Category: CBSE Sample Papers

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 5 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 5 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Express the relation among cell constant, resistance of the solution in the cell and conductivity of the solution. How is molar conductivity of a solution related to its conductivity? (2)
Answer:
κ = \(\frac{1}{R} \times\left(\frac{l}{A}\right)\)
Where κ = Conductivity
\(\frac{l}{A}\) = Cell Constant
R = Resistance
A_m = \(\frac{\kappa \times 1000}{M}\)
Where A_m = Molar conductivity
κ = Conductivity
M = Molarity of Solution

Related theory-Conductivity: Conductivity of a solution is defined as the conductance of a solution of 1 cm length and having 1 sq. cm as the area of cross-section. It is represented by K. Its unit is S cm^-1

Molar conductivity: Molar conductivity of a solution at a dilution V is the conductance of all the ions produced from one mole of the electrolyte dissolved in V cm³ of the solution when the electrodes are 1 cm apart and the area of the electrodes is so large that the whole of the solution is contained between them. It is represented by ∧_m. Its unit is S cm² mol^-1

Conductivity and molar conductivity of electrolytes increase with increasing temperature.

Question 2.
Study the change below after addition of Tollen’s reagent to the test tube and report the kind of compound present in the test tube ? (2)

Answer:
The compound in the test tube is an aldehyde. When an aldehyde is introduced to the Tollen’s reagent,

Aldehyde is oxidized by the Tollen’s reagent and forms a carboxylic acid. The silver ions present in the Tollen’s reagent are reduced into metallic silver. This is because the reduction of the silver ions into metallic silver form a silver mirror on the test tube This reaction can be written as follows:

Question 3.
Write short notes on the following (any two) (2)
(A) Gabriel phthalimide synthesis
Answer:
Gabriel phthalimide synthesis:
It is used for mainly aliphatic primary amine preparation. Phthalimide reacts with ethanolic potassium hydroxide and produces potassium salt of phthalimide which on heating with alkyl halide followed by alkaline hydrolysis gives corresponding primary amine.

(B) Hoffmann’s bromamide degradation
Answer:
Hoffmann’s Bromamide synthesis:
When amide is treated with Br₂ in aqueous or ethanolic solution of NaOH, degradation of amide results in formation of primary amine.
Primary amine formed contains one carbon less than the number of carbon atoms in amide.

(C) Carbyl amine reaction
Answer:
When aliphatic/aromatic primary amines are heated with chloroform and ale KOH, foul-smelling alkyl isocyanides or carbyl amines are obtained. Secondary or tertiary amines do not give this test.

Related Theory:
Students should write the chemical reaction involved in every name reaction.

SECTION – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
(A) Give simple tests to distinguish between the following pairs of compounds:
(i) Pentan-2-one and Pentan-3-one
(ii) Benzaldehyde and Acetophenone
Answer:
(A) (i) Pentan-2-one and Pentan -3 -one:
Iodoform Test – Pentan – 2- one contains CH₃CO^– group and as such it will give iodoform test with NaOl [NaOH + I₂] while no such group is there in pentan -3- one and so it will not give iodoform test.

(ii) Benzaldehyde is an aromatic aldehyde while acetophenone is a methyl ketone. These may be distinguished as follows.
Iodoform test: Acetophenone, due to the presence of CH₃CO^– group, will give iodoform test with NaOl (NaOH + I₂), while benzaldehyde will not give this test.

C₆H₅CHO + NaOl → No yellow ppt

(B) There are two – NH₂ groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones. Why? (3)
Answer:
Due to resonance one – NH₂ group undergoes or involved in resonance and hence can’t participate in the formation of semicarbazone. Long pair of – NH₂ group is not involved in resonance and is available for nucleophillic attack.

Question 5.
A copper-silver cell is set up. The copper ion concentration in it is 0.10 M. The concentration of silver ion is not known. The cell potential is measured 0.422 V. Determine the concentration of silver ion in the celL
Given : E°_Ag⁺/Ag = + 0.80 V, E°_Cu²⁺/cu = + 0.34 V.
OR
A current was passed for 5 hours through two electrolytic cells connected in series. The first cell contains AuCl₃ and second cell CuSO₄ solution. If 9.85 g of gold was deposited in the first cell, what amount of copper gets deposited in the second cell? Also calculate magnitude of current in ampere.
Given: Atomic mass of Au = 197 amu and Cu = 63.5 amu (3)
Answer:

Related Theory:
Students should not forget to put powers of stoichiometric coefficients on the concentrations of products and reactants in Nernst equation.
OR
According to Faraday’s second law
Equivalent of gold formed = Equivalent of Cu formed

= 0.804 A

Related Theory:
Students often forget to divide Molar mass by F (Faraday) along with n (no. of electrons) for the calculation of Z (gram equivalent).

Question 6.
Give reasons for the following:
(A) pK_b value for aniline is more than that for methylamine.
(B) Ethylamine is soluble in water whereas aniline is not soluble in water.
(C) Primary amines have higher boiling points than tertiary amines.
OR
Three test tubes labelled as A, B and C contain three types of amines i.e., primary, secondary and tertiary. Describe a method which can be used for the identification of the amines. Also write chemical equations of the reaction involved. 3
Answer:
(A) Aniline undergoes resonance and as a result the electrons on the N-atom are delocalized over the benzene ring. Therefore, the electrons on the N-atom are less available to donate whereas in methylamine (due to the +1 effect of methyl group), the electron density on the N-atom is increased. As a result, aniline is less basic than methylamine. Thus, pK_b of aniline is more than that of methylamine.

Explanation:

(B) Ethylamine when added to water forms intermolecular H—bonds with water. Hence, it is soluble in water but aniline can form H-bonding with water to a very small extent due to the presence of a large hydrophobic -C₆H₅ group. Hence. aniline is insoluble in water.

(C) The extent of inter-molecular hydrogen bonding in primary amines is higher than the secondary amines. Secondary amine has only one hydrogen, and tertiary amines have no hydrogen atom for inter-molecular hydrogen bonding whereas primary amine has two hydrogen atom available for inter-molecular hydrogen bonding.
Explanation: Higher the extent of hydrogen bonding, higher ¡s the boiling point.

OR
Hinsberg test is used identication of primary. secondary ad tertiary amine. Benzenesulphonyl chloride (C₆H₅SO₂Cl). which is also known as Hinsbergs reagent, reacts with primary and secondary amines to form sulphonamides.

The reaction of benzenesulphonyl chloride with primary amine yields N-ethylbenzenesulphonyl amide. The hydrogen attached to nitrogen in sulphonomide is strongly acidic due to the presence of strong electron withdrawing sulphonyl group. Hence, it is soluble in alkali.

In the reaction with secondary amine, N, N- diethylbenzenesulphonamide is formed. Since N, N-diethylbenzene sulphonamide does not contain any hydrogen atom attached to nitrogen atom, it is not acidic and hence insoluble in alkali.

Tertiary amines do not react with benzenesulphonyl chloride due to the absence of replaceable hydrogen.

Question 7.
Show that In case of first order reaction, the time required for 99.9% completion is nearly ten times the time required for half completion of the reaction. (3)
Answer:
For first order reaction.

Question 8.
An organic compound with molecular formula C₉H₁₀O forms 2, 4, – DNP derivative, reduces Tollen’s reagent and undergoes Cannizzaro’s reaction. On vigorous oxidation it gives 1, 2-benzene-di- carboxylic acid. Identify the compound.
OR

Answer:
It is on aLdehyde or ketone os it forms 2,4-DNP derivative.
As the compound reduces Tollen’s reagent and undergoes cannizzaro reaction, it is an aldehyde and not a ketone.
On vigorous oxidation, it gives 1, 2-benzenedicarboxylic acid. So. it must have an aLkyL group at o-ortho position with respect to CHO group on the benzene ring.
Motecular formula suggests that it should be 2-ethyl benzoldehyde.

OR

Question 9.

Observe the table and give reasons for the following trends
(A) The second ionization enthalpy vaLues of Cr and Cu are unusually high.
Answer:
Due to extra stability of half filled d orbitals in Cr⁺ (3d₅) and full filled orbitals in Cu⁺ (3d₁₀), second ionization enthalpies for Cr and Cu are very high.

(B) The second Ionization enthalpy of Zn is comparatively low.
Answer:
Second electron in Zn⁺ is to be removed from 4s¹ because it is very less stable so second ionization enthalpy of Zn is low.

(C) The third ionization enthalpy of Mn and Zn are unsually high.
Answer:
Electronic configuration of Mn²⁺ is [Ar]3d¹⁰ 4s⁰ and of Zn²⁺ is [Ar]3d¹⁰ 4s⁰. Due to extra stability of half filled and full filled orbitals, third ionization enthalpies for both the elements are very high.

Question 10.
Explain the following terms with help of one example of each: (3)
(A) Ambidentate ligands
Answer:
Ambidentate ligands: The monodentate ligands which can coordinate with the central atom through more than one site are called ambidentate ligands.
These ligands contain more than one coordinating atoms in their molecules.
For example, NO₂ can coordinate to the metal atom through N or O as follows –

Explanation: Other examples are as follows: coordinate through C or N as

(B) Spectra chemical series
Answer:
Spectrochemical series: The arrangement of Ligands in the increasing order of crystal field splitting is called spectrochemical series. It is an experimentally determined series based on the absorption of light by complexes with different ligands. This is shown below.
I^– < Br^– < SCN^– < Cl^– < F^– < OH^– < Ox^2- < O^2- < H₂O < NCS^– < Py = NH₃ 2- < CN^– < CO.

Related Theory:
Weak Held ligands are those which cause less crystal field splitting. These form high spin complexes. For example, Cl^–, F^–, etc.
Strong field ligands are those which cause greater crystal field splitting. These form low spin complexes. For example, CN^–, NO₂^–, CO.

(C) Heteroleptic complexes
Answer:
Heteroleptic complexes: The complexes in which the metal is bound to more than one kind of donor groups (ligands) are called heteroleptic complexes. Some common examples of heteroleptic complexes are [NiCl₂(H₂O)₄], [CoCl₂(NH₃)₄]⁺, etc

Question 11.
(A) Describe the general trends in the following properties of the first series (3d) of the transition elements:
(i) Number of oxidation states exhibited
(ii) Formation of oxometal ions
(B) The transition metals and many of their compounds act as good catalysts. Give reason.
OR
(A) Assign reasons for the following:
(i) Copper(l) ion is not known to exist in aqueous solutions.
(ii) Both O₂ and F₂ stabilize high oxidation states of transition metals but the ability of oxygen to do so exceeds that of fluorine.
(B) Which of the following cations are coloured in aqueous solutions and why?
Sc³⁺, V³⁺, Ti⁴⁺, Mn²⁺
(At. Nos. Sc = 21, V = 23, Ti = 22, Mn = 25) (3)
Answer:
(A) (i) The number of oxidation states increases upto middle of series i.e. unto +7 and then decreases.
Explanation: The numbers of the oxidation states increase on moving from Sc to Mn. On moving from Mn to Zn, the number of the oxidation states decrease due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the d orbital.
(ii) Oxometal ions are polyatomic ions with oxygen. Example: VO₂⁺, VO₂⁺, TiO²⁺

Explanation: The elements of the first series of transition metals form a variety of oxides of different oxidation states having general formulae MO, M₂O₃, M₃O₆, MO₂, MO₃, etc. All metals except Sc form MO oxides which are ionic in nature. Beyond group 7, no higher oxides except Fe₂O₃ are known. The oxocations stabilise as VO₂⁺, VO₂⁺ and TiO²⁺.

(B) Transition metals and their compounds act as good catalysts due to its ability to show variable oxidation state and form complexes and they provide a suitable surface for reaction to take place.
OR
(A) (i) Cu²⁺(aq) is much more stable than Cu⁺_(aq). This is because although second ionization enthalpy of copper is large but Ahyd (hydration enthalpy) for Cu²⁺_(aq) is much more negative than that for Cu⁺_(aq) and hence it more than compensates for the second ionization enthalpy of copper. Therefore, many copper (I) compounds are unstable in aqueous solution and undergo disproportionation as follows:
2Cu⁺ → Cu²⁺ + Cu
(ii) The ability of O₂ to stabilize higher oxidation states exceeds that of fluorine because oxygen can form multiple bonds with metals.

(B) V³⁺ and Mn²⁺ are coloured, due to the presence of unpaired (d³ in V³⁺ and d⁵ in Mn²⁺) electrons, they can undergo d-d transitions. Others are colourless due to the absence of unpaired electrons and cannot undergo d-d transitions.

Explanation: The absorption of visible light and hence coloured nature of the transition metal cations is due to the promotion of one or more unpaired d-electron from a lower to a higher level within the same d-subshell. This promotion requires small amount of energy available in the visible light. This is called d-d transition.

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the questions that follow:
Colloids occupy an intermediate place between [particulate] suspensions and solutions, both in terms of their observable properties and particle size. In a sense, they bridge the microscopic and the macroscopic. As such, they possess some of the properties of both, which makes colloidal matter highly adaptable to specific uses and functions. Colloid science is central to biology, food science and numerous consumer products.

Colloidal dispersions appear to be homogeneous, and the colloidal particles they contain are small enough (generally between 1-1000 nm) to exhibit Brownian motion, cannot be separated by filtration, and do not readily settle out. But these dispersions are inherently unstable and under certain circumstances, most colloidal dispersions can be “broken” and will “flocculate” or settle out.

Particles composed of ionic or ionizable substances usually have surface charges due to adsorption of an ion (usually an anion) from the solution, or to selective loss of one kinds of ion from the crystal surface. For example, Ag+ ions on the surface of a silver iodide crystal go into solution more readily than the Br~ ions, leaving a negatively – charged surface.

Charged colloidal particles will attract an excess of oppositely – charged counter-ions to their vicinity from the bulk solution, forming a localized “cloud” of compensating charge around each particle. The entire assembly is called an electric double layer. Electric double layers of one kind or another exist at all phase boundaries, but those associated with colloids are especially important
(A) What is the size of colloidal particles?
(B) Name the property of colloidal solutions due to which their particles do not settle down.
(C) What is the cause of charge on colloidal particles?
(D) How is the electrical double layer formed in colloidal solutions? What is the other name given to the double layer?
OR
How is electrokinetic potential is produced between the two layers of charges? Write the term used for this potential. (5 )
Answer:
(A) Size of colloidal particles lies between the size of particle of true solutions and suspensions and their particles size range is generally between 1-1000 nm.

(B) Brownian movement
Explanation: Colloidal particles in a sol are continuously bombarded by the molecules of the dispersion medium on all sides. As a result the sol particles show random or zig-zag movements. This random or zig¬zag motion of the colloidal particles in a sol is called Brownian motion or Brownian movement

(C) The charge on the colloidal particles is due to adsorption of common ions of the electrolyte on the surface of the colloidal particles, e.g., Fe³⁺ from FeCl₃ on the surface of Fe(OH)₃ particles.

(D) One type of the ions (common ion) of the electrolyte are adsorbed on the surface of colloidal particles due to preferential adsorption, it forms a “Fixed layer”. It attracts the opposite ions to form another layer called “diffused layer”.
The double layer of opposite charge thus formed is called Helmholtz electrical double layer.
OR
As the separation of charge is a seat of potential, the charges of opposite signs on the fixed and diffused parts of the double layer result in a difference in potential between these layers.
This potential difference between the fixed layer and the diffused layer of opposite charges is called the electrokinetic potential or zeta potential.

Students can access the CBSE Sample Papers for Class 12 Geography with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Geography Term 2 Set 3 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35
Roll No. ___________

General Instructions:

• Question paper is divided into 5 sections A, B, C, D & E
• In section A, question number 1 to 3 are Very Short Answer type questions. Attempt any 3 questions.
• In section B, question number 4 is Source based question.
• In section C, question number 5 & 6 are Short Answer type questions.
• In section D, question number 7 to 9 are Long Answer type questions.
• In section E, question number 10 is a Map based question.

Section – A
Very Short Answer Questions

Question 1.
State about mineral based industries. (2)
Or
Highlight the characteristics of traditional large scale industrial regions.
Answer:
Mineral based industries use minerals as a raw material. Some industries use ferrous metallic minerals which contain ferrous (iron), such as iron and steel industries but some use non-ferrous metallic minerals, such as aluminium, copper and jewellery industries. Many industries use non-metallic minerals such as cement and pottery industries.
Or
Traditional large scale industrial regions are based on heavy industry. Their characteristics are

• High proportion of employment in manufacturing industry.
• High-density housing, often of inferior type, and poor services.
• Unattractive environment, for example, pollution, waste heaps, and so on.
• Problems of unemployment, emigration and derelict land areas caused by closure of factories because of a worldwide fall in demand.

