way to connect to the Divine by remembering and repeating the Divine Name.
2. The ways in which the major teachings of Kabir and Baba Guru Nanak were transmitted are as given below –
   • Compositions of the sants were sung by roadside musicians.
   • Baba Guru Nanak would sing his compositions in various ragas while his attendant Mardana played the rabab.
   • Baba Guru Nanak organised his followers into a community. He set up rules for congregational worship (sangat) involving collective recitation.

Question 6.
Discuss the major beliefs and practices that characterised Sufism.
Solution :
After the advent of Islam in the early, middle ages , it saw a new movement in later part. The movement has had great impact and reach in the Indian subcontinent. It is called Sufi movement. The Sufi saints were mystics. Their preachings included:
1. Sufi saints did not subscribe to the theological and rigid interpretations of religious scriptures of Islam. They believed that the interpretation have to be based on individual experiences. This way the theological interpretations became flexible. Further the control of the orthodox religious leaders got weakened. This was a people centric move.
2. They rejected the high sounding rituals. They also emphasised on simplicity in religious traditions and rites.
3. Sufi saints prescribed devotion to Almighty as path to salvation. They even approved of singing and dancing as part of devotion. It is notable that classical Islam has forbidden singing, dancing and any music.
4. The most important theme of Sufi philosophy was that serving people is the true religion. With the objective of serving the poor people they also held Langar. Today also one can go to Ajmer and can partake in the Langar organised on the tomb of Nijammudin Auliya, the great Sufi saint.
5. Sufi saints also emphasised on the equality among people and oneness among all.

Question 7.
Examine how and why rulers tried to establish connections with the traditions of the Nayanars and the sufis.
Solution :
The rulers tried to establish connections with the traditions of the Nayanars and the sufis to claim divine support and proclaim their own power and status.
(a) Rulars and Sufis :
(i) The Turks who had set up Delhi Sultanate also required legitimation from the sufis because they resisted the insistence of the ulama on imposing sharva as state law because they anticipated opposition from their non-Muslim subjects. The sultans, therefore, sought the support of the Sufis who derived their authority directly from God and did not depend on jurists to interpret the sharva.

(ii) The sultans wanted to have the blessings of Sufi saints because it was believed that the Auliya could intercede with God in order to improve the material and spiritual conditions of ordinary human beings. The Sultans of Delhi set up charitable trusts as endowments for hospices and granted tax-free land.

(iii) Akbar maintained connections to get their blessings for new conquests, fulfilment of vows and birth of sons. He used to visit the Dargah of Muinuddin Chishti. He went there 14 times and gave generous gifts, which were recorded in imperial documents. For example, in 1568, he offered a huge cauldron (degh) to facilitate cooking for pilgrims. He also had a mosque constructed within the compound of the dargah.

(b) Rulers and Nayanars :
(i) The Chola kings tried to establish connections with Nayanars and Alvars by building splendid temples that were adorned with stone and metal sculpture to recreate the visions of these popular saints who sang in the language of the people.

(iii) Those kings also introduced the singing of Tamil Shaiva hymns in the temples under royal patronage taking initiative to collect and organise them into a text.

(iii) An inscription suggests that Chola ruler Parantaka I had consecrated metal images of Appar, Sambandar and Sundarar in a Shiva temple. These were carried in processions during the festivals of these saints.

Question 8.
Analyse, with illustrations, why bhakti and sufi thinkers adopted a variety of languages in which to express their opinions.
Solution :
The bhakti and sufi thinkers adopted a variety of languages to express their opinions due to the following reasons :

1. The local languages were used so that people might understand their verses easily.
2. The language of the Vedas was Sanskrit. The Bhakti traditions were against the beliefs and practices of the Vedas. So, it was necessary to use local languages which people might understand and follow them in practice. That is why Kabir’s poems have survived in several languages. Similarly, Baba Guru Nanak expressed his ideas through hymns called “shabad” in Punjabi, the language of the region.
3. The Chishtis too adopted local languages such as Hindavi because poetic compositions were recited in hospices, usually during ‘sama’. In Bijapur, sufi poetry was composed in Dakhani because these were probably sung by women while performing household chores like grinding grain and spinning.

