Raju

Students can access the CBSE Sample Papers for Class 12 Physics with Solutions and marking scheme Term 2 Set 6 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 6 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.
How is an n-type semiconductor formed?
Draw the energy band diagram n-type semiconductor. Name the majority charge carriers in it. (2)
Answer:
If pentavalent impurity atoms of Bi, Sb or P are doped in a tetravalent crystal of silicon or Germanium, we get a n-type semiconductor. Electrons are the majority charge carriers in a n-type semiconductor.

Question 2.
Name the characteristics of electromagnetic waves that:
(A) increases
(B) remains constant in the electromagnetic spectrum, as one moves from radioactive region towards ultraviolet region. (2)
Answer:
As one moves from radioactive region towards ultraviolet region,
(A) the frequency of the electromagnetic waves increases.
(B) The speed of the electromagnetic waves remains constant.

Related Theory
Speed of electromagnetic wave remains constant in the vacuum. Every massless object will travel with the speed of light.

Question 3.
Explain with the help of Einstein’s photoelectric equation any two observed features in photoelectric effect which cannot be explained by wave theory.
OR
Consider a metal exposed to light of wavelength 600 nm. The maximum energy of the electron doubles when light of wavelength 400 nm is used. Find the work function in eV. (2)
Answer:
Features of photoelectric equation which cannot be explained by wave theory :
(A) The wave theory could not explain the instantaneous process of photoelectric effect.
(B) ‘Maximum kinetic energy’ of the emitted photo electrons is independent of intensity of incident light.

Related Theory
Einstein gives the correction in photoelectric effect and give Einstein’s photoelectric equation.
OR
Let the maximum energies of emitted electrons are K₁ and K₂ when 600 nm and 400 nm visible light are used according to question
K₁ = 2K₁

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

Question 4.
In compound microscope, length of microscope is 21.5 cm, focal length of objective is 1.6 cm, focal length of eye piece is 2.1 cm and if final image is formed at infinity, then find the distance of the object from objective lens? (3)
Answer:
Here,
v = 21.5 – 2.1
= 19.4 cm
because the image formed by the objective should lie at the focus of the eyepiece, from relation,
\(\frac{1}{f}=\frac{1}{v}-\frac{1}{u}\)
⇒ \(\frac{1}{u}=\frac{1}{v}-\frac{1}{f}\)
= \(\frac{f-v}{v f}\)
⇒ u = \(\frac{v f}{f-v}\)
= \(\frac{1.6 \times 19.4}{1.6-19.4}\)
= -1.74 cm

Question 5.
In a single slit diffraction pattern, how does the angular width of central maxima change when
(A) Silt width is decreased
(B) Distance between the slit and screen is increased and
(C) Light of short visible wavelength is used? Justify your answer. (3)
Answer:
(A) Angular width of centrol ,maximum
2θ = \(\frac{2 \lambda}{d}\)
where d is slit width.
if slit width decreases, angular width increases.

(B) It does not depend upon D. So, it will not have any effect.

(C) Angular width
2θ = \(\frac{2 \lambda}{d}\)
∴ If Light of smaller wavelength is used then angular width decreases.

Question 6.
Spectra are produced due to electronic transitions from higher energy level to lower energy level.
(A) Name the series spectra produced by hydrogen.
(B) Which series lies in the visible region.
(C) Write down the Balmer formula for the wavelength of H_α line. (3)
Answer:
(A) Lyman, Balmer, Paschen, Brackett and Pfund series.
(B) Balmer series.
(C) Wavelength of Ha line \(\frac{1}{\lambda}\) = R \(\frac{1}{2^{2}}-\frac{1}{n^{2}}\)

Question 7.
Two convex lenses of an astronomical telescope of focal length 5 cm and 20 cm respectively, are shown in the given figure.

(A) Which of the given lenses will be used as an objective lens?
(B) How should the lenses adjusted to have the telescope in its normal adjustment position?
(C) What will be the magnifying power of the telescope in its normal adjustment position? (3)
Answer:
(A) Because of the larger focal length and larger radius, the lens B should be used as an objective Lens.
(B) In normal adjustment, the distance between objective and eyepiece will be
f_o + f_e = 20 + 5
= 25 cm
The distance required to be increased between the two lenses.
= 25 – 15 = 10 cm

(C) Magnifying power of the telescope in normal adjustment will be,
m = \(\frac{f_{0}}{f_{e}}\) = \(\frac{20 \mathrm{~cm}}{5 \mathrm{~cm}}\) = 4