Question 2.
Explain the concept of medical tourism by providing relevant examples to support your answer. (2)
Answer:
When medical treatment is combined with international tourism activity, it lends itself to what is commonly known as medical tourism. For example, India has emerged as the leading country of medical tourism in the world. World class hospitals located in metropolitan cities cater to patients all over the world. Medical tourism brings abundant benefits to developing countries like India, Thailand, Singapore and Malaysia. Beyond medical tourism, is the trend of outsourcing of medical tests and data interpretation.

Question 3.
Identify some popular tourist regions of the world. (2)
Answer:
Some popular tourist regions of the world are

• The warmer places around the Mediterranean Coast and the West Coast of India.
• Others include winter sports regions, found mainly in mountainous areas, and various scenic landscapes and national parks, which are scattered.
• Historic towns also attract tourists because of the monument, heritage sites and cultural activities.

Section – B
Source Based Question

Question 4.
Read the source given below and answer the following questions by choosing the correct option.
Ferrous minerals such as iron ore, manganese, chromite, etc., provide a strong base for the development of metallurgical industries. Our country is well-placed in respect of ferrous minerals both in reserves and production. India is endowed with fairly abundant resources of iron ore. It has the largest reserve of iron ore in Asia. The two main types of ore found in our country are haematite and magnetite.

It has great demand in international market due to its superior quality. The iron ore mines occur in close proximity to the coal fields in the North-Eastern Plateau region of the country which adds to their advantage. About 95 per cent of total reserves of iron ore is located in the States of Odisha, Jharkhand, Chhattisgarh, Karnataka, Goa, Telangana, Andhra Pradesh and Tamil Nadu. In Odisha, iron ore occurs in a series of hill ranges in Sundergarh, Mayurbhanj and Jhar.

The important mines are Gurumahisani, Sulaipet, Badampahar (Mayurbhaj), Kiruburu (Kendujhar) and Bonai (Sundergarh). Similar hill ranges, Jharkhand has some of the oldest iron ore mines and most of the iron and steel plants are located around them. Most of the important mines such as Noamundi and Gua are located in Poorbi and Pashchimi Singhbhum districts. This belt further extends to Durg, Dantewara and Bailadila.

Dalli, and Rajhara in Durg are the important mines of iron ore in the country. In Karnataka, iron ore deposits occur in Sandur -Hospet area of Ballari district, Baba Budan hills and Kudremukh in Chikkamagaluru district and parts of Shivamogga, Chitradurg and Tumakuru districts. The districts of Chandrapur, Bhandara and Ratnagiri in Maharashtra, Karimnagar and Warangal district of Telangana, Kurnool, Cuddapah and Anantapur districts of Andhra Pradesh, Salem and Nilgiris districts of Tamil Nadu are other iron mining regions. Goa has also emerged as an important producer of iron ore.

i. Which country has the largest reserves of Haematite and Magnetite in Asia? (1)
Answer:
India has the largest reserves of Haematite and Magnetite.

ii. Which states collectively has the highest iron ore reserves in India? (1)
Answer:
Odisha, Jharkhand, Chhattisgarh, Karnataka, Goa, Telangana, Andhra Pradesh and Tamil Nadu collectively has the highest iron ore reserves in India.

iii. Gurumahisani, Sulaipet, Badampahar (Mayurbhaj), Kiruburu (Kendujhar) and Bonai (Sundergarh) are iron ore mines located in which state of India? (1)
Answer:
Gurumahisani, Sulaipet, Badampahar (Mayurbhaj) (Kendujhar) and Bonai (Sundergarh) are iron ore mines located in Odisha.

Section – C
Short Answer Questions

Question 5.
Comment on the emergence of the notion of sustainable development in the world. (3)
Or
Discuss the process of development of the tribal area of Bharmaur region of Himachal Pradesh
Answer:
The notion of sustainable development emerged in the wake of general rise in the awareness of environmental issues in the late 1960s in Western World. It reflected the concern of people about undesirable effects of industrial development on the environment. The publication of ‘The Population Bomb’ by Ehrlich in 1968 and “The Limits to Growth’ by Meadows and others in 1972 further raised the level of fear among environmentalists in particular and people in general.

This sets the scenario for the emergence of new models of development under a broad phrase ‘sustainable development. Concerned with the growing opinion of world community on the environmental issues, the United Nations established a World Commission on Environment and Development (WCED) headed by the Norwegian Prime Minister Gro Harlem Brundtland.

The Commission gave its report (also known as Brundtland Report) entitled ‘Our Common Future’ in 1987. The report defines sustainable development as a “development that meets the needs of the present without compromising the ability of future generations to meet their own needs.”

The process of development of tribal area of Bharmaur of Himachal Pradesh, inhabited by the Gaddi tribe started in 1970s when Gaddis were included among’scheduled tribes’. Under the Fifth Five Year Plan, the tribal sub-plan was introduced in 1974 and Bharmaur was designated as one of the five Integrated Tribal Development Projects (ITDP) in Himachal Pradesh.

This area development plan was aimed at improving the quality of life of the Gaddis and narrowing the gap in the level of development between Bharmaur and other areas of Himachal Pradesh. This plan laid the highest priority on development of transport and communications, agriculture and allied activities, and social and community services.

The most significant contribution of tribal sub-plan in Bharmaur region is the development of infrastructure in terms of schools, healthcare facilities, potable water, roads, communications and electricity. But the villages located along the river Ravi in Holi and Khani areas are the main beneficiaries of infrastructural development. The remote villages in Tundah and Kugti areas still do not have sufficient infrastructure.

Question 6.
Explain how the Indian railways is the largest government undertaking in the country and highlight any of its achievement. (3)
Answer:
Indian Railways is the largest government undertaking in the country as it is one of the longest network in the world. It facilitates the movement of both freight and passengers and contributes to the growth of the economy. Its large size puts a lot of pressure on a centralised railway management system. Thus, in India, the railway system has been divided into 16 zones.

One of the important achievements of the Indian Railways has been the construction of Konkan Railway in 1998. It is 760-km long rail route connecting Roha in Maharashtra to Mangalore in Karnataka. It is considered an engineering marvel. It crosses 146 rivers, streams, nearly 2000 bridges and 91 tunnels. Asia’s largest tunnel which is nearly 6.5 km long, also lies on this route. The states of Maharashtra, Goa and Karnataka are partners in this undertaking.

Section – D
Long Answer Questions

Question 7.
Explain tertiary activities with relevant examples. (5)
Or
Communication is an integral part of exchanging information. Explain.
Answer:
Tertiary activities in an economy are related to the service sector where the service provider has theoretical knowledge and practical training. For example, in need of a medical assistance, an individual can visit a professional such as doctors.

In educational institutions, trained teachers impart education to the students. These services are provided against a payment. Manpower is an important component of the service sector as most of the tertiary activities are performed by skilled labour, professionally trained experts and consultants. In the initial stages of economic development, larger proportion of people works in the primary sector.

However, in a developed economy, the majority of workers get employment in tertiary activity and a moderate proportion is employed in the secondary sector. Tertiary activities include both production and exchange. The production involves the ‘provision’ of services that are ‘consumed’. The output is indirectly measured in terms of wages and salaries. Tertiary activities involve the commercial output of services rather than the production of tangible goods.

They are not directly involved in the processing of physical raw materials. Common examples are the work of a plumber, electrician, technician, launderer, barber, shopkeeper, driver, cashier, teacher, doctor, lawyer and publisher etc.
Or
Communication services involve the transmission of words and messages, facts and ideas. It is an integral part of exchanging information in the following ways

→ Telecommunications is a form of communication services. It has revolutionised communications because of the speed with which messages are sent. The time reduced is from weeks to minutes. Besides, the recent advancements like mobile telephony have made communications direct and instantaneous at any time and from anywhere.

→ Radio and television also help to relay news, pictures, and telephone calls to vast audiences around the world and hence they are termed as mass media. They are vital for advertising and entertainment.

→ Newspapers are able to cover events in all comers of the world. Satellite communication relays information of the earth and from space.

→ The internet has truly revolutionised the global communication system.

Question 8.
What are highways and border roads? Provide relevant examples of highways from India and other countries. (5)
Answer:
Highways are metalled roads connecting distant places. They are constructed in a manner for unobstructed vehicular movement. Highways are usually 80 m wide, with separate traffic lanes, bridges, flyovers and dual carriageways to facilitate uninterrupted traffic flow. In developed countries, every city and port town is linked through highways.

Border roads are the roads laid along international boundaries. They play an important role in integrating people in remote areas with major cities and providing defence. Almost all countries have such roads to transport goods to border villages and military camps. Examples of highways from India and other countries are

→ In India, there are many highways linking the major towns and cities. For example, National Highway No. 7 (NH 7), connecting Varanasi with Kanya Kumari, is the longest in the country.

→ The Trans Canadian Highway links Vancouver in British Columbia (west coast) to St. John’s City in Newfoundland (east coast) and the Alaskan Highway links Edmonton (Canada) to Anchorage (Alaska).

→ The Trans Continental Stuart Highway connects Darwin (north coast) and Melbourne via Tennant Creek and Alice Springs in Australia.

Question 9.
State two important man-made canals which are important for global commerce. (5)
Answer:
The Suez and the Panama Canals are two vital man-made navigation canals or waterways which serve as gateways of commerce for both the eastern and western worlds. The Suez Canal This canal had been constructed in 1869 in Egypt between Port Said in the North and Port Suez in the South linking the Mediterranean Sea and the Red Sea.

It gives Europe a new gateway to the Indian Ocean and reduces direct sea-route distance between Liverpool and Colombo compared to the Cape of Good Hope route. It is a sea-level canal without locks which is about 160 km and 11 to 15 m deep. A railway follows the canal to Suez, and from Ismailia there is a branch line to Cairo.

A navigable fresh-water canal from the Nile also joins the Suez Canal in Ismailia to supply fresh-water to Port Said and Suez. The Panama Canal This canal connects the Atlantic Ocean in the east to the Pacific Ocean in the west. It has been constructed across the Panama Isthmus between Panama City and Colon by the US government which purchased 8 km of area on either side and named it the Canal Zone. The Canal is about 72 km long and involves a very deep cutting for a length of 12 km.

It has a six lock system and ships cross the different levels (26 m up and down) through these locks before entering the Gulf of Panama. It shortens the distance between New York and San Francisco by 13,000 km by sea.

Section – E
Map Based Question

Question 10.
(a) On a political map of India, mark and indicate the following features (Attempt any 5) (1 × 5 = 5)
(i) Hazaribagh copper mine
(ii) Manganese mine in Karnataka
(iii) Bilaspur bauxite mine
(iv) Bailadila iron-ore mine
(v) Bokaro coal mine
(vi) Jamnagar oil refineries
Answer:

Students can access the CBSE Sample Papers for Class 12 Geography with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Geography Term 2 Set 1 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35
Roll No. ___________

General Instructions:

• Question paper is divided into 5 sections A, B, C, D & E
• In section A, question number 1 to 3 are Very Short Answer type questions. Attempt any 3 questions.
• In section B, question number 4 is Source based question.
• In section C, question number 5 & 6 are Short Answer type questions.
• In section D, question number 7 to 9 are Long Answer type questions.
• In section E, question number 10 is a Map based question.

Section – A
Very Short Answer Questions

Question 1.
Explain the prominent features of traditional large scale industrial regions. (2)
Answer:
The prominent features of traditional large scale industrial regions are

• High proportion of employment in manufacturing industry.
• Problems of unemployment, emigration and ruined land areas caused by closure of factories.

Question 2.
Define Tertiary Activities. (2)
Answer:
Tertiary activities involve the commercial output of services rather than the production of tangible goods. Here skilled people are needed who provide services to people against payment of fees. Common examples of tertiary activities are the work of a plumber, electrician, technician, launderer, barber shopkeeper, driver, cashier, teacher, doctor, lawyer, publisher, etc.

Question 3.
Differentiate between retail trading and wholesale trading. (2)
Or
Differentiate between departmental stores and chain stores.
Answer:
The business activities concerned with the sale of goods directly to the consumers are known as retail trading, while wholesale trading constitutes bulk business through numerous intermediary merchants except retail stores. Or In departmental system, there is one store with many departments while in chain system, there are several shops and the shops are scattered over several places. Departmental stores deals in a large variety of goods while chain stores deals only in one commodity.

Section – B
Source Based Question

Question 4.
Read the source given below and answer the following questions by choosing the correct option.
‘Manufacturing Industry’
Manufacturing literally means ‘to make by hand. However, now it includes goods ‘made by machines’. It is essentially a process which involves transforming raw materials into finished goods of higher value for sale in local or distant markets. Conceptually, an industry is a geographically located manufacturing unit maintaining books of accounts and records under a management system.

As the term industry is comprehensive, it is also used as synonymous with ‘manufacturing’. When one uses terms like ‘steel industry’ and chemical industry’, one thinks of factories and processes. But there are many secondary activities which are not carried on in factories such as what is now called the ‘entertainment industry’ and ‘Tourism industry’, etc. So for clarity, the longer expression ‘manufacturing industry’ is used.

i. What do you understand by the term Manufacturing? (1)
Answer:
The conversion of raw material into finished goods through application of power is called manufacturing

ii. “Manufacturing involves value addition.” Explain with any one suitable example. (1)
Answer:
Manufacturing is a process which involves transforming raw materials into finished goods of higher value. Example, raw cotton is transformed to more valuable threads and then threads are further transformed into more valuable fabric.

iii. Name two industries which provide services rather than production of goods. (1)
Answer:
The ‘entertainment industry’ and ‘tourism industry’ are two industries which provides services rather than production of goods.

Section – C
Short Answer Questions

Question 5.
Road transport plays a vital role in the promotion of trade and tourism in the world. Support this statement with three suitable arguments. (3)
Answer:
Road transport plays a vital role in the promotion of trade and tourism in the world. It can be supported with the following examples

1. Modern society requires speedy and efficient road transport system to assist in the production, distribution and consumption of goods.
2. The opening-up of tourist areas has been aided by improvement in transport facilities. Road transport provides link between the origin and destination of tourist places.
3. Road transport facilitates the movements of holiday makers, business travellers, people visiting relatives and those undertaking educational and health tourism.

Question 6.
Satellites and computers have brought revolutionary changes in the present life of the people. Elaborate the statement with three suitable examples. (3)
Or
Evaluate the role and importance of roads in the economic development of India?
Answer:
Satellites and computers have brought revolutionary changes in the present life of people. It can be elaborated with the help of following examples

i. Satellites are a part of human lives in many ways. Every time we use a cell phone to call a friend, send SMS or watch a popular programme on cable television, we are using satellite communication.

ii. Today weather forecasting through television is helpful in saving the loss of life and property. It is possible through artificial satellites which are successfully installed in the earth’s orbit and these can connect even the remote corners of the globe with limited on-site verification.

iii. As many people use the Internet via computers each year, cyberspace will expand the contemporary economic and social space of humans through email, e-commerce, e-leaming and e-governance.

Or
Importance and role of roadways in the economic development in India are

→ Roads provides better connectivity between rural and urban areas and hence advancement of rural India occurs.
→ Road connectivity is very important in linking intra-state urban areas and inter-states for better transportation of various goods and services and hence for overall development in terms of saving time, money etc and reduction in pollution and CO₂ emission level.
→ Rural economy would be connected with urban economy through a dense road network.
→ A good road network helps to improve import and export in the country.

Section – D
Long Answer Questions

Question 7.
How do quaternary services differ from tertiary services? Give three reasons why the service sector in developing and developed countries is growing faster than the manufacturing sector? (5)
Answer:
The activities concerning knowledge such as education, information, research and development and more intellectual activities where the major task is to think, research and develop ideas are quaternary activities. For example, as a medical transcription and a CEO of an MNC work under quaternary sector. Tertiary activities are related to service sector.

These activities include the segment of economy that provides services to consumers such as transport, communication, education, health, etc. Here skilled people are needed who provide services to people against payment of fees. Such as doctors, lawyers, publishers, electrician, technician and so on. The service sector in developing and developed countries is growing faster than the manufacturing sector due to following reasons

• The rising per capita income in both, developed and developing countries has generated proportionately larger increase in the many kinds of services such as trade, transport, real estate financial banking etc.
• There is also demand for educational services at all levels with the increase in the demand for literacy and computer skills at the workplace.
• The increasing living standard of people has resulted in rising demand for services such as transport, tourism, sports, etc.