Question 9.
Read any five of the sources included in this chapter and discuss the social and religious ideas that are expressed in them.
Solution :
The period of the Bhakti Movement and Sufi Movement also has many sources that contribute to the history of those days. Some of the major social and religious ideas expressed in the various sources of history are as follows:
1. The first is the architecture. The different types of stupas, temple, monasteries all symbolise different types of religious belief system and practices. Some of them exist as it is and enable us to look into the annals of history of those days. Some of them are in the form of ruins but they also throw light on the, religion and society of those days alike.
2. The next important source of history is the composition of the saints both Bhakti
and Sufi. In terms of content they are religious but they are not the divine textbooks of religion that are sacrosanct. The compilation throws light on the life of common men and village lifestyle. They also impact the music and art of those days.
3. Another very important source of the history of those days is the biographies of the Saints. The biographies include the description of the society and prevalent beliefs and practices. It is notable that such biographies may not be in the written form still they can give insight into the life of those days. It is the story prevalent that I when Kabirdas died, both Hindus and Muslims fought for his dead body later on his body turned into flowers. Some were taken by Muslims and others by Hindus.
This represents that there conflict and collaboration between both Hindus and Muslims of those days.
4. This was also the period of rise of religious leaders who were intermediaries between common men and God. Earlier it was only the Brahmins who got this role. Now many people from other background also joined in. To some extent it acted as the the force that idolised equality and fraternity.
5. The other source is the folklore. They are described in our art forms. It may be dance, paintings, and sculpture and so on. They all talk about the universal brotherhood of mankind and love for one and all.

We hope the NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 6 Bhakti-Sufi Traditions Changes in Religious Beliefs and Devotional Texts, drop a comment below and we will get back to you at the earliest.

NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments are part of NCERT Solutions for Class 12 History. Here we have given NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments.

Board	CBSE
Textbook	NCERT
Class	Class 12
Subject	History
Chapter	Chapter 4
Chapter Name	Thinkers, Beliefs and Buildings Cultural Developments
Number of Questions Solved	9
Category	NCERT Solutions

NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments

Question 1.
Were the ideas of the Upanishadic thinkers different from those of the fatalists and materialists ? Give reasons for your answer.
Solution :
The ideas of the Upanishadic thinkers were different from those of the fatalists and materialists. Many ideas found in the Upanishadas show that people were curious about the meaning of life, the possibility of life sifter death and rebirth. There were debates on these issues. The thinkers were concerned with understanding and expressing the nature of the ultimate reality. The ideas such as the nature of the self and the true sacrifice were discussed in Upanishads.

On the other hand, the fatalists believed that everything was pre-determined. Pleasure and pain could not be altered, lessened or increased in the course of Samsara (transmigration). The materialists did not believe in alms or sacrifice or offerings. For them there was no such thing as this world or the next. A human being is made of four elements. When a human being dies the four elements; return to earth, water, fire and air.

Question 2.
Summarise the central teachings of Jainism.
Solution :
The main teachings of Jainism are as follows:
(i) The entire world is animated. Life exists even in rocks and stones normally considered non-living.
(ii) The principle of non-violence is practised in extreme form in Jainism. No harms should be caused to animals, plants and insects and any other living beings that may include rocks and stones too. This is notable that Jains are forbidden to eat late night lest they kill insects by mistake.
(iii) The cycle of birth and rebirth is shaped through Karma. If one is to escape this cycle of Karma, one must practise ascetism and penance. It is possible when one renounces the world. So one has to live in monastery to attain salvation.
(iv) Jain monks have to take vows to observe the following:
(a) Not to kill anyone
(b) Not to steal anything
(c) Not tell lies
(d) Not to possess property
(e) To observe celibacy.

Question 3.
Discuss the role of the Begums of Bhopal in preserving the stupa at Sanchi.
Solution :
The role of the Begums of Bhopal in preserving the stupa at Sanchi was as given below:

• The French sought Shahjehan Begum’s permission to take away the eastern gateway for a museum in France. Englishmen too wanted to do the same. They were not allowed to do so. The French and the English were allowed to prepare plaster-cast copies only.
• Shahjehan Begum and her successor Sultan Jehan Begum, provided money for the preservation of the ancient site.
• Sultan Jehan funded the building of museum as well as guesthouse where persons like John Marshall lived and wrote his important volumes.
• She funded the publication of the volumes.

Question 4.
Read this short inscription and answer :
In the year 33 of the maharaja Huvishka (a Kushana ruler), in the first month of the hot season on the eighth day, a Bodhisatta was set up at Madhuvanaka by the hhikkhuni Dhanavati, the sister’s daughter of the bhikkhuni Buddhamita, who knows the Tipitaka, the female pupil of the hhikkhu Bala, who knows the Tipitaka, together with her father and mother.
(a) How did Dhanavati date her inscription ?
(b) Why do you think she installed an image of the Bodhisatta ?
(c) Who were the relatives she mentioned ?
(d) What Buddhist text did she know ?
(e) From whom did she learn this text ?
Solution :
(a) Dhanavati dated her inscription as in the year 33 of the maharaja Huvishka (a Kushan ruler), in the first month of the hot season on the eighth day.
(b) During the period of Kanishka, the Buddhism was divided into two branches i.e., Hinayana and the Mahayana. The worship of images of the Buddha and Bodhisattas had become an important part of this branch. So, she installed an image of the Bodhisatta in order to enable its followers to worship it.
(c) The relatives, she mentioned, were her father, mother, sister of her mother bhikkhuni Buddhamita.
(d) Tipitaka.
(e) She learnt this text from bhikkhu Bala.