Question 8.
A projectile of mass m, charge Ze, initial speed v and impact parameter b is scattered by a heavy nucleus of charge Ze. Obtain a relation between minimum distance s of particle from nucleus an impact parameter using conservation of angular momentum and energy. Show that if b = 0, s becomes distance of closest approach. (Ignoring size of nucleus and recoil motion).
OR
Considering that the protons and neutrons have equal masses, calculate how many times nuclear matter is denser than water. Take the value of nuclear radius R = 1.2 × 10^-15 A^1/3 metre and mass of the nucleus = 1.67 × 10^-27 kg. (3)
Answer:

Angular momentum of the projectile at α from the nucleus = mvb
At point of minimum distance (s) from nucleus velocity v’ of projectile is normal to radius vectors.
Angular momentum of projectile at minimum nucleus = mv’s
By Law of Conservation of angular momentum, mv’s = mvb
⇒ V’ \(\frac{v b}{s}\)
By Conservation of Energy
(KE + PE) act as = (KE + PE) at distance s

Dividing by \(\frac{1}{2}\) mv² equation (i)
b² + \(\frac{1}{2 \pi \varepsilon_{0}} \frac{Z Z^{\prime} e^{2}}{m v^{2}}\) = s²
s = \(\frac{1}{2 \pi \varepsilon_{0}} \frac{Z Z^{\prime} e^{2}}{m v^{2}}\)
which is distance of closest approach.
OR
Density pf nucleus
ρ_n = \(\frac{m}{\frac{4}{3} \pi R^{3}}\) = \(\frac{3 m}{4 \pi R^{3}}\)
ρ_n = \(\frac{3 \times 1.67 \times 10^{-27}}{4 \times 3.14 \times\left(1.20 \times 10^{-15}\right)^{3}}\)
p_n = 2.307 × 10¹⁷ kg m^-3
Density of water ρ_w = 10³ kg m^-3
> \(\frac{\rho_{n}}{\rho_{w}}\) = \(\frac{2.307 \times 10^{17}}{10^{3}}\) = 2.307 × 10¹⁴

Question 9.
Draw a circuit diagram showing the biasing of an LED and state the factors which control wavelength of light and intensity of light emitted by a diode. (3)
Answer:

Wavelength of light is controlled by the band gap of semiconductor material.
Intensity of ammeter by the diode concentration of impurity in the junction diode.

Related Theroy
The p-n junctions can be designed so that current through them changes either by causing electron excitation by light photons or conversely through electron excitation by a suitable bias voltage resulting in emission of light photons.

Question 10.
You are given three lenses of power 0.5 D, 4 D and 10 D to design a telescope.
(A) Which lenses should be used as objective and eyepiece? Justify your answer.
(B) Why is the aperture of the objective preferred to be large?
OR
A beam of light consisting of two wavelengths, 800 nm and 600 nm is used to obtain the interference fringes in a Young’s double slit experiment on a screen placed 1.4 m away. If the two slits are separated by 0.28 mm, calculate the least distance from the central bright maximum where the bright fringes of the two wavelengths coincide. (3)
Answer:
(A) Magnification,
m = \(\frac{f_{0}}{f_{e}}\)
= \(\frac{P_{e}}{P_{0}}\)
Therefore, the lens of 0.5 D should be used as objective and the lens of 10 D should be used as eye-piece in order to achieve higher magnification.

(B) The aperture of the objective lens is made larger so, that it receives as much light as coming from the distant object and the resolving power of the telescope increases.
OR
Given:
λ₁ = 800 nm
= 800 × 10^-9 m
λ₂ = 600 nm
= 600 × 10^-9 m
D = 1.4 m
d = 0.28 mm
= 0.28 × 10^-3m
Suppose n₁^thaximum corresponds to wavelength λ₁ and it coincides with n₂^th maximum corresponding to wavelength λ₂.
∴ n₁\(\frac{\lambda_{1} D}{d}\) = n₂\(\frac{\lambda_{2} \mathrm{D}}{d}\)
or \(\frac{n_{1}}{n_{2}}=\frac{\lambda_{2}}{\lambda_{2}}\) = \(\frac{600}{800}=\frac{3}{4}\)

Thus, 3^rd maximum corresponding to wavelength 800 nm coincides with 4^th maximum corresponding to wavelength 600 nm.
And the minimum distance is given by,
X_min = n₁\(\frac{\lambda_{1} D}{d}\)
\(\frac{3 \times 800 \times 10^{-9} \times 1.4}{0.28 \times 10^{-3}}[ /latex]
X_min = 12mm

Question 11.
Write three characteristic features in photoelectric effect which cannot be explained on the basis of wave theory of light, but can be explained only by using Einstein’s equation. (3)
Answer:
(A) Existence of threshold frequency: According to wave theory, there should not exist any threshold frequency but Einstein’s theory explains the existence of threshold frequency.