Question 8.
How is the use of plastic bags harmful for environmental degradation? Evaluate it by citing suitable reasons. (5)
Or
The disposal of urban waste has become a serious concern for the local authorities. Analyze the statement with suitable examples.
Answer:
Use of plastic bags is harmful for environmental degradation in the following ways

• The major impact of plastic bags on the environment is that it takes many years to for them to decompose.
• Waste from plastic bags poses serious environmental danger to human and animal health. If plastic bags are not properly disposed of, they can impact the environment by causing littering and stormwater drain blockages.
• Animals often confuse the plastic bags for food and consume them, therefore blocking their digestive processes.
• Plastic buried deep in landfills can drain harmful chemicals that spread into groundwater.
• If plastic bags are burned, they release a toxic substance into the air causing ambient air pollution. Or The disposal of urban waste has become a serious concern for the local authorities in the following ways
• Tons of waste come out daily in metropolitan cities and are burnt. The smoke released from the waste pollutes the air.
• Lack of sewers or other means to dispose of human excretes safely and the inadequacy of garbage collection sources adds to water pollution.
• The concentration of industrial units in and around urban centres gives rise to a series of environmental problems.
• Dumping of industrial waste into rivers is the major cause of water pollution. The solid waste generation continues to increase in both absolute and per capita in cities.

The improper disposal of solid waste attracts rodents and flies which spread diseases. The thermal plants release a lot of smoke and ash in the air. For example, a plant producing 500MW electricity releases 2000 tons of ash which is difficult to manage.

Question 9.
Attaining sustainable development in the command area requires major thrust upon the measures to achieve ecological sustainability. Highlight the measures proposed to promote sustainable development in the command area of Indira Gandhi Rajasthan Canal (5)
Answer:
The measures proposed to promote sustainable development in the Indira Gandhi Rajasthan Canal Command Area are

→ Strict implementation of water management policy is required. The canal project envisages protective irrigation in Stage-1 and extensive irrigation of crops and pasture development in Stage-II.

→ In general, the cropping pattern shall not include water intensive crops. It shall be adhered to and people shall be encouraged to grow plantation crops such as citrus fruits. The CAD (Common Area Development) programmes such as lining of water courses, land development and leveling and warabandi system (equal distribution of canal water in the command area of outlet) shall be effectively implemented to reduce the conveyance loss of water.

→ The areas affected by water logging and soil salinity shall be reclaimed.

→ The eco-development through afforestation, shelterbelt plantation and pasture development is necessary particularly in the fragile environment of Stage II.

→ The social sustainability in the region can be achieved only if the land allottees having poor economic background are provided adequate financial and institutional support for cultivation of land.

Section – E
Map Based Question

Question 10.
On the outline map of India indicate and mark the following features (Attempt any 5). (1 × 5 = 5)
(a) Iron ore mines of Chhattisgarh
(b) Oil refinery located in Uttar Pradesh
(c) The largest lignite coal mines
(d) Easternmost terminal of East-West Corridor
(e) Southernmost metropolitan city connected by Golden Quadrilateral
(f) Oldest copper mines of Rajasthan
Answer:

Students can access the CBSE Sample Papers for Class 12 Economics with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Economics Term 2 Set 2 with Solutions

Time allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This is a Subjective Question Paper containing 13 questions.
• This paper contains 5 questions of 2 marks each, 5 questions of 3 marks each and 3 questions of 5 marks each.
• 2 marks questions are Short Answer Type Questions and are to be answered in 30-50 words.
• 3 marks questions are Short Answer Type Questions and are to be answered in 50-80 words.
• 5 marks questions are Long Answer Type Questions and are to be answered in 80-120 words.
• This question paper contains Case/Source Based Questions.

Question 1.
Distinguish between stock and flow variables.
OR
Distinguish between consumption goods and capital goods. (2)
Answer:
Any economic variable which is measured at a point of time is known as stock, e.g. capital, etc. Whereas, any economic variable measured during a period of time is known as flow, e.g. income, etc.
OR
Goods purchased for satisfaction of wants are consumer goods whereas Final goods that are used for producing other goods are capital good.

Question 2.
If the Marginal Propensity to Save is 20% and is constant at all levels of income and the autonomous consumption is ₹ 100 crores, construct consumption function of the given hypothetical economy.
OR
S = – 100 + 0.2 Y is the saving function in an economy. Investment expenditure is ₹ 5,000. Calculate the equilibrium level of income. (2)
Answer:
MPS = 20% = 0.2
and C = ₹100 crore
MPC = 1- MPS = 0.8
C = C + MPC(Y)
C = 100 + 0.8Y
OR
S = – 100 + 0.2 Y
At equilibrium S = I
I = 5000
5000 = – 100 + 0.2 Y
0.2 Y = 5100 Y
= ₹25500

Question 3.
Justify the following statement with valid reason.
“Ex-ante Aggregate Demand is always equal to Ex-ante Aggregate Supply.”
Answer:
Ex-ante Aggregate Demand is equal to Ex-ante Aggregate Supply only when the economy is in equilibrium. At under-employment equilibrium level, when Ex-ante Aggregate Demand falls short of Ex-ante Aggregate Supply, it will lead to accumulation of unplanned inventories. Hence the producer will reduce employment leading to reduction in output and income till the two forces becomes equal to each other and vice versa.

Question 4.
Discuss briefly the concept of informalisation of workforce’, in the context of Indian economy.
OR
The following table shows the population and worker population ratio for India in 1999-2000. Can you estimate the workforce (urban and total) for India? (2)

Region	Estimates of Population (in crores)	Worker Population Ratio	Estimated No. of Workers (in crores)
Rural	71.88	41.9	71.88/100 x 41.9 = 30.12
Urban	28.52	33.7	?
Total	100.40	39.5	?

Answer:
In the recent years, India has witnessed an unprecedented shift of the workforce from the formal sector to informal sector. This process whereby, the proportion of informal worker in the total workforce increases is known as informalisation of workforce.
OR
Estimated number of workers (in crores) for urban = (28.52 x 33.7)/100 = 9.61 crores Total workforce = 30.12 + 9.61 = 39.73 crores

Question 5.
“It is necessary to create employment in the formal sector rather than in the informal sector.” Defend or refute the given statement with valid arguments. (2)
Answer:
The given statement is true and can be defended on the following arguments:

• Formal sector of employment provides greater job security as compared to informal sector of employment.
• Under the formal sector of employment people are entitled to better social security benefits.

Question 6.
Giving reason, explain how the following should be treated in estimation of national income:
(A) Expenditure by a firm on payment of fees to a chartered accountant
(B) Payment of corporate tax by a firm
(C) Purchase of refrigerator by a firm for own use.
OR
If the Nominal Gross Domestic Product = ₹4,400 and the Price Index (base = 100) = 110. Calculate Real Gross Domestic Product. (3)
Answer:
(A) Payment of fees to chartered accountant by a firm is intermediate cost to the firm and, therefore not included.

(B) Payment of corporate tax by a firm is a transfer payment and thus not included.

(C) Purchase of a refrigerator by a firm for own use is investment expenditure and thus included.
OR

Question 7.
Analyse the trends in sectoral distribution of workforce in India on the basis of data:
Trends in Employment Pattern (Sector-wise), 1972-2012 (in %)
Read the following text carefully and answer question number 8 and 9 given below:

A comparison of progress by India and China since their founding

Although India has made progress in every field in the last seven decades of its independence, the truth is that the neighboring country China is now far ahead of India on most social indicators, such as health, education, safe drinking water, electricity, livable housing conditions, and sanitation facilities. The whole world marvels at the unprecedented economic and social advancement China has achieved after the introduction of the landmark economic reform and opening-up policy in the late 1970s.

In the 1980s, India and China were on the same level as per the standard of the significant economic development scale, GDP. But now 40 years later, where is China and where is India? Although the Indian political establishment at the center always claims the country is a superpower, the truth is that what India has achieved pales in comparison with what China has accomplished in its poverty alleviation programme, iconic infrastructure projects and landmark innovations in science and technology over 70 years.

Since the outbreak of the novel coronavirus last year, the pandemic has been disturbing the economic activities in India, more and more people are falling below the poverty line every day. According to estimates by Pew Research, more than 75 million Indians have plunged back into poverty and more than 25 million people have lost their jobs since the start of 2021. As per the survey of the Centre for Monitoring Indian Economy published on July 8, 2021, in The Indian Express, around 150-199 million additional people will fall into poverty this year. It means an overall increase in poverty by 15¬20 percent, making around half of the country’s population poor.

On the other hand, despite being the first victim of the COVID-19 pandemic, China added 11.86 million new urban jobs across the nation in 2020. Over 60 million urban jobs were created in the past five years in China, according to Premier Li Keqiang’s 2021 Government Work Report.

https://news.cgtn.com/news/2021-08-19/A- comparison-of-progress-by-India-and-China- since-their-founding–12PM4eJjMlw/index.html
Answer:
The given data indicates that over the given period, the proportion of workforce in primary sector has gone down rapidly. Whereas, the employment share of both secondary sector and the services sector has increased. While the share of secondary sector (between 1993-94 and 2011-12) has gone up by approximately 9%, the corresponding figure for services sector has gone up by approximately 7%.

Question 8.
Is China growing faster than India? Elucidate. (3)
Answer:
Currently, China’s economy is estimated to be five times larger than India’s. China also continues to have higher economic growth rates each year as well as a higher overall GDP. Although India has made progress in every field in the last seven decades of its independence, the truth is that the neighboring country China is nowfarahead of India on most social indicators, such as health, education, safe drinking water, electricity, liveble housing conditions, and sanitation facilities. Although the Indian political establishment at the center always claims the country is a superpower, the truth is that what India has achieved pales in comparison with what China has accomplished in its poverty alleviation programme, iconic infrastructure projects and landmark innovations in science and technology over 70 years.

Question 9.
Analyse the impact of COVID-19 pandemic on India vis-a-vis the China in terms of poverty. (3)
Answer:
India has been hard-hit by the Covid-19 pandemic. The pandemic has been disturbing the economic activities in India, more and more people are falling below the poverty line every day. According to estimates by Pew Research, more than 75 million Indians have plunged back into poverty and more than 25 million people have lost their jobs since the start of 2021. An overall increase in poverty by 15-20 percent, making around half of the country’s population poor.

On the other hand, despite being the first victim of the COVID-19 pandemic, China added 11.86 million new urban jobs across the nation in 2020. Over 60 million urban jobs were created in the past five years in China

Question 10.
Sale of petrol and diesel cars is rising particularly in big cities. Analyse its impact on gross domestic product and welfare. (3)
Answer:
Final sale of cars raises GDP, because final sales are final products. Cars provide convenience in transportation but at the same time it causes traffic jams, air pollution and noise pollution reducing the welfare of the people. Pollution has bad effects on the health of the people.

Question 11.
Discuss the adjustment mechanism in the following situations:
(A) Aggregate demand is lesser than Aggregate Supply.
(B) Ex-Ante Investments are greater than Ex-Ante Savings. (5)
Answer:
(A) When Aggregate Demand is lesser than Aggregate Supply (AD<AS), buyers are planning to buy lesser goods and services than what producers are planning to produce.

It will lead to rise in planned inventories above the desired level. The producers in turn will produce less, which will reduce the income level i.e. AS. This process will continue till AD becomes equal to AS.

(B) Ex-ante investments are greater than ex-ante saving (I > S) means buyers are planning to buy more output as to what producers are planning to produce. It will lead to fall in planned inventories below the desired Level. As a result the producers will raise production, leading to increase in income till savings becomes equal to investments.

Question 12.
(A) Calculate sales from the following:

(B) State the four major sectors in an economy according to the macroeconomic point of view.
OR
(A) From the following data, calculate net value added at factor cost.

(B) What are the four factors of production and what are the remuneration to each of these called? (5)
Answer:
(A) Sales = Net value added at factor cost + Depreciation + Indirect taxes + Intermediate cost – Change in stock
= 560 + 60 + 60 + 1,000 – (-) 30
= ₹ 1710 lakh

(B) An economy is classified into the following four categories:

• Household sector
• Producing sector
• Government sector
• Rest of the world

OR

(A) NVA_FC = Sales + Change in stock – Intermediate consumption – Depreciation – Net indirect taxes
= 300 + 20 – 120 – 30 – 15
= ₹ 155 crores

(B)

Sr.	Factors of Production	Remuneration
(i)	Land	Ren
(ii)	Labour	Wages
(iii)	Capital	Interest
(iv)	Entrepreneurship	Profit

Question 13.
(A) “Recently Indians have drifted away from the traditional knowledge and practices and caused large scale damage to environment”. Explain how, adopting the traditional practices can be helpful in achieving the objective of sustainable development?

(B) Children below 15 years of age and the old people beyond 60 years are not considered while counting the number of unemployed. Do you agree? (5)
Answer:
(A) The given statement is quite appropriate. Indian traditional practices were environment friendly and worked as complementary to the system and not its controller. The traditional agriculture system, healthcare system, housing, transport etc. were intrinsically environment friendly. The traditional practices used natural products and processes which are more or less free from side effects. For example by adopting medicinal plants/products we can conserve the resources and achieve the objective of sustainable development.

(B) Yes, I agree that children below 15 years of age and the old people beyond 60 years are not considered while counting the number of unemployed because these persons cannot become part of labour force even when they are able and willing to work.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 11 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 11 for Practice

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Define the following terms – (Any two):
(A) Molar conductivity
(B) Inert electrolyte in salt bridge
(C) Cell constant (2)

Question 2.
Aldol condensation of a ketone in the presence of a dilute alkali gives 4-Hydroxy – 4 – methyl pentan – 2 – one. Write the structure of the ketone and its IUPAC name. (2)

Question 3.
Give reasons for the following:
(A) Iodoform is obtained when methyl ketones react with hypoiodite but not with iodide.
(B) Hydrazones of aldehydes and ketones are not prepared in highly acidic medium. (2)

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
Write the products formed when ethanal reacts with the following reagents:
(i) CH₃MgBr and then H₃O⁺
(ii) Zn-Hg/conc. HCl
(iii) C₆H₅CHO in the presence of dilute NaOH
OR
Write the chemical equations to illustrate each of the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro reaction
(iii) Hell-Volhard-Zelinsky reaction (3)

Question 5.
Answer the following questions –
(A) Why is the third ionisation energy of Manganese (Z = 25) is unexpectedly high?
(B) Silver (Ag) has completely filled d-orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element?
(C) Mention the name of the element among lanthanoids known to exhibit a +4 oxidation state. (3)

Question 6.
For the first-row transition metals the enthalpy of atomisation values are:

Assign reason for the following:
(A) Transition elements have higher values of enthalpies of atomisation.
(B) The enthalpy of atomisation of zinc is the lowest in 3d – series.
(C) Transition metals have high melting points.
OR
Answer the following questions:
(A) Explain briefly how +2 state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?
(B) In what way is the electronic
configuration of the transition elements different from that of the non-transition elements? 3
(a) A chelating agent has two or more than two donor atoms to bind to a single metal ion. Give one example of such type of ligand. Write its structure also.
(B) What kind of isomerism exists between [Cr(H₂O)₆]Cl₃ (violet) and [Cr(H₂O)5Cl)Cl₂. H₂O (greyish- green)?
(C) Why are low spin tetrahedral complexes not formed? (3)

Question 8.
Consider the following figure and answer the following questions.

(A) Cell ‘A’ has E_cell = 2 V and cell ‘B’ has E_cell= 1.1 V which of the two cells ‘A’ or ‘B’ will act as an electrolytic cell. Which electrode reactions will occur in this cell?
(B) If cell ‘A’ has E_cell = 0.5 V and cell ‘B’ has E_cell = 1.1 V, what will be the reactions at anode and cathode?
OR
Calculate E°_cell and ∆_rG° for the following reaction at 25 °C:
A²⁺_(aq) + B⁺_(aq) → A⁵⁺_(aq) + B_(s)
Given K_c = 10₁₀, 1F = 96500 C mol_-1 (3)
(A) Why is Fe(OH)₃ colloid positively charged, when prepared by adding FeCl₃ to hot water?
(B) On the basis of Hardy-Schulze rule, explain why the coagulating power of phosphate is higher than chloride?
(C) Why does leather get hardened after tanning? (3)

Question 10.
Identify the missing reagent/product in the following reactions:

OR
An organic compound A (C₂H₃N) is used as a solvent of choice for many organic reactions because it is not reactive in mild acidic and basic conditions. Compound A on treatment with Ni/H₂ forms B. When B is treated with nitrous acid at 273 K, ethanol is obtained. When B is warmed with chloroform and NaOH, a foul-smelling compound C formed. Identify A, B and C. Write all the chemical reactions involved. (3)

Question 11.
Account for the following:
(A) Aniline cannot be prepared by Gabriel phthalimide synthesis,
(B) The C-N-C bond angle in trimethyl amine is 108°.
(C) The pKb value of benzeneamine is 9.33 while that of ammonia is 4.75. (3)

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the following questions:
Number of molecules which must collide simultaneously to give product is called molecularity. It is equal to sum of coefficient of reactants present in stoichiometic chemical equation.
For reaction, m₁A + m₂B → product
Molecularity = [m₁ + m₂]
In complex reaction each step has it own molecuiarity which is equal to the sum of coefficients of reactants present in a particular step. Molecuiarity is a theoretical property. Its value is any whole number. Number of concentration terms on which rate of reaction depends is called order of reaction or sum of powers of concentration terms present in the rate equation is called order of reaction.