Question 5.
Why do you think women and men joined the sangha?
Solution :
Women and men joined the sangha because within the sangha, all – workers, slaves, wealthy men, gahapatis and kings – were regarded as equal. It was an organisation of monks who could become teachers of dhamma. They shed their earlier social identities on becoming bhikkhus and bhikkhunis. Not only this, the internal functioning of the sangha was based on the traditions of ganas and sanghas where consensus was arrived at through discussions. If that failed, decisions were taken by a vote on the subject.

Question 6.
To what extent does knowledge of Buddhist literature help in understanding the sculpture at Sanchi?
Solution :
Buddhist literature help us upto some extent in understanding the sculpture at Sanchi. It is important that the sculptures at Sanchi depict the teachings of Buddha only. The teachings of Buddha are captured in the literature.

It is notable that Buddha used to roam around among people , preaching them on his teachings. However, he did not claim supernatural power. He told us that the world is ever changing. It is full of sorrows. Sorrow flows out of desire. Buddha asked the followers to take the middle path, not too much of penance, nor too much of indulgence. The literature of Buddhism is useful for the interpretation of the sculpture at Sanchi. People are shown in different moods and in sorrow. Different stages of life are depicted and so on. Hence, it can be stated that Buddhist literature throws valuable light on the sculptures of the Sanchi.

Question 7.
Figs A and B are two scenes from Sanchi. Describe what you see in each of them, focusing on the architecture, plants and animals, and the activities. Identify which one shows a rural scene and which an urban scene, giving reasons for your answer.

Solution :

1. In the figure A, depiction of animals, dead and alive, hunters with bow and arrow, trees have been made. These are one of the finest depictions. Several animal stories have been depicted at Sanchi, the above scene may be one of them. In the second figure B, the images are being worshipped.
2. The scene in the figure A is a rural scene because it contains animals, hunters with bow and arrow and dead animals too. The second scene is an urban scene because it depicts worshipping of images.

Question 8.
Discuss how and why stupas were built. Describe the structure of stupa with example.
Solution :
About 200 years after the time of Buddha King Asoka erected a pillar at Lumbini. This was to announce the visit of Buddha to this place.Stupas were the mounds put on the bodily remains of the body of Lord Buddha or of any object that was used by him. At the place of stupas such objects were buried. These were places of great respect under the tradition of Buddhism, as they had the relics of Buddha. As per the description of Asokavadana winch a famous Buddhist book, Emperor Asoka gave Buddha’s relic to all major cities. Later on such places stupas were put. The most important stupas are at Sanchi, Bharhut and Saranath.

The structure of a stupa was like a dome and hemisphere. On the top of it, there would be a balcony called harmik. This balcony represented the abode of God. The harmik was covered with an umbrella. There used to be railings around the balcony.

The construction of the stupas was made possible by the contribution of many. On the forefront were the monarchs. The Satvahan Kings offered huge amount for the construction of the stupqs. Apart from the monarchs, merchants, artisans and common men and women also contributed to the construction of the stupas.

Question 9.
Discuss how and why stupas were built.
Solution :
(a) The stupas were built to bury the relics of Buddha such as his bodily remains or objects used by him. These were mounds known as stupas. Since they contained relics regarded as sacred, the entire stupa came to be venerated as an emblem of both the Buddha and Buddhism. According to Ashokavadana, Asoka distributed portions of the Buddha’s relics to every important town and ordered the construction of stupas over them.
(b)

• Inscriptions found on the railing and pillars of stupas record donations made for building and decorating them. Some donations were made by kings such as the Satavahanas.
• Some donations were made by guilds such as that of the ivory workers who financed part of one of the gateways at Sanchi.
• Hundreds of donations were made by women and men who mention their names, sometimes adding the name of the place from where they came, as well as their occupations and names of their relatives.
• Bhikkhus and bhikkhunis also contributed towards building the stupas.

We hope the NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments help you. If you have any query regarding NCERT Solutions for Class 12 History Chapter 4 Thinkers, Beliefs and Buildings Cultural Developments, drop a comment below and we will get back to you at the earliest.