(B) Dependence of kinetic energy on frequency of incident light: According to wave theory, the maximum kinetic energy of emitted electrons should depend on intensity of incident light and not on frequency whereas Einstein’s equation explains that it dependence on frequency and not on intensity of the incident light.

(C) Instantaneous emission of electrons: According to wave theory there should be time lag between emission of electrons and incident of light whereas Einstein’s equation explains why there is no time lag between incident of light and emission of electrons.

Related Theory
Some features cannot be explained in photoelectric current. So, Einstein gave his correction to explain the effect.

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

Question 12.
The wave theory of light was first put forward by Christian Huygen. During that period, everyone believed in Newton’s corpuscular theory, which had satisfactorily explained the phenomenon of reflection, refraction, the rectilinear propagation of light and the fact that light could propagate through vacuum. Huygen’s model was accepted when Thomas Young performed the interference experiment.

(A) Which one of the following phenomena is not explained by Huygen’s construction of wave front?
(i) Refraction
(ii) Reflection
(iii) Diffraction
(iv) Origin of spectra

(B) When light waves suffer reflection at the interface from air to glass, the change in phase of reflected waves is equal to:
(i) Zero
(ii) [latex]\frac{\pi}{2}\)
(iii) π
(iv) 2π

(C) In the given diagram a wavefront AB moving in air is incident on a plane glass surface xy. Its position CD after refraction through the glass slab is shown also along with normal drawn at A and D.

The refractive intex of glass will be equal
(i) \(\frac{\mathrm{BD}}{\mathrm{AC}}\)
(ii) \(\frac{A B}{C D}\)
(iii) \(\frac{B D}{A D}\)
(iv) \(\frac{A C}{A D}\)

(D) Select the right options in the following:
(i) Christian Huygen’s a contemporary
of Newton established the wave theory of light by assuming that light waves were transverse.
(ii) Maxwell provided the compelling theoretical evidence that light is transverse wave.
(iii) Thomas Young experimentally proved the wave behaviour of light and Huygen’s assumption.
(iv) All the statements given above, correctly answers the question.

(E) The diagram below shows two sources, A and B, vibrating in phase in the same uniform medium and producing circular wave fronts.

Which phenomenon occurs at point P?
(i) Destructive interference
(ii) Constructive interference
(iii) Reflection
(iv) Refraction (5)
Answer:
(A) (d) Origin of spectra
Explanation: The Huygen’s construction of wavefront does not explain the phenomena of origin of spectra.

(B) (c) π
Explanation: When a light wave suffers reflection at the interface from air to glass, it experiences a change in phase because the refractive index of glass is more than air, as we know that air is a rarer medium and glass is denser medium. So, the change in phase of the reflected wave is π.

Caution
Students are often confused about the phase change in reflection. When a light wave travels from rarer to denser than the phase of the incident wave change with angle it and when it travels from denser to rare then the phase of the incident wave does not change.

(C)(a) \(\frac{B D}{A C}\)
Explanation: BD = V₁t,
AC = V₂t
\(\frac{\sin \theta}{\sin \phi}\) = \(\frac{\left(\frac{B D}{A D}\right)}{\left(\frac{A C}{A D}\right)}\)
\(\frac{\mathrm{BD}}{\mathrm{AC}}\) = \(\frac{v_{1} t}{v_{2} t}\)
= \(\frac{v_{1}}{v_{2}}=\frac{\mu_{2}}{\mu_{1}}\)
= \(\frac{\mu_{2}}{\mu_{1}}=\frac{\mathrm{BD}}{\mathrm{AC}}\)

Caution
Students are often confused in the geometry of the diagram. They should observe the angles and the common arm of the triangle.

(D) (b)
Maxwell provided the compelling theoretical evidence that light is transverse wave
Explanation: In Huygens principle the wave theory was established by assuming that the waves were longitudinal. So, (a) is wrong.

In Maxwell’s theory of light, he stated that light is a transverse wave and conducted multiple experiments to prove. So, (b) is correct.

Thomas Young conducted his double slit experiment and tried to prove the wave nature of light by proving that light undergoes interference which is a phenomenon showed by waves. He never proved the type of wave light is, i.e., transverse or longitudinal. So, (c) is wrong.

These statements mostly highlight the wave nature of light. They do not comment on the particle nature of light or the dual nature of light established. Hence, they clearly do not answer the question “what is light” correctly. So, option (D) is also wrong.

(E) (b) Constructive interference Explanation: Wave fronts for a source represents locus of points at same phase. A and B are vibrating in phase. Hence, all the wave fronts shown in the diagram are in the same phase. At P, there are wave fronts of both A and B. Hence, the waves of A and B are at same phase at P. Hence, constructive interference occurs at P.

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)

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)

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)

(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