If rate equation of reation is:
Rate = k × C_A^m1. C_B^m2
Then order of reaction = m₁ + m₂
In simple reaction, order and molecuiarity are same.
In complex reaction, order of slowest step is the order of over all reaction. This step is known as rate determining step. Order is an experimental property. Its value may be zero, fractional or negative.
(A) Write two differences between order and molecuiarity of a reaction?
(B) What will be the molecuiarity of the following reaction?
6FeSO₄ + 3H₂S0₄ + KClO₃ → KCl + 3Fe₂(S0₃)₂ + 3H₂O
(C) The rate of reaction, A + 2B -> products, is given by – d[A] / dt = k[A] [B]². If B is present in large excess, what will be the order the reaction?
(D) The rate of a certain reaction is given by , rate = k [H⁺]ⁿ. The rate increases 100 times when the pH changes from 3 to 1. What will be the order (n) of the reaction?
OR
In a chemical reaction A + 2B → products, when concentration of A is doubled, rate of the reaction becomes 4 times and concentration of B alone is doubled rate continues to be the same. What will be the order of the reaction? (5)

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 10 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 10 for Practice

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
Calculate ∧⁰_m for acetic acid –
Given that ∧° (HCl) = 426 S cm² mol^-1
∧° (NaCl) = 126 S cm² mol^-1
∧° (CH3COONa) = 91 S cm² mol^-1 (2)

Question 2.
Define the following terms – (Any two)
(A) Order of a reaction
(B) Rate law
(C) Elementary reaction (2)

Question 3.
What products will be formed on reaction of propanal with 2-methyl propanal in the presence of NaOH? Write the name of the reaction also. (2)

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
Complete the following reaction sequence:

OR
Compound ‘A’ was prepared by oxidation of compound ‘B’ with alkaline KMnO₄. Compound ‘A’ on reduction with lithium aluminium hydride gets converted back to compound ‘B’. When compound ‘A’ is heated with compound ‘B’ in the presence of H₂S0₄ it produces fruity smell of compound ‘C to which family the compounds ‘A’, ‘B’ and ‘C’ belong to? (3)
(A) Redraw the diagram to show the direction of flow of current.
(B) Write the reactions taking place at anode and cathode.
(C) Construct the Nernst equation for the given Electro chemical cell. (3)

Question 5.
Observe the diagram given below and answer the questions that follow:

(A) Redraw the diagram to show the direction of flow of current.
(B) Write the reactions taking place at anode and cathode.
(C) Construct the Nernst equation for the given Electrochemical cell. (3)

Question 6.
For the reaction: 2A + B → A₂B the rate = klA] [B]² with k = 2.0 x 10^-6 mol L^-1 s. Calculate the initial rate of the reaction when [A] = 0.1 mol L^-1, [B] = 0.2 mol L^-1. CalcuLate the rate of reaction after [A] is reduced to 0.06 mol L^-1.
OR
The decomposition of NH₃ on platinum surface is zero order. What are the rates of production of N₂ and H₂ If k = 2.5 × 10³ mol L^-1 s^-1? (3)

Question 7.
What happens when:
(A) A colloidal solution undergoes persistent dialysis.
(B) River water meets sea water.
(c) Alum is applied on cuts during bleeding
OR
Answer the folLowing questions:
(A) Why is it important to have clean surface in surface studies?
(B) Why is chemisorption referred to as activated adsorption?
(C) What type of solutions are formed on dissolving different concentrations of soap in water? (3)

Question 8.
(A) Out of the following elements, identify the element which does not exhibit a variable oxidation state?
Chromium (Cr), Cobalt (Co), Zinc (Zn)
(B) Calculate the magnetic moment of Cu²⁺ (Z = 29 ) on the basis of “spin-only” formula.
(C) Out of Fe and Cu, which one would exhibit higher melting point? (3)

Question 9.
(A) Arrange the following complexes in the increasing order of conductivity of their solution:
[Co(NH₃)₃Cl₃], [CO(NH₃)4Cl₂]Cl,
[CO(NH₃)₆]Cl₃, [CO(NH₃)₅Cl]Cl₂
(B) On the basis of crystal field theory explain why Co (III) forms paramagnetic octahedral complex with weak field ligands whereas it forms diamagnetic octahedral complex with strong field ligands. (3)

Question 10.
Using crystal field theory, draw energy level diagram and write electronic configuration of the central metal ion of [Co(H₂0)]²⁺.
OR
Using valence bond theory, explain the following in relation to the complex [Mn(CN)₆]^3-:
(i) Type of hybridisation.
(ii) Inner or outer orbital complex.
(iii) Magnetic behaviour. (3)

Question 11.
(A) Alkenes (>C = <) and carbonyl compounds (>C = O), both contain a π bond but alkenes show electrophilic addition reactions whereas carbonyl compounds show nucleophilic addition reactions. Explain.
(B) Why is there a large difference in the boiling points of butanal and butan-1 -ol? (3)

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the following passage and answer the questions that follow: (5)
Amines are classified as primary, secondary and tertiary amines. Primary amines cannot be obtained by ammonolysis of alkyl halide because we will get mixture of 1° 2° and 3° amines. Cyanides, on reduction give primary amines where as isocyanides on reduction give seconday amines. Nitro compounds, on reduction also give primary, amines. Primary amines react with CHCl₃ and KOH to form foul smelling isocyanide. They react with HNO₂ and liberate N₂ gas. They react with Hinsberg’s reagent to form salt soluble in KOH.

Secondary amine form yellow oily compounds with HN0₂ and salt formed with C₆H₅SO₂Cl is insoluble in KOH. 3° amines form salt soluble in water with HNO₂ but does not react with C₆H₅SO₂Cl.
(A) Name two reducing agents which can be used to prepare primary amines from nitro compounds.
(B) Write the isomer of C₃H₉N which does not react with Hinsberg’s reagent.
(C) Primary amines cannot be prepared by ammonolysis of alkyl halides. Give reason.
(D) Write structures along with the IUPAC names:
(i) the amide which gives propanamine by Hoffmann bromamide reaction.
(ii) the amine produced by the Hoffmann degradation of benzamide.
OR
(i) What is the product when C₆H₅CH₂NH₂ reacts with HNO₂?
(ii) Why does ethanol have higher boiling point than ethanamine?

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 4 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 4 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
(A) Hydrazones of aldehydes and ketones are not prepare in highly acidic medium.
Answer:
Hydrazine becomes protonated in the very acidic media and hence is unable to serve as a nucleophile. As a result, extremely acidic media is not used to prepare aldehydes and ketones.

(B) Benzaldehyde gives a positive test with Tollens reagent but not with Fehlings solution. (2)
Answer:
Under normal circumstances, aldehydes that lack alpha hydrogens cannot form an enolate do not produce a positive test using Fehling’s Solution, which is a weaker oxidising agent than Tollen’s reagent.

Question 2.
Rate constant k for a first order reaction has been found to be 2.54 × 10^-3 sec^-1. Calculate its \(\frac{3}{4}\) th life, (log 4 = 0.6020)
OR
The rate constant for a reaction of zero order is A is 0.0030 mol L^-1 s^-1. How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M? (2)
Answer:
For first order reaction:
t = \(\frac{2.303}{k}\)log \(\frac{a}{a-x}\)

For \(\frac{3}{4}\)th life: a – x = a – \(\frac{3}{4}\)a
On substitution these values,

OR
For a zero order reaction,
K = 0.0030 mol L^-1 s^-1
[A]₀ = 0.10M, [A] = 0.075M
K = \(\frac{[A]_{0}-[A]}{t}\)
t = \(\frac{0.10-0.075}{0.0030}\) = 8.33 s

Question 3.

Observe the two solutions (a) and (b), solution (b) scatters light while solution (a) does not. What do you conclude about the nature of the two solutions? (2)
Answer:
When a beam of light passes through a colloid, the colloidal particles present in the solution do not allow the beam to completely pass through. The light collides with the colloidal particles and is scattered (it deviates from its normal trajectory, which is a straight line). This scattering makes the path of the light beam visible, hence Solution (b) is a colloidal solution so that path of light becomes visible while solution (a) is true solution which allows the light to pass through it.

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
Answer the following questions:
(A) How may methyl bromide be preferentially converted to methyl isocyanide?
Answer:
By carbylamine reaction:

Explanation:
The carbylamine reaction is also known as Hoffmann’s isocyanide synthesis. The carbylamine reaction involves synthesis of an isocyanide when primary amine, chloroform, and a base are reacted. The reaction involves conversion of dichlorocarbene which acts to be an intermediate in the reaction.

(B) Arrange the following compounds in increasing order of solubility in water:
C₆HSNH₂, (C₂H₅)₂NH, C₂H₅NH₂
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂
Explanation:
Primary amines are less soluble than tertiary amines as primary amines can form hydrogen bonds with water but tertiary amines cannot.

(C) Why is an alkylamine more basic than ammonia? (3)
Answer:
Due to electron releasing inductive effect (+1) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

Question 5.
(A) On the basis of crystatfield theory, write the electronic configuration for d⁴ ion is ∆₀ < P.
Answer:
t³ _2g e¹_g ∆₀< P, the configuration will be t_{2g, eg} and it is in the case of weak field Ligands and high spin complex will be formed.

(B) Write the hybridization and shape of [Ni(CN)₄]^2-. (Atomicnumber of Ni = 28)
Answer:
dsp², square planar.

Related Theory:
The Crystal Field Stabilization Energy (CFSE) derived from the splitting of the d orbital energies is greater for square planar arrangement. In a tetrahedral arrangement, no pair of ligands is pointing at one specific d orbital. Thus, for first-row atoms such as Ni²⁺ (a efi species), tetrahedral arrangement will be pre-ferred if the ligands are large and weak-field. If the ligands are small (more rod-like) and are strong-field, the planar arrangement will be preferred.

(C) Which of the following is more stable complex and why?
[Co(NH₃)₆]³⁺ and [Co(en)₃]³⁺ (3)
Answer:
[Co(en)₃]³⁺ is more stable than [Co(NH₃)₆]³⁺ as [Co(en)₃]³⁺ is a metal chelate due to presence of bidentate ligand ethylene diamine ligand. Metal chelates are more stable than complexes containing only monodentate ligands.

Related Theory:
When a ligand attaches to the metal ion in a manner that forms a ring, then the metal- li-gand association is found to be more stable. In other words, complexes containing chelate rings are more stable than complexes without rings. This is known as the chelate effect.

Question 6.
Give reason for the following:
(A) Sky appears blue
(B) Colloidal medicines are more effective
(C) Alum is used in purfication of water
OR
Give factors governing colour of colloidal solutions. (3)
Answer:
(A) Sky appears blue in colour due to scattering of light by colloidal particles. It is called Tyndall effect. It is the phenomenon in which the particles of gas or liquid get uniformly distributed throughout the body of the solid.

(B) Medicines are more effective in their colloidal state because colloids have a larger surface area. Thus, they get easily assimilated, absorbed and digested .and thus medicines in colloidal form are easily absorbed by the body tissues.

(C) Alum helps in purifying water by aluminum ion which coagulates the mud particles. Because of the high charge on the aluminum ion, it helps the small particles of suspended solids to group together until they are of such size that they sink and can be removed.
OR
The colour of the colloidal solution depends on

• the wavelength of the light scatter by the dispersed particles,
• size of the dispersed particles
• nature of the dispersed particles and the manner in which it is viewed.

Example:
Finest gold sol is red in colour and as the size of the particle keeps increasing its colour changes to blue, then purple and finally gold.

Question 7.
Explain the following observations giving an appropriate reason for each:
(A) The enthalpies of atomization of transition elements are quite high.
(B) There occurs much more frequent metal- metal bonding in compounds of heavy transition metals (i.e.) 3rd series).
(C) Mn²⁺ is much more resistant than Fe²⁺ towards oxidation.
OR
Observe the table below:

(A) How is the variability in oxidation states of transition metals different from that of the non transition metals? Illustrate with examples.
(B) What may be the stabLe oxidation state of the transition element with the following d electron configurations in the ground state of their atoms: 3d³,3d⁵3d⁸ and 3d⁴? (3)
Answer:
(A) In transition elements, there are large number of unpaired electrons in their atoms, thus they have a stronger inter atomic interaction and thereby stronger bonding between the atoms. So they have high enthalpies of atomization.

(B) It is due to presence of unpaired electrons in the (n-1)d shell have tendency to make strong metallic bond by covalent nature of bonds.

(C) The 3d orbital in Mn²⁺ is half-filled and is more stable compared to Fe²⁺ having 6 electrons in the 3d orbital. Mn²⁺ prefer to lose an electron or get oxidised whereas Fe²⁺ will readily lose one electron or get oxidised. Therefore, Mn²⁺ is much more resistant than Fe²⁺ towards oxidation.
OR
(a) In case of transition elements, the variability in the oxidation state is due to participation of (n-1) d orbitals and ns orbitals.
The oxidation states differ by unity. Thus, V shows +2, +3, +4 and +5 oxidation states and Mn shows +2, +3, +4, +5, +6 and +7 oxidation states.
On the other hand, the variable oxidation states shown by some p block elements differ by two units. Thus, tin shows +2 and +4 oxidation state and indium shows +1 and +3 oxidation states.

(b) Up to Mn, the maximum stable oxidation state corresponds to sum of s and d electrons. After Mn, there is decrease in the stability of higher oxidation states. The stable oxidation states of the transition elements with the following d electron configurations in the ground state of their atoms are as shown.
3d³4s²: +5
3d⁵4s¹: +1, +6
3d⁵4s²: +2, +7
3d⁸4s²: +2 .
3d⁴4s²: +1, +6
The reactions can be represented at anode and at cathode in the following ways:

Question 8.
Two half-reactions of an etectrochemical cell are given below:
MnO^–_4(aq) + 8H⁺_(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O
(I), E° = 1.51 V
Sn²⁺_(aq) → Sn⁴⁺_(aq) + 2e^–, E° = + 0.15 V.
Build a redox reaction equation from the two half-reactions and calculate the cell potential from the standard potentials also predict if the reaction is reactant or product favoured.
OR
Conductivity of 0.00250 M acetic acid is 8.00 × 10^-5 cm^-1. Calculate its molar conductivity. If ∆₀ for acetic acid is 300.0 S cm² mol^-1 what ¡s its dissociation constant? (3)
Answer:
At anode (oxidation):
[Sn²⁺ → Sn⁴⁺_(aq) + 2e^–] × 5 (E° = 0.15 V)

At cathode (reduction):
MnO^–_4(aq)(aq) + 8H⁺ _(aq) + 5e^– → Mn²⁺_(aq) + 4H₂O _(l)] × 2 (E° = + 1.51 V)

Cell reation:
MnO^–_4(aq)(aq) + 16H⁺ + 5Sn²⁺ → 2Mn²⁺ + 5Sn⁴⁺ + 8H₂O

NOW,
E°_cell = E°_cathode – E°_anode
= 1.51 – 0.15 = + 1.36 V

∴ Positive value of E°cell favours formation of product.
OR
Molar Conductivity

Question 9.
Give the structure of A, B and C in the following reactions:

OR
How will you convert the following:
(A) Nitrobenzene into aniline
(B) Ethanoic acid into methanamine
(C) Aniline into N-phenylethanamide (Write the chemical equations involved) (3)
Answer:

OR
(A) Nitrobenzene into aniline

(B) Ethanoic acid into methanomine

(C) Aniline into N-phenytethanamide

Question 10.
For a reaction

The proposed mechanism is as given below:
1. H₂O₂ + I^– → H₂O + IO^– (slow)
2. H₂O₂ + IO^– → H₂O + I^– + O₂ (fast)
(i) Write rate law for the reaction.
(ii) Write the overall order of reaction.
(iii) Out of steps (1) and (2), which one is rate-determining step? (3)
Answer:
(i) To find the rate expression, slow step is to be considered as rate determining step, which is step (l) for this reaction. Then according to the rate law, Rate = k[H₂O₂] [I^–]

(ii) Since, two reagents are involved in the rate-limiting step, thus overall rate of reaction is proportional to the concentration of those reagents, that is, 2.

(iii) The slowest step in this reaction is called the rate-determining step, that is step (1).

Question 11.
(A) Give one chemical test as evidence to show that [Co(NH₃)₅Cl] S0₄ and [CO(NH₃)₅(S0₄)]Cl are ionisation isomers.
(B) [NiCl₄]^2- is paramagnetic while [Ni(CO)₄] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)
OR
(A) Draw figure to show the splitting of d orbitals in an octahedral crystal field.
(B) What will be the correct order for the wavelengths of absorption in the visible region for the following:
[Ni(N0₂)₆]^4-, [Ni(NH₃)₆]²⁺, [Ni(H₂0)₆]^2- (3)
Answer:
(A) When [Co(NH₃)₅(S0₄)]Cl is treated with silver nitrate solution, a white precipitate of AgCl is formed. But [Co(NH₃)₅Cl] S0₄ does not give white ppt with AgN0₃ solution.
[Co(NH₃)₅S0₄]Cl + AgN0₃ → Agd (White ppt)
[CO(NH₃)₅ Cl]S0₄ + AgN0₃ → No (white ppt)

(B) In [NiCl₄]^2- complexion, nickel is in +2 oxidation state and the configuration is 3d⁸. Since the molecule is tetrahedral, it involves sp³ hybridisation as shown below:

The molecule is paramagnetic because it contains two unpaired electrons.
In [Ni(CO)₄], nickel is in 0 oxidation state and has the configuration 4s² 3d⁴ or 3d¹⁰. The molecule is tetrahedral and involves sp3-hybridisation as given below:

Each CO donates a pair of electrons forming four Ni-CO bonds. The compound is diamagnetic since it contains no unpaired electron.
OR

The splitting of the d orbitals in an octahedral field takes place in such a way that d_x²-y², d_z² experience a rise in energy and form the e_g level, while d_xy d_zy and d_zx experience a fall in energy and form the t_2g level

(B) The wavelengths of absorption in the visible region will be in the order:
[Ni(N0₂)₆]^4- > [Ni(NH₃)₆]²⁺ > [Ni(H₂0)₆]^2-

Explanation:
The central metal ion in all three complexes is the same. Therefore, absorption in the visible region depends on the ligands. The order in which the CFSE values of the ligands increases in the spectrochemical series is as follows:
H²O < NH³ < NO^2-

Thus, the amount of crystal-field splitting observed will be in the following order:
∆⁰(H²O) < ∆⁰(NH³) < ∆⁰(N0² )
(More the energy, less the wavelength)

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the questions that follow:
Functional groups related to the carbonyl group include the -CHO group of an aldehyde, the -CO- group of a ketone, the -CO₂H group of a carboxylic acid, and the -C0₂R group of an ester. The carbonyl group, a carbon-oxygen double bond, is the key structure in these classes of organic molecules: Aldehydes contain at least one hydrogen atom attached to the carbonyl carbon atom, ketones contain two carbon groups attached to the carbonyl carbon atom, carboxylic acids contain a hydroxyl group attached to the carbonyl carbon atom, and esters contain an oxygen atom attached to another carbon group connected to the carbonyl carbon atom. All of these compounds contain oxidized carbon atoms relative to the carbon atom of an alcohol group.

Aldehydes and ketones are commonly used in nature in conjunction with other functional groups. They are mainly present in microorganisms or plants and substances such as cinnamaldehyde (cinnamon bark), vanillin (vanilla bean), Citra (lemongrass), helminthosporium (a fungal toxin), carvone (spearmint and caraway), camphor (camphor trees).
For the manufacture of acetic acid and pyridine derivatives, acetaldehyde is used to a great degree.
In perfumes, beauty products, and colourants, benzaldehyde is used. It is applied and often used as a bee repellent to give almond flavour to food items.

Acetone is used in the kitchen as a nail polish remover and paint thinner.
It is used for chemical peeling and acne procedures in medicine.
(A) Name two naturally occurring aldehydes with their sources.
(B) How are ketones and aldehydes different?
(C) Write one use each of benzaldehyde and propanone.
(D) Explain the mechanism of a nucleophilic attack on the carbonyl group of an aldehyde or a ketone.
OR
Aldehydes are more reactive than ketones towards nucleophilic reagents. Explain. (5)
Answer:
(A) Cinnamaldehyde (cinnamon bark), vanillin (vanilla bean),

(B) Hydrogen atom is bound to an aldehyde’s carbonyl carbon, while that of a ketone is attached to two alkyl or aryl groups. The aldehyde C-H bond makes them readily oxidisable (they are strong reducing agents).

(C) Benzaldehyde is used, in perfumes, beauty products, and colourants. Propanone is used as a nail polish remover and paint thinner.

(D) Mechanism of a nucleophilic attack on the carbonyl group:

is a polar group in which carbon acquires positive charge and O acquires negative charge due to more electronegativity of oxygen. The Nu” attacks on carbon and forms a tetrahedral intermediate and then electrophile attacks on oxygen and forms a compound.

OR
+I effect: Due to smaller +I effect of one alkyl group in aldehydes as compared to larger +I effect of two alkyl groups, the magnitude of positive charge on the carbonyl carbon is more in aldehydes thdn in ketones. As a result nucleophilic addition reactions occur more readily in aldehydes than in ketones.

Steric hinderance: Due to presence of a H-atom on the carbonyl group, aldehydes can be more easily oxidised than ketones. Ketones are more sterically hindered due to the presence of two bulky alkyl groups.

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 3 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

• There are 12 questions in this question paper with internal choice.
• SECTION A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• SECTION B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• SECTION C- Q. No. 12 is case based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed.

Section – A
(Section A-Question No 1 to 3 are very short answer questions carrying 2 marks each.)

Question 1.
(A) Devise the correct cell representation for the given cell reaction?
Zn + H₂SO₄ → ZnS0₄ + H₂
Answer:
The Correct representation of the cell is , Zn|Zn²⁺ || H⁺, H₂ | Pt

Caution:
In cell representation oxidation is represented in left part and the reduction in night. It has to be remembered that H2 electrode has Pt too in it.

(B) Why is it not possible to measure single electrode potential? (2)
Answer:
It is not possible to measure single electrode potential because the half cell containing a single electrode cannot exist independently, as the charge cannot flow on its own in a single electrode.

Question 2.
Answer the following questions- (any two)
(A) You see a small packet inserted in a vitamin bottles or leather products, what is its role?
Answer:
Little packets of silica gel are found in all sorts of products because silica gel is a desiccant it adsorbs and holds water vapour. In leather products and foods like pepperoni, the lack of moisture can limit the growth of mold and reduce spoilage, If a bottle of vitamins contained any moisture and were cooled rapidly, the condensing moisture would ruin the pills hence silica gel is added to adsorb moisture.

(B) How is a delta created?
Answer:
Rivers are negatively charged colloidal solutions of silicates dispersed in water and sea water contain positively charged calcium and magnesium ions. When the two meet coagulation occurs and a delta is formed by deposition of particles.

Related Theory:
Coagulation it is a process of aggregating together the colloidal particle so as to change them into large sized particles which ultimately setties as precipitate. The coagulation is generally brought about by the addition of electrolytes.

(C) Critical temperatures of N₂, SO₂ and CH₄ are 126, 230,110K respectively. Arrange them in increasing order of adsorption on the surface of activated charcoal. (2)
Answer:
Extent of adsorption is directly proportional to critical temperature. Hence the order of extent of adsorption of the given gases is: SO₂ > CH₄ > N₂

Related Theory:
The critical temperature of a substance can be defined as the highest temperature at which the substance can exist as a liquid. At temperatures above the critical temperature, the substance in question (in its vapour/gaseous state) can no longer be liqufied, regardless of the amountt pressure applied to it.

Question 3.
(A) Give the product for:

Answer:
NaBH₄ does not reduce ester group. It is mild reducing reagent which reduces aldehyde and ketone into their respective alcohols.

(B) Which among the following will undergo connizaro reciction?
Benzaldehyde, Acetone, Formaldehyde (2)
Answer:
Benzaldehyde and formaldehyde undergo cannizaro reaction as these compounds do not contain alpha hydrogen

Related Theory:
An alpha carbon is the first carbon that is joined to the functional group, in the case of aldehydes and ketones, a functional group is a carbonyl group. The functional group is a carbonyl group, the functional group is responsible for the formation of alpha hydrogen.

Section – B
(Section B-Question No 4 to 11 are short answer questions carrying 3 marks each.)

Question 4.
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 × 10³ ohm. Calculate its resistivity, conductivity and molar conductivity. (3)
Answer:
A = πr² = 3.14 × (0.5)² cm² = 0.785 cm2 = 0.785 × 10^-4 m²
l = 50 cm = 0.5 m

If we want to calculate the values of different quantities in terms of ‘m’ instead of ‘cm’.

Question 5.
The following graphs represents concentration of reactants versus time for a first-order reaction.
(A) What is the value of slope and intercept?
(B) Calculate the rate constant if the half-life of a first-order reaction was found to be 10 min at a certain temperature.

OR
Show that time required for 99% completion is twice the time required for the completion of 90% reaction. (3)
Answer:
(A) In [A] – In [A]₀ = -kt
Rearrange to solve for [A] to obtain one form of the rate law; ln[A]₀ = ln[A]₀ – kt

This can further be arranged into y = mx +b form;
ln [A] = -kt + ln[A]₀

The equation is a straight line with slope m will be equal: to negative of rate constant mx = -kt so slope = -k and y-intercept b will be equal to b= ln[A]₀

(B) For first order reaction
k = \(\frac{0.693}{t_{1 / 2}}\), where t_1/2 = 600 s = 10 min

Question 6.
Observe the experimental set up below and answer the following questions:

(A) What happens when an electric current is passed through a colloidal solution?
Answer:
On passing electric current through a sol, colloidal particles start moving towards oppositely charged electrode where they lose their charge and get coagulated (electrophoresis).

(B) What will happen if this movement of particles is prevented?
Answer:
When the movement of particles is prevented, it is observed that the dispersion medium starts to move in the electric field. This is called electroosmosis.

Related Theory
Electrophoresis, which is one of the electrokinetic phenomena observed in colloidal systems, is the motion of charged colloidal particles in a liquid medium under an applied electric field.

(C) Name the process and write its one application. (3)
Answer:
Process is electrophoressis by which lyophobic colloids can be coagulated.

Question 7.
Answer the following question:
(A) What is Tollen’s reagent? Write one usefulness of this reagent.
(B) An organic compound A has the molecular folmula C₈H₁₆O₂. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produced B. C on dehydration reaction gives but-l-ene. Write equations for the reactions involved. (3)
Answer:
Ammonical silver nitrate solution is called Tollen’s reagent. [Ag (NH3)2]OH Uses: It is used to test aldehydes. Both aliphatic and aromatic aldehydes reduce Tollen’s reagent to shining silver mirror. It is also used to distinguish aldehydes from ketones.

(B) The compound A with molecular formula is Butyl butanoate.

Question 8.Compare the following complexes with respect to structural shapes of units, magnetic behaviour and hybrid orbitals involved in units:
[CO(NH₃)₆]⁺³, [Cr(NH₃)₆]⁺³, Ni(CO)₄ (At. nos.: Co = 27, Cr = 24, Ni = 28
OR
What is meant by crystal field splitting energy? On the basis of crystal field theory, write the electronic configuration of d4 terms of t2g and eg in an octahedral field when (3)
(i) ∆₀ > p
(ii) ∆₀ < p
Answer:
(i) [Co(NH₃)₆]⁺³ → Octahedral shape, d²sp³ hybridisation, diamagnetic
Formation of [Co(NH₃)₆]⁺³ → oxidation state of Co is +3.

(ii) [Cr(NH₃)₆]⁺³
Octahedral shape, d²sp³ hybridisation poramagnetic (3 unpaired electrons)

(iii) [Ni(CO)₄]
Tetrahedral shape, sp³ hybridisation. diamagnetic (no unpaired eLectrons)

Or

Crystal field splitting energy: When ligands approach the central metal ion, the degenerate d-orbitals split into two sets, one with lower energy (t_2g) and the other with higher energy (e_g). The difference of energy between these two sets of orbitals is called crystal field splitting energy. (∆₀ for octahedral complexes). The magnitude of ∆₀ decides the actual configuration of d-orbitals by the help of mean pairing energy.

(i) If P > ∆₀ then pairing of electrons does not occur and electrons enter in the higher energy eg orbitals and thus form high spin complexes due to weak field ligands.

(ii) If P < ∆₀ then pairing of electrons occurs within the same set and form low spin complexes due to strong field ligands.

Question 9.
(A) What kind of isomerism is shown by
[Cu(NH₃)₄] [PtCl₄] and [Pt(NH₃)₄] [CuCl₄]?
(B) A coordination compound has the formula CoCl₃. 4NH₃. It does not liberate ammonia but forms a precipitate with AgNO₃. Write the structure and IUPAC name of the complex compound.
(C) State whether the following is a cationic or anionic complex [Co(H₂NCH₂ CH₂NH₂)₃]₂(SO₄)₃ (3)
Answer:
Coordination Isomers

Explanation:
In a coordination isomer the total ratio of ligand to metal remains the same, but the ligands attached to a specific metal ion change both the cation and anion are complex ions. In the first isomer, NH3 is attached to the copper and the Cl’ are attached to the platinum. In the second isomer, they have swapped. Hence [Cu(NH₃)₄] [PtCl₄] and [PtCl₄] and [Pt(NH₃)₄] [CuCl₄] are coordination isomers

(B) Formula of the compound is [Co(NH₃)₄Cl₃]Cl and IUPAC name is Tetraamminedichloridocobalt (III) chloride.

Explanation:
Since, it does not liberate ammonia, the 4NH₃ groups are coordinated to Co. In order to satisfy the coordination number of Co, two chlorides are also coordinated. Therefore, the formula of the complex compounds is [Co(NH₃)₄Cl₂]Cl and its IUPAC name is tetraamminodichloridocobalt (III) chloride.

(C) Since, a complex in which the complex ion carries a net positive charge is called a cationic complex. Since in the given complex the complex ion [Co(H₂NCH₂CH₂NH₂)₃]³⁺ Carries a positive charge hence it is a positive complex.

Caution
The sulphate is the counter anion in this molecule. Since, it takes 3 sulphates to bond with two complex cations, the charge on each complex cation must be +3.

Question 10.
Identify compounds B, C and D

OR
Give suitable chemical reactions for the following:
(A) Methanal gives methanol on catalytic hydrogenation in the presence of catalysts such as Ni, Pt or Pd.
(B) Methanal reacts with hydrogen cyanide to form methanal cyanohydrin.
(C) When methanal reacts with blue coloured Fehling’s solution, red precipitates of cuprous oxide (Cu₂0) are formed and the colour of Fehling’s solution changes from blue to red. (3)
Answer:
Cyclopentanone reacts with methyl magnesium iodide (Grignard reagent) to form 1-methylcyclopentanoL The dehydration of 1-methylcyclopentanol with cone, sulphuric acid gives 1-methylcyclopent- 1-ene.

Reductive ozonolysis of 1-methylcyclopent-l- ene gives 5-oxohexanal.

OR
(A) Methanal gives methanol on catalytic hydrogenation in the presence of catalysts such as Ni, Pt or Pd.
Ni/Pt
HCHO → CH3OH

(B) Methanal reacts with hydrogen cyanide to form methanal cyanohydrin.

(C) Methanal on reaction with Fehlings solution gives red precipitate and itself oxidises to sodium salt of methanoic acid.

Question 11.
Answer the following questions:
(A) Write the formula of an oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number.
(B) Why do transition elements exhibit higher enthalpies-of atomization?
(C) Name the element showing the maximum number of oxidation states among the first series of transition metals from Sc (Z = 21) to Zn (Z = 30).
OR
The standard electrode potential for the M3+/M2+ half-cell gives the relative stability between M3+ and M2+. The reduction potential values are tabulated as below:
Reaction-Standard reduction potential (V)

Reaction	Standard reduction potential (V)
Ti3+ + e– → Ti2+	-0.37
V3+ + e– → V2+	-0.26
Cr3+ + e– → Cr2+	-0.41
Mn3+ + e– → Cr2+	+1.51
Fe3+ + e– → Mn2+	+0.77
Co3+ + e– → Co2+	+1.81

(A) What does the negative values for titanium, vanadium and chromium indicate?
(B) What does reduction potentail of Mn³⁺/ Mn²⁺ indicate about the stability of these oxidation states?
(C) The E° value for the Mn3+/ Mn2+couple is much more positive than that for Cr³⁺/ Cr²⁺ couple or Fe³⁺/Fe²⁺ couple. (3)
Answer:
(A) An oxo-anion of chromium (Cr) in which it shows the oxidation state equal to its group number is Cr₂ O₇^2-. Here, oxidation number of Cr is + 6 which is equal to its group number.

(B) The high enthalpies of atomization are due to a large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

(C) Mn (Z=25) has the maximum number of unpaired electrons present in the d-subshell so it shows maximum oxidation states (+2 to +7)
OR
(A) The negative values for titanium, vanadium and chromium indicate that the higher oxidation state is preferred.

(B) The high reduction potential of Mn³⁺/ Mn²⁺ indicates Mn²⁺ is more stable than Mn³⁺ Mn³⁺ has a 3d⁴ configuration while that of Mn²⁺ is 3d⁵. The extra stability associated with a half filled d sub shell makes the reduction of Mn³⁺ very feasible (E° = + 1.51V).

(C) Because Mn3+ has the outer electronic configuration of 3d⁴ and Mn²⁺ has the outer electronic configuration of 3d⁵. Thus, the conversion of Mn³⁺ to Mn²⁺ will be a favourable reaction since 3d⁵ is a very stable configuration as it is half filled configuration. Hence, E° value for Mn³⁺/ Mn²⁺ couple is positive.
Fe³⁺ to Fe²⁺ undergoes a change in outer electronic configuration from 3d⁵ to 3d⁶. The configuration of the resultant are not stable

Related Theory:
Low reduction potential means there is a greater tendency for reduction to occur, while a lower one means there is a greater tendency for oxidation to occur.

Section – C
(Section C-Question No 12 is case-based question carrying 5 marks.)

Question 12.
Read the passage given below and answer the questions that follow:
Amine, any member of a family of nitrogen- containing organic compounds that is derived, either in principle or in practice, from ammonia (NH₃).
Naturally occurring amines include the alkaloids, which are present in certain plants; the catecholamine neurotransmitters (i.e., dopamine, epinephrine, and norepinephrine); and a local chemical mediator, histamine, that occurs in most animal tissues.
Most of the numerous methods for the preparation of amines may be broadly divided into two groups: (1) chemical reduction (replacement of oxygen with hydrogen atoms in the molecule) of members of several other classes of organic nitrogen compounds and (2) reactions of ammonia or amines with organic compounds.
Amines are classified as primary, secondary, or tertiary depending on whether one, two, or three of the hydrogen atoms of ammonia have been replaced by organic groups. In chemical notation these three classes are represented as RNH₂, R₂NH, and R₃N, respectively. A fourth category consists of quaternary ammonium compounds, which are obtained by replacement of all four hydrogen atoms of the ammonium ion, NH₄+; an anion is necessarily associated (R₄N⁺X^–). Amines are also classified as aliphatic, having only aliphatic groups attached, or aromatic, having one or more aryl groups attached. They may be open-chain, in which the nitrogen is not part of a ring, or cyclic, in which it is a member of a ring (generally aliphatic). The carbylamine reaction, also known as Hofmann’s isocyanide test, is a chemical test for the detection of primary amines.
(A) Name two alkaloids present in neurotransmitters.
(B) Write the structure of n-methyl ethanamine.
(C) What are the two methods generally used for the preparation of amines?
(D) Write chemical equations for the following conversions: (5)
(i) Nitrobenzene to benzoic acid.
(ii) Benzyl chloride to 2-phenylethanamine
OR
Account for the following :
(i) Primary amines (R-NH₂) have higher boiling point than tertiary amines (R₃N).
(ii) Aniline does not undergo Friedel – Crafts reaction.
Answer:
(A) Dopamine, Epinephrine

(B) H₃C—H₂C—NH—CH₃

(C) (i) Chemical reduction (replacement of oxygen with hydrogen atoms in the molecule) of members of several other classes of organic nitrogen compounds

(ii) reactions of ammonia or amines with organic compounds.

(D) (i) Nirtobenzene to benzoic acid:

(ii) Benzyl chloride to 2-phenylethanamine

OR
(i) Due to the presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo H- bonding. As a result, primary amines have higher boiling points than 3° amines.

(ii) Aniline being a Lewis base reacts with Lewis acid AlCl3 to form a salt.

As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Friedel Craft reaction.

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 7 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 7 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• There are 12 questions in all. All questions are compulsory.
• This question paper has three sections: Section A, Section B and Section C.
• Section A contains three questions of two marks each, Section B contains eight questions of three marks each, Section C contains one case study-based question of five marks.
• There is no overall choice. However, an internal choice has been provided in one question of two marks and two questions of three marks. You have to attempt only one of the chokes in such questions.
• You may use log tables if necessary but use of calculator is not allowed.

SECTION – A
(Section A contains 3 questions of 2 marks each.)

Question 1.
A student claims that she is going to eject electrons from a piece of metal by placing a radio transmitter antenna adjacent to the metal and sending a strong AM radio signal into the antenna. The work function of a metal is typically a few electron volts. Will this work? (2)
Answer:
Work function= few electron volt
AM radio signal= 1600 kHz
= 1.6 × 10⁶ Hz
Energy of photon= \(\frac{h c}{\lambda}\)
= 6.6 × 10^-9 eV

Ejection of electron is not possible by AM radio signal, as energy of its photon is less than the work function.

Caution
Students are often confused when photoetectrons eject from the metal. For it the energy of the photon must be more than the work function.

Question 2.
How does an air bubble in water behaves as a diverging lens? (2)
Answer:
If a parallel beam of light is incident on an air bubble in water, then the refracted rays diverge from each other.

Question 3.
Write two drawbacks of Rutherford’s atomic model.
OR
Write main use of the:
(A) Solar Cell
(B) Photodiode (2)
Answer:
(A) It does not explain the stability of atom.

(B) In Rutherford’s model, an electron can revolve in orbits of all possible radii. So, it should emit a continuous spectrum. But an atom like hydrogen always emits a discrete line spectrum.
OR
(A) A solar cell is a junction diode which converts solar energy to electrical energy. The generation of voltage is due to constant bombardment or light photons.

(B) In demodulation and direction of optical signals, in light operated switches and an electronic counter, photodiodes are used.

Related Theory
Photodiodes are preferably operated in reverse bias condition. Even fractional increase in majority charge carriers is much less than the fractional change increases in minority charge carriers. The fractional change due to the photo effects on the minority carriers dominated reverse bias current is more easily measurable than the fractional change in the majority charge carriers dominated forward bias current.

SECTION – B
(Section B consists of 8 questions of 3 marks each.)

Question 4.
What is a hole? Give their important characteristics. (3)
Answer:
Vacancy or absence of electron in the bond of a covaiently bonded crystal are called holes. Whenever an electron is removed from the completely filled valence band of a semiconductor, a vacancy is left Behind in the valence band. A hole is not a physical entity but just a convenient way for describing charge motion.

(A) A hole just a vacancy created by removal of electron from a covalent bond of a semiconductor.
(B) It has the same mass as the removed electron.
(C) It is associated with the positive charge of magnitude e.

Question 5.
Light from a point source in air falls on a convex spherical glass surface of refractive index 1.5 and radius of curvature 20 cm. The distance of light source from the glass surface is 100 cm. At what position is the image formed? (3)
Answer:
Given, refractive index,
n₂ = 1.5,
n = 1 (air)
Radius of curvature,
R = 20 cm
Object distance,
u = – 100 cm
To find, Image distance, v
\(\frac{n_{2}}{v}-\frac{n_{1}}{u}\) = \(\frac{n_{2}-n_{1}}{\mathrm{R}}\)
\(\frac{1.5}{v}+\frac{1}{(100)}\) = \(\frac{1.5-1}{20}=\frac{1}{40}\)
\(\frac{1.5}{v}\) = \(\frac{1}{40}-\frac{1}{100}\)
= \(\frac{5-2}{200}=\frac{3}{200}\)
v = \(\frac{200}{3}\)
= 100 cm
The image is formed at 100 cm in denser medium.

Question 6.
(A) What is the relation between the
intensity and amplitude of wave?
(B) Two coherent light beams of intensities I and 41 are superposed. Find the maximum and minimum possible intensities in the resulting beam? (3)
Answer:
(A) Intensity ∝ (Amplitude)²
(B) The net intensity due to interference of two beams of intensities I₁ I₂ with a phase difference of Φ) is:
I_net = I₁ + I₂+ 2I₁I₂ cos Φ
Thus, I_netmax occurs for cos Φ = 1
= I_netmax = (\(\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)\))² = 9I
Similarly,_Inetmin inetmin occurs for cos = -1
I_netmin = (\(\left(\sqrt{I_{1}}+\sqrt{I_{2}}\right)\))²

Question 7.
In a Geiger-Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 80, when α-particle of 8 MeV energy impinges on it before it comes momentarily to rest and reverses its direction. How will the distance of closest approach be affected when the kinetic energy of the α-particle is doubled?.
OR
What do you mean by coherence? How are coherent sources produced in laboratory? (3)
Answer:
Let r_o be the centre to centre distance between the alpha-particle and nucleus.
Given Z = 80 E_k = 8MeV

= 2.88 × 10^-14 m
= 28.8 fm
Since, ro ∝ \(\frac{1}{E_{k}}\)
So when kinetic energy is doubled the distance of closest approach ro is halved.
OR
Coherence is an ideal property of waves that enables stationary interference.

Methods to produce them:
(A) Division of wavefront: Wavefront from a point source of light illuminated by an ordinary source of light can be divided into two parts by allowing it to fall on a pair of slits.
(B) Division of amplitude: Light from an ordinary source may be divided into two or more beams by partial reflection. This way the amplitude of the incident beam is divided.

Question 8.
(A) Under what conditions, total internal
reflection occurs?
(B) State, with the help of a ray diagram, the working principle of optical fibres. Write one important use of optical fibres. (3)
Answer:
(A) (i) Light must go from denser to rarer medium.
(ii) The angle of incidence in the denser medium must be greater than the critical angle of the pair of media.

Optical fibre works on principle of total internal reflection. When angle of incidence is greater than critical angle then incident rays are totally reflected back in same media.
When, θ_i > θ_c , Total internal reflection occurs and if θ_i < θ_c refraction occurs.
Application: Optical fibres are used for communication due to very high bandwidth of media.

Question 9.
Explain briefly how V-l characteristics of a p-n junction diode obtained in:
(A) Forward bias
(B) Reverse biased
And draw shape of curves obtained. (3)
Answer:
(A) In forward bias characteristics, it is observed that initially no current flows up to knee voltage and after knee voltage current increases rapidly.

(B) In reverse bias characteristics it is observed that initially very little current flows due to minority charge carriers up to breakdown voltage after which the current becomes independent of voltage.

Question 10.
(A) How are electromagnetic waves
produced by oscillating charges?
(B) State clearly how a microwave oven works to heat up a food item containing water molecules.
(C) Why are microwaves found useful for the radar systems in aircraft navigation? (3)
Answer:
(A) If a charged particle oscillate with some frequency, produces an oscillating electric field in space, which produces an oscillating magnetic field, which in turn, is a source of electric field, and so on. Thus, oscillating electric fields and magnetic fields regenerate each other, and an electromagnetic wave propagates in the space.

(B) In microwave oven the frequency of the microwaves is selected to match the resonant frequency of water molecules so that energy from the waves get transferred efficiently to the kinetic energy of the molecules. This kinetic energy raises the temperature of any food containing water.

(C) Microwaves are short waveleangth radio waves with frequency of order of GHz. Due to short wavelength, they have high penetrating power with respect to atmosphere and less diffraction in the atmospheric layers. So, these waves are suitable for the radar systems used in aircraft navigation.

Question 11.
Define the terms:
(A) ‘cut-off voltage’ and
(B) ‘threshold frequency’ in relation to the phenomenon of photoelectric effect.

Using Einstein’s photoelectric equation show how the cut-off voltage and threshold frequency for a given photosensitive material can be determined with the help of a suitable \(\frac{\text { plot }}{\text { graph }}\)
OR
In the case of plastic lens, the thickness of the lens increased in comparison of the glass lens. Justify the statement. (3)
Answer:
When light of suitable frequency is incident on a metal surface, electrons are ejected from the metal. This phenomenon is called the photoelectric effect.

(A) The cathode is illuminated with light of some fixed frequency v and fixed intensity li. A small photoelectric current is observed due to few electrons that reach anode just because they have sufficiently large velocity of emission. If we make the potential of the anode negative with respect to cathode then the electrons emitted by cathode are repelled. Some electrons even go back to the cathode so that the current decreases. At a certain value of this negative potential, the current is completely stopped. The least value of this anode potential which just stops the photocurrent is called cut off potential or stopping potential.

(B) For a given material, there is a certain minimum frequency that if the incident radiation has a frequency below this threshold, no photoelectric emission will take place, howsoever intense the radiation may be falling. This minimum frequency is called threshold frequency.
According to Einstein’s photoelectric equation, maximum K.E. is given as
K.E_max = \(\frac{h c}{\lambda-\phi}\)
= hv – Φ
Where λ, is wavelength of light and v is corresponding frequency and Φ is work function. We expose a material to lights of various frequencies and thus photoelectric current is observed and cut off potential needed to reduce this current to zero is noted. A graph is plotted and that is straight line.

According to Einstein’s phottoelectric equation
K.E_max = \(\frac{h c}{\lambda}\) – Φ
= hv – Φ
> eV₀ = hv – Φ
V₀ = (\(\frac{h}{e}\))V – \(\frac{\phi}{e}\) ………… (i)

We can read the value of threshold frequency from graph. From equation (i), we can find the value of stopping potential (V₀).
OR
According to lens maker’s formula,
\(\frac{1}{f}\) = (n – 1) (\(\frac{1}{R_{1}}-\frac{1}{R_{2}}\))
where f is the focal length, n is the refractive index and R₁ and R₂ is the radius of curvature of the lens.
Since, n_g > n_p
where, g stands for glass and p stands for plastic.
Therefore, we get
(n_g -1) > (n_p – 1)
Now, using the lens maker’s formula, we see that focal length is inversely proportional to
(μ – 1).
Hence, f_p > f_g

Thus, in the case of plastic lens the thickness of the lens should be increased to keep the same focal length as that of the glass lens to give the same power.

Caution
Students are often confused about the plastic and glass lens. Due to the different lenses of different refractive index plastic lens will be thicker than the glass lens.

SECTION – C
(Section C consists one case study-based question of 5 marks.)

Question 12.
The main objectives of the Indian Atomic Energy programme are to provide safe and reliable electric powerfor the country’s social and economic progress and to be self- reliant In all aspects of nuclear technology. U²³⁵, Pu²³⁹ and U²³³ are called ‘fissile’

materials because they immediately start fission reaction once a neutron hits them but U²³⁸and Th²³² are called ‘fissionable’ material because adding a neutron makes them fissile and then only, they are ready for fission reaction.

(A) The control rod in a nuclear reaction is made of:
(i) Plutonium
(ii) Cadmium
(iii) Graphite
(iv) Uranium

(B) Solar energy is mainly caused due to:
(i) Burning of hydrogen in the oxygen
(ii) Fission of uranium present in the oxygen
(iii) Fusion of protons during synthesis of heavier elements
(iv) Gravitational contraction

(C) When (92²³⁵)U undergoes fission by absorbing (o¹, (56¹⁴⁴)Ba and (36⁸⁸)Kr are formed. The number of neutrons produced will be:
(i) 0
(ii) 1
(iii) 2
(iv) 3

(D) Energy is released in nuclear fission is due to:
(i) Few mass is converted into energy
(ii) Total binding energy of fragments is more than the binding energy of parental element
(iii) Total B.E. of fragments is less than the binding energy of parental element
(iv) Total B.E. of fragments is equals to the binding energy of parental element

(E) During the fission of U²³⁵2 00 MeV is released. How many acts of fission must occur per second to produce a power of 1 kW?
(i) 3.1 × 10¹³
(ii) 3.1 × 10¹³
(iii) 3.1 × 10¹³
(iv) 3.1 × 10¹³
Answer:
(A) (b) Cadmium
Explanation: Control rods are used to control the rate of fission of the nuclear fuel. Cadmium is used as materival to make control rods because it is capable of absorbing many neutrons without fissioning itself.

(B) (c) Fusion of protons during synthesis of heavier elements
Explanation: Fusion reaction in the sun is the fusion of hydrogen nuclei.

(C) (d)3
Explanation: When (92²³⁵)U undergoes fission by absorbing (0¹)n, (56¹⁴⁴)Ba and (36⁸⁸)Kr are formed with emission of 3 neutrons.
o¹n + 92²³⁵U → 92²³⁶U 56¹⁴⁴Ba+ 36⁸⁸Kr + 30¹n

(D) (b) Total binding energy of fragments is more than the energy of parental element Explanation: Nuclear fission occurs in heavy nuclei and leads to the formation of lighter nuclei. It is well known from the binding energy curve that for heavier nuclei, a decrease in the atomic mass lead to increase of binding energy per nucleon in a nucleus.

(E) (a) 3.1 × 10¹³
Explanation: Each fission 200 MeV is released or energy released = 200 × 10⁶ × 1.6 × 10^-19 Power needed = 1000 W
Number of fission = \(\frac{1000}{200 \times 10^{6} \times 1.6 \times 10^{-19}}\)
= 3 × 10¹³

Students can access the CBSE Sample Papers for Class 12 Geography with Solutions and marking scheme Term 2 Set 2 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Geography Term 2 Set 2 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 35
Roll No. ___________

General Instructions:

• Question paper is divided into 5 sections A, B, C, D & E
• In section A, question number 1 to 3 are Very Short Answer type questions. Attempt any 3 questions.
• In section B, question number 4 is Source based question.
• In section C, question number 5 & 6 are Short Answer type questions.
• In section D, question number 7 to 9 are Long Answer type questions.
• In section E, question number 10 is a Map based question.

Section – A
Very Short Answer type Questions 

Question 1.
Differentiate between tertiary activities and quaternary activities. (2)
Or
Differentiate between trade and trading centres.
Answer:
Tertiary activities are related to the service sector. Tertiary activities include both production and exchange. The production involves the provision of services that are ‘consumed’. On the other hand, the service sector which is knowledge oriented comes under quaternary activities. These activities centre around research, development and may be seen as an advance form of services involving specialised knowledge and technical skills.
Or
Trade is essentially buying and selling of items produced elsewhere. All the services in retail and wholesale trading or commerce are specifically intended for profit. On the other hand, the towns and cities where all these works take place are known as trading centres.

Question 2.
Among the non-metallic minerals produced in India, mica is one of the most important minerals. Name the states where mica is produced. (2)
Answer:
Mica is one of the most important non-metallic minerals produced in India. In India, it is produced in Jharkhand, Andhra Pradesh, Telangana and Rajasthan followed by Tamil Nadu, West Bengal and Madhya Pradesh.

Question 3.
Outline the distribution of sources of manganese in two major states in India. (3)
Answer:
The distribution of sources of manganese in two major states in India are

• Major mines in Odisha are located in the central part of the iron ore belt of India, particularly in Bonai, Kendujhar, Sundergarh, Gangpur, Koraput, Kalahandi and Bolangir.
• Karnataka is another major producer and here the mines are located in Dharwar, Ballari, Belagavi, North Canara, Chikkmagaluru, Shivamogga, Chitradurg and Tumakuru.

SECTION – B
Source Based Question

Question 4.
Read the source given below and answer the following questions by choosing the correct option.
Land Transport
Most of the movement of goods and services takes place over land. In early days, humans themselves were carriers. Have you ever seen a bride being carried on a palanquin (palki/doli) by four persons (Kahars in North India). Later animals were used as beasts of burden. Have you seen mules, horses and camels, carrying loads of cargo in rural areas? With the invention of the wheel, the use of carts and wagons became important. The revolution in transport came about only after the invention of the steam engine in the eighteenth century.

Perhaps the first public railway line was opened in 1825 between Stockton and thern England and then onwards, railways became the most popular and fastest form of transport in the nineteenth century. It opened up continental interiors for commercial grain farming mining and manufacturing in U.S.A. The invention of the internal combustion engine revolutionised road transport in terms of road quality and vehicles (motor cars and trucks) plying over them.

Among the newer developments in land transportation are pipelines, ropeways and cableways. Liquids like mineral oil, water, sludge and sewers are transported by pipelines. The great freight carriers are the railways, ocean vessels, barges, boats and motor trucks and pipelines. In general, the old and elementary forms like the human porter, pack animal, cart or wagon are the most expensive means of transportation and large freighters are the cheapest. They are important in supplementing modern channels and carriers which penetrate the interiors in large countries. In the densely populated districts of India and China, overland transport still takes place by human porters or carts drawn or pushed by humans.

i. Which were the earliest mode of land transportation? (1)
Answer:
Humans and animals were the earliest mode of land transportation.

ii. Which event can be considered as a revolution in transport? (1)
Answer:
Invention of steam engine can be considered as a revolution in transport.

iii. Which place in Europe witnessed the first public railway line? (1)
Answer:
England witnessed the first public railway line in Europe.

Section – C
Short Answer Questions

Question 5.
How does access to labour supply and transportation and communication facilities influence industrial locations? (3)
Or
Explain the difference between household industries and small scale manufacturing.
Answer:
Access to labour supply and transportation and communication facilities influence industrial location in the following ways

• Labour supply is an important factor in the location of industries. Some types of manufacturing require skilled labour.
• Speedy and efficient transport facilities to carry raw materials to the factory and to move finished goods to the market are essential for the development of industries.
• The cost of transport plays an important role in the location of industrial units. For example, Western Europe and Eastern North America have a highly-developed transport system which has always induced the concentration of industries in these areas.
• Modern industry is inseparably tied to transportation systems. Improvements in transportation led to integrated economic development and regional specialisation of manufacturing.

Or
Household industries is the smallest manufacturing unit. The artisans use local raw materials and simple tools to produce everyday goods in their homes with the help of their family members or part time labour. Finished products may be for consumption in the same household or, for sale in local (village) markets, or, for barter.

Capital and transportation do not wield much influence as this type of manufacturing has low commercial significance and most of the tools are devised locally. Manufacturing of jewellery, artefacts and crafts made of wood or bamboo, fabrics, mats tools etc are examples of household industries.

Small scale manufacturing is distinguished from household industries by its production techniques and place of manufacture. This type of manufacturing uses local raw material, simple power-driven machines and semi-skilled labour. It provides employment and raises local purchasing power. Examples of small scale industry can include the manufacturing of spices, biscuits, soaps and oils etc.

Question 6.
What is the role of the National Highways Authority of India? Highlight some major projects undertaken by the National Highways Authority of India. (3)
Answer:
The National Highways Authority of India (NHAI) was operationalised in 1995. It is an autonomous body under the Ministry of Surface Transport. It is entrusted with the responsibility of development, maintenance and operation of National Highways. This is also the apex body to improve the quality of the roads designated as National Highways.

NHAI has taken up some major projects in the country under different phases which are

i. Golden Quadrilateral It comprises construction of 5,846-km long 4/6 lane, high density traffic corridor, to connect India’s four big metro cities of Delhi-Mumbai- Chennai-Kolkata. With the construction of Golden Quadrilateral, the time, distance and cost of movement among the mega cities of India will be considerably minimised.

ii. North-South and East-West Corridors North-South Corridor aims at connecting Srinagar in Jammu and Kashmir with Kanniyakumari in Tamil Nadu (including Kochchi-Salem Spur) with 4,076-km long road. The East-West Corridor has been planned to connect Silchar in Assam with the port town of Porbandar in Gujarat with 3,640-km of road length.

Section – D
Long Answer Questions 

Question 7.
What are secondary economic activities and the role of manufacturing in secondary activity? (5)
Or
The locations of industries are dependent on several factors including access to raw material, transport and energy. Discuss these dependent factors for industry locations.
Answer:
Secondary economic activities are the activities which add value to natural resources by transforming raw materials into valuable products. It refers to those activities which produce a finished and usable product. For instance, cotton in the boll has limited use but after it is transformed into yarn, becomes more valuable and can be used for making clothes. Iron ore, cannot be used; directly from the mines, but after being converted into steel it gets its value and can be used for making many valuable machines, tools, etc. The same is true of most of the materials from the farm, forest, mine and the sea.

Secondary activities, therefore, are concerned with manufacturing processing and construction (infrastructure) industries. Manufacturing which is an essential first step in a secondary activity can involve a full array of production from handicrafts to moulding iron and steel and stamping out plastic toys to assembling delicate computer components or space vehicles. The role of manufacturing in secondary activities may entail the following common characteristics (i) The application of power. (ii) The mass production of identical products. (iii) Specialised labour in factory settings for the production of standardised commodities.
Or
The location of industries are dependent on the following factors

→ Access to Market The existence of a market for manufactured goods is the most important factor in the location of industries.

→ Access to Raw Material Raw material used by industries should be cheap and easy to transport. Industries based on cheap, bulky and weight-losing material (ores) are located close to the sources of raw material such as steel, sugar, and cement industries.

→ Access to Labour Supply Labour supply is an important factor in the location of industries. Increasing mechanisation, automation and flexibility of industrial processes have reduced the dependence of industry upon the labours.

→ Access to Sources of Energy Industries which use more power are located close to the source of the energy supply such as the aluminium industry.

→ Access to Transportation and Communication Facilities Speedy and efficient transport facilities to carry raw materials to the factory and to move finished goods to the market are essential for the development of industries. Access to Agglomeration Economies Links between Industries Many industries benefit from nearness to a leader industry and other industries. These benefits are termed as agglomeration economies. Savings are derived from the linkages which exist between different industries.

Question 8.
What is a mineral? State its characteristics and types with relevant example. (5)
Answer:
A mineral is a natural substance of organic or inorganic origin with definite chemical and physical properties, e.g. gold, silver, iron ore, bauxite, etc. Minerals have certain characteristics such as

→ They are unevenly distributed over space. In India, bulk of the valuable minerals are associated with metamorphic and igneous rocks of the peninsular India. The vast alluvial plain tract of North India is devoid of minerals of economic use.

→ There is an inverse relationship in quality and quantity of minerals i.e. good quality minerals are less in quantity as compared to low quality minerals.

→ All minerals are exhaustible over time. These take long to develop geologically and they cannot be replenished immediately at the time of need. Thus, they have to be conserved and not misused as they do not have the second crop. Minerals may be grouped under two main categories of metallic and non-metallics based on chemical and physical properties

(i) Metallic minerals are the sources of metals. Iron ore, copper, gold produce metal and are included in this category. Metallic minerals are further divided into ferrous and non-ferrous metallic minerals. Ferrous refers to iron. All those minerals which have iron content are ferrous such as iron ore itself and those which do not have iron content are non-ferrous such as copper, bauxite, etc.

(ii) Non-metallic minerals are either organic in origin such as fossil fuels also known as mineral fuels which are derived from the buried animal and plant life such as coal and petroleum. Other type of non-metallic minerals are inorganic in origin such as mica, limestone and graphite, etc..

Question 9.
Non-conventional sources of energy are considered to be an environmental friendly and sustainable alternative to conventional fossil fuels. Discuss non-conventional sources of energy available in India along with their current status and examples. (5)
Answer:
Non-conventional sources of energy such as solar and wind energy, are renewable sources. They are equitably distributed and environment friendly as there are no carbon or pollution emissions upon their use. They are not exhaustible as they do not have fixed deposits or reserves.

These sources can provide sustained, cheap and environment friendly energy and hence they are considered as a sustainable alternative to fossil fuels. Some of the commonly used nonconventional sources of energy in India are

→ Solar Energy Sun rays tapped in photovoltaic cells can be converted into energy, known as solar energy. The two effective processes considered to be very effective to tap solar energy are photovoltaics and solar thermal technology, The Western part of India has greater potential for the development of solar energy in Gujarat and Rajasthan.

→ Wind Energy This form of energy is absolutely pollution free, inexhaustible source of energy. The kinetic energy of wind, through turbines is converted into electrical energy. The permanent wind systems such the trade winds, westerlies and seasonal wind like monsoon have been used as source of energy. Besides these, local winds, land and sea breezes can also be used to produce electricity. In Rajasthan, Gujarat, Maharashtra and Karnataka, favourable conditions for wind energy exist.

→ Tidal and Wave Energy Ocean currents are the store house of infinite energy. Large tidal waves are known to occur along the West Coast of India. Hence, India has great potential for the development of tidal energy along the coasts but so far these have not yet been utilised.

→ Geothermal Energy When the magma from the interior of earth, comes out on the surface, tremendous heat is released. This heat energy can successfully be tapped and converted to electrical energy. Apart from this, the hot water that gushes out through the geyser wells is also used in the generation of thermal energy. It is popularly known as geothermal energy. In India, a geothermal energy plant has been commissioned at Manikaran in Himachal Pradesh.

Section – E
Map Based Question

Question 10.
On a political map of India, mark and indicate the following features. (Attempt any 5) (1 × 5 = 5)
(i) Khetri copper mine
(ii) Ratnagiri iron ore mine
(iii) Mathura oil refinery
(iv) Important node of North connidor
(v) Katni bauxite mine
(vi) Jharia coal mine
Answer:

Students can access the CBSE Sample Papers for Class 12 Economics with Solutions and marking scheme Term 2 Set 1 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Economics Term 2 Set 1 with Solutions

Time allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This is a Subjective Question Paper containing 13 questions.
• This paper contains 5 questions of 2 marks each, 5 questions of 3 marks each and 3 questions of 5 marks each.
• 2 marks questions are Short Answer Type Questions and are to be answered in 30-50 words.
• 3 marks questions are Short Answer Type Questions and are to be answered in 50-80 words.
• 5 marks questions are Long Answer Type Questions and are to be answered in 80-120 words.
• This question paper contains Case/Source Based Questions.

Question 1.
Distinguish between Rnal Goods and Intermediate Goods.
OR
Distinguish between positive externalities and negative externalities. (2)
Answer:
Final goods refer to those goods which are used either for final consumption or for investment purposes. Whereas Intermediate goods refer to those goods which are used either for resale or for further production purposes.

Explanation:

Final Goods	Intermediate Goods
Goods which are used for either consumption or investment is known as final goods.	Goods which are used for further production as raw material or purchased for resale is known as intermediate goods.
Final goods included in estimation national income.	Intermediate goods the are not included in of the estimation of national income.

OR
Positive externalities refer to benefits caused by one entity to another, without being paid for it. Whereas Negative externalities refer to the harms caused by one entity to another without being penalized for it.

Explanation:

Positive Externalities	Negative Externalities
Positive externalities refer to the benefits of an economy activity to the others without involving any price paid for that.	Negative externalities refer to the negative impact of an economy activity on the others without involving any penalty.
Employment generated due to establishment of a factory is an example of positive externality.	Smoke omitted by factories causes air pollution is an example of negative externality.

Question 2.
Calculate equilibrium level of income for a hypothetical economy, for which it is given that:
(1) Autonomous Investments = ₹500 crores, and
(2) Consumption function, C = 100 + 0.80 Y
OR
Calculate Change in Income (ΔY) for a hypothetical economy. Given that
(1) Marginal Prospensity to Consume (MPC) = 0.8, and
(2) Change in Investment (Δl) = ₹1,000 crores (2)
Answer:
Given Consumption function is,
C = 100 + 0.8 Y
Autonomous investments = ₹500 crores We know, at equilibrium level
Y = C + l
Y = 100 + 0.8 Y + 500
Y – 0.8Y = 600 0.2 Y = 600
Y = ₹3,000 crores

Explanation:
Given,
Consumption function (C) = 100 + 0.8Y
Autonomous investments (I) = 500 crores
Y = C + l
Y = 100 + 0.8Y + 500
Y – 0.8Y = 600
0.2Y = 600
Y = ₹ 3,000 crores
OR
P Given ΔI = ₹1,000 crores
MPC = 0.8
As we know,
Multiplier (K) = \(\frac{1}{1-M P C}=\frac{1}{1-0.8}=\frac{1}{0.2}\)
= 5 times
We know K = \(\frac{\Delta Y}{\Delta l}\)
5 = \(\frac{\Delta Y}{1000}\)
ΔY = ₹ 5,000 crores

Explanation: Given,
Change in Investment
(Δl) = ₹ 1,000 crores
Marginal Propensity to Consume (MPC) = 0.8
Multiplier (K) = 1/1 – MPC
= 1/1 – 0.8
= 1/0.2 = 5
Multiplier (K) = Change in Income (ΔY)/
Change in Investment (Δl) =
5 = ΔY/1,000 ΔY = ₹ 5,000 crores

Question 3.
‘As the income increases, people tend to save more’. Justify the given statement. (2)
Answer:
At a lower level of income, a consumer spends a larger proportion of his/her income on consumption expenditure (basic survival requirements). As the income increases, owing to the psychological behavior of a consumer (rational), people tend to consume less and save more for future uncertainty.

Explanation:
It is true that as the income increases, people tend to save more. At a lower Level of income, a person spends most of his income on the basic survival requirements. People on very low income cannot afford the Luxury of saving. But as the income increases, households tend to have a higher marginal propensity to save as they save more for future uncertainty.

Question 4.
State and discuss any two indicators that help in measuring the health status of a country.
OR
Compare and analyze the ‘Women Worker Population Ratio’ in Rural and Urban areas based on following information: Worker-Population Ratio in India, 2017-2018. (2)

Sex	Worker-Population Ratio
	Total	Rural	Urban
Men	52.1	51.7	53.0
Women	16.5	17.5	14.2
Total	34.7	35.0	33.9

Answer:
Two indicators that help to measure the health status of a country are:

1. Infant Mortality Rate – Infant Mortality Rate (or IMR) indicates the number of children that survive first year of their life.
2. Literacy Rate – Literacy rate measures the proportion of literate population in the age group of seven years or above.

Explanation:
Two indicators that help to measure the health status of a country are:

1. Life expectancy: It refers to the number of years a person is expected to live based on the statistical average.
2. Infant Mortality Rate: The infant mortality rate is the number of infant deaths for every 1,000 live births.

OR

The data in the given table reveals that:
(1) Women constitute 16.5% of the total worker population in the economy.

(2) The number of women workers in rural areas (17.5%) are relatively higher than the women workers in urban areas (14.2%). Due to the abject poverty in rural areas, rural women are compelled to work more than their urban counterparts.

Explanation:
Women constitute 16.5% of the total worker population in the economy. The difference in participation rates is large as in urban areas, for every 100 urban females, only about 14 are engaged in some economic activities. In rural areas, for every 100 rural women about 18 participate in the employment market. As poverty in the rural areas is more widespread than in the urban areas, so, the rural women engage themselves in low productive jobs just to support the livelihood of their families.

Question 5.
‘Investment in infrastructure contributes to the economic development of a country’. Justify the given statement with a valid argument. (2)
Answer:
The given statement is true; infrastructural development in an economy increases productivity, induces higher investment, facilitates employment, and generates more income. With the rise in income the quality of life of the people improves. Thus, conclusively we may say that infrastructure contributes to the economic development of a country.

Explanation:
Infrastructure contributes to economic development of a country both by increasing the productivity of the factors of production and improving the quality of life of its people.

Question 6.
Giving valid reasons explain which of the following will not be included in estimation of National Income of India?
(A) Purchase of shares of X. Ltd. by an investor in the National Stock Exchange.
(B) Salaries paid by the French Embassy, New Delhi to the local workers of the housekeeping department.
(C) Compensation paid by the Government of India to the victims of floods.

OR
Estimate the value of Nominal Gross Domestic Product for a hypothetical economy, the value of Real Gross Domestic Product and Price Index are given as ₹500 crores and 125 respectively. (2)
Answer:
Following will not be included in estimation of National Income of India:
(A) As such transactions are mere paper claims and do not lead to any value addition.
(B) Compensation paid by the Government of India is mere transfer payment and does not lead to any flow of goods and services in an economy.

Explanation:
(A) Purchase of shares of X. Ltd. by an investor in the National Stock Exchange will not be included in the estimation of national income because it is just a financial transaction, leading to change of ownership of financial asset. There is no flow of goods or services in the economy.
(B) This will be included in estimation of the National Income of India.
(C) Compensation paid by the Government of India to the victims of floods will not be included in the estimation of national income because it is a transfer payment. There is no flow of goods or services in the economy.
OR

Question 7.
Study the following information and compare the Economies of India and Singapore on the grounds of investment in Infrastructure as a percentage of GDP’. (2)

Some Infrastructure in India and Other Countries, 2018

Sources: World Development Indicators 2019, World Bank website: www.worldbank.org.:
BP Statistical Review of World Energy 2019, 69th Edition Note: (*) refers to Gross Capital Formation.
Answer:
“Investment in infrastructure as a percentage of GDP’ is that proportion of Gross Domestic Product which is invested for the development of infrastructural facilities in a country. According to the given data it is evident that India is contributing 30% of its total GDP on infrastructural progress, which is just a notch above the corresponding figure of 28% for Singapore.

Considering the vast geography of India this is a relatively lower proportion in this direction. If India wants to grow at a faster rate, she must concentrate on higher judicious investment on development of infrastructure.

Explanation:
The data given in the table reveals that India is contributing 30% of its GDP on infrastructure, whereas Singapore is contributing 28% of its GDP. So it can be concluded that India is investing in infrastructural facilities more than Singapore but this gap is not that large. Some economists have projected that India will become the third biggest economy in the world a few decades from now. Considering this fact, this is a relatively lower proportion in this direction. For that to happen, India will have to boost its infrastructure investment.

Read the following text carefully and answer questions number 8 and 9 given below:

SINO-PAK FRIENDSHIP CORRIDOR
The China-Pakistan Economic Corridor (CPEC) relationship between the two nations. But it has also sparked criticism for burdening Pakistan with mountains of debt and allowing China to use its debt strategic assets of Pakistan.

The foundations of CPEC, part of China’s Belt and Road Initiative, were laid in May 2013. At the time, Pakistan was reeling under weak economic growth. China committed to play an integral role in supporting Pakistan’s economy.

Pakistan and China have a strategic relationship that goes back decades. Pakistan turned to China at a time when it needed a rapid increase in external financing to meet critical investments in hard infrastructure, particularly power plants and highways. CPEC’s early harvest projects met this need, leading to a dramatic increase in Pakistan’s power generation capacity, bringing an end to supply-side constraints that had made rolling blackouts a regular occurrence across the country.

Pakistan leaned into CPEC, leveraging Chinese financing and technical assistance in an attempt to end power shortages that had paralyzed its country’s economy. Years later, China’s influence in Pakistan has increased at an unimaginable pace.

China As Pakistan’s Largest Bilateral Creditor: China’s ability to exert influence on Pakistan’s economy has grown substantially in recent years, mainly due to the fact that Beijing is now Islamabad’s largest creditor. According to documents released by Pakistan’s finance ministry, Pakistan’s total public and publicly guaranteed external debt stood at $44.35 billion in June 2013, just 9.3 percent of which was owed to China. By April 2021, this external debt had ballooned to $90.12 billion, with Pakistan owning 27.4 percent -$24.7 billion – of its total external debt to China, according to the International Monetary Fund (IMF). Additionally, China provided financial and technical expertise to help Pakistan build its road infrastructure, expanding north-south connectivity to improve the efficiency of moving goods from Karachi all the way to Gilgit-Baltistan (POK). These investments were critical in better integrating the country’s ports, especially Karachi, with urban centers in Punjab and Khyber- Pakhtunkhwa provinces.

Despite power asymmetries between China and Pakistan, the latter still has tremendous agency in determining its own policies, even if such policies come at the expense of the longterm socioeconomic welfare of Pakistani citizens.

(https://www.usip.org/publications /2021/ 05/ pakistans-growing-problem-its-china-economic-corridor – Modified)

Question 8.
Outline and discuss any two economic advantages of China Pakistan Economic Corridor (CPEC) accruing to the economy of Pakistan. (3)
Answer:
Economic advantages of China Pakistan Economic Corridor (CPEC) to the economy of Pakistan are:

• China provided financial and technical expertise to help Pakistan build its road infrastructure, supporting employment and income in the economy
• CPCE has led to a massive increase in power generation capacity of Pakistan. It has brought an end to supply-side constraints in the nation, which had made blackouts a regular phenomenon across the country.

Explanation: Economic advantages of China Pakistan Economic Corridor (CPEC) accruing to the economy of Pakistan:

1. CPEC’s early harvest projects met investment needs of Pakistan in power plants, leading to a dramatic increase in Pakistan’s power generation capacity, bringing an end to supply-side constraints that had made rolling blackouts a regular occurrence across the country.
2. China provided financial and technical expertise to help Pakistan build its road infrastructure, expanding north-south connectivity to improve the efficiency of moving goods.

Question 9.
Analyse the implication of bilateral ‘debt-trap’ situation of Pakistan vis-a-vis the Chinese Economy. (3)
Answer:
China has become famous for its ‘Debt Trap Diplomacy’ in recent times. Under this China provides financial and technical expertise/ assistance to help various nations to bring them under its direct or indirect influence.

The first and the foremost implication of the diplomacy is that Beijing has now become Islamabad’s largest creditor. According to documents released by Pakistan’s finance
ministry, its total public external debt stood at $44.35 billion in June 2013, just 9.3 percent of which was owed to China. By April 2021, this external debt had ballooned to $90.12 billion, with Pakistan owing 27.4 percent -$24.7 billion – of its total external debt to China, according to the IMF.

Explanation:
China’s ability to exert influence on Pakistan’s economy has grown substantially in recent years, mainly due to the fact that Beijing is now Islamabad’s largest creditor. It is the first and foremost implication of this ‘Debt=trap’ diplomacy. According to documents released by Pakistan’s finance ministry, Pakistan’s total public and publicly guaranteed external debt stood at $44.35 billion in June 2013, just 9.3 percent of which was owed to China. By April 2021, this external debt had ballooned to $90.12 billion, with Pakistan owing 27.4 percent -$24.7 billion – of its total external debt to China, according to the International Monetary Fund (IMF).

Question 10.
Explain how ‘Non-Monetary Exchanges’ impact the use of Gross Domestic Product as an index of economic welfare. (3)
Answer:
Non-monetary exchange transactions are not included in the estimation of Gross Domestic Product on account of practical difficulties like non-availability of reliable data. Although these activities enhance public welfare which may lead to underestimation of GDP.

For example: kitchen gardening, services of homemaker etc.

Explanation:
Such exchanges which don’t involve the use of money as a medium of exchange are known as non-monetary exchanges. In non-monetary exchanges, goods are exchanged for goods.

As these transactions are outside the monetary system of exchange, thus, are not included in gross domestic product of the economy. Due to this, GDP remains underestimated. Thus, non-monetary exchanges make GDP an inappropriate index of welfare.

Question 11.
‘Monetary measures offer a valid solution to the problem of Inflationary gap in an economy’. State and discuss any two monetary measures to justify the given statement. (5)
Answer:
Two measures which may be used to solve the problem of inflation are:
(i) An increase in Cash Reserve Ratio (CRR) may reduce the credit creation capacity of the commercial banks in the economy. This may lead to a fall in the borrowings from
banks causing a fall in Aggregate Demand in the economy, and helps to correct the inflationary gap in the economy.

(ii) Sale of Government Securities in the open market by the Central Bank will adversely affect the ability of the Commercial Banks to create credit in the economy. As a result Aggregate Demand in the economy may fall and correct the inflationary gap in the economy.

Explanation:
Two measures which may be used to solve the problem of inflation are:
(i) Repo Rate: The rate at which central bank lends money to the commercial banks is known as repo rate. To correct the situation of inflationary gap, repo rate should be increased. A rise in repo rate is expected to cause a rise in market rate of interest, leading to a fall in demand for funds in the money market. Implying a fall in expenditure or AD, correcting inflationary gap.

(ii) Open Market Operations: In the situation of inflationary gap, cash balances need to be reduced. In this situation, Central bank starts selling securities. It reduces purchasing power from the money market. Consequently, aggregate demand is decreased and inflationary gap is corrected.

Question 12.
(A) From the following data calculate the value of Domestic Income:

Items	Amount (in crores)
Compensation Employees of	2,000
Rent and Interest	800
Indirect Taxes	120
Corporate Tax	460
Consumption Fixed Capital of	100
Subsidies	20
Dividend	940
Undistributed Profits	300
Net Factor Income from Abroad	150
Mixed-Income of Self Employed	200

(B) Distinguish between ‘Value of Output’ and ‘Value Added’.
OR
(A) Given the following data, find Net Value Added at Factor Cost by Sambhav (a farmer) producing Wheat:

Items	Amount (in crores)
Sale of wheat by the farmer in the local market	6,800
Purchase of Tractor	5,000
Procurement of wheat by the Government from the farmer	200
Consumption of wheat by the farming family during the Year	50
Expenditure on the maintenance of existing capital stock	100
Subsidies	20

(B) State any two components of‘Net Factor Income from Abroad’. (5)
Answer:
(A) Domestic Income (NDP@fc)
= (i) + (ii) + (iv) + (vii) + (viii) + (x)
= ₹ 2000 + ₹ 800 + ₹ 460 + ₹ 940 + ₹ 300 + ₹ 200
= ₹4,700 crore

Explanation:
Domestic Income (NDPFC) = Compensation of Employees + Rent and Interest + Corporate Tax + Dividend + Undistributed Profits + Mixed Income of Self Employed
= ₹ 2000 + ₹ 800 + ₹ 460 + ₹ 940 + ₹ 300 + ₹ 200
= ₹ 4,700 crores

(B) Value of output is the estimated money value of all the goods and services, inclusive of change in stock and production for self-consumption. Whereas, Value added is the excess of value of output over the value of intermediate consumption.

Explanation:
Monetary value of all the goods and services produced in a given period of time is known as value of output, whereas, value-added is the excess of value of output over the value of intermediate consumption.
OR
(A) Net Value Added at Factor Cost (NVA @ FC)
= (i) + (iii) + (iv) + (vi) – (v)
= ₹ 6800 + ₹200 + ₹ 50 + ₹20 – ₹100
= R6,970crore

Explanation:
Net Value Added at Factor Cost (NVA_FC) = Sale of wheat by the farmer in the local market+ Procurement of wheat by the Government from the farmer+ Consumption of wheat by the farming family during the Year + Subsidy – Expenditure on the maintenance of existing capital stock
= ₹ 6800 + ₹ 200 + ₹ 50 + ₹ 20 – ₹ 100
= ₹ 6,970 crores

(B) Component of net factor income from abroad are:

• Net compensation of employees
• Net income from property and entrepreneurship
• Net retained earnings of resident companies abroad (any two)

Explanation:
Components of Net Factor Income from Abroad are:

• Net compensation of employees
• Net income from property and entrepreneurship
• Net retained earnings of resident companies abroad.

Question 13.
(A) ‘Pesticides are chemical compounds designed to kill pests. Many pesticides can also pose health risks to people even if exposed to nominal quantities. ‘

In the light of the above statement, suggest any two traditional methods for replacement of the chemical pesticides.

(B) ‘In recent times the Indian Economy has experienced the problem of Casualisation of the workforce. This problem has only been aggravated by the outbreak of COVID-19.’

Do you agree with the given statement? Discuss any two disadvantages of casualisation of the workforce in the light of the above statement. (5)
Answer:
(A) The traditional practices can help in controlling contamination without the use of chemical fertilizers, as follows:

1. Neem trees and its by products are a natural pest-controller, which has been used since ages in India. Recently, the government promoted the sale Neem coated urea as a measure of natural pest control.
2. Large variety of birds should be allowed to dwell around the agricultural areas, they can clear large varieties of pests including insects

Explanation: Traditional methods for replacement of the chemical pesticides are:
(i) Neem: Neem is a naturally occurring pesticide that comes from the seeds of the neem tree. Neem seeds and leaves contain many compounds which are useful for pest control. Neem repels or reduces the feeding of many species of pest insects as well as some nematodes.

(ii) Animals and Birds: Animals and birds are also one of the traditional methods of pest control. For example, snakes are one of the prime group of animals which prey upon rats, mice and various other pests. Similarly, large varieties of birds, for example, owls and peacocks, prey upon vermin and pests. If these are allowed to dwell around the agricultural areas, they can clear large varieties of pests including insects. Lizards are also important in this regard.

(B) The given statement is quite appropriate with reference to the ‘casualisation of labour’ in India.

(i) For casual workers, the rights of the labour are not properly protected by labour laws. Particularly, during pandemic times, as demand for goods and services fell the casual workers were left jobless, without any compensation or support.

(ii) During the COVID-19 lockdown millions of casual workers lost their jobs, raising the question of their survival. Also, additional health expenditure added to their troubles. Had such workers been working under the formal sector, it would have given them some respite in their difficult times.

Explanation:
It is true that in recent times the Indian Economy has experienced the problem of Casualisation of the workforce. This problem has only been aggravated by the outbreak of COVID-19. India lacks of opportunities in the organised sector, thus people start working as casual workers. The coronavirus pandemic comes as huge shock to the labour market in India.

Disadvantages to casual workers during Covid-19:

• The Indian labor market is largely informal, lacks a social security net, which increases labor vulnerability and distress during lockdowns.
• Thousands of casual workers and self-employed in cities like Mumbai and Pune have left for their native places in the last few months after the Covid-19 outbreak.