Prasanna

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

CBSE Sample Papers for Class 12 Biology Standard Term 2 Set 12 for Practice

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• All questions are compulsory.
• The question paper has three sections and 13 questions. All questions are compulsory.
• Section-A has 6 questions of 2 marks each; Section-B has 6 questions of 3 marks each; and Section-C has a case-based question of 5 marks.
• There is no overall choice. However, internal choices have been provided in some questions. A student has to attempt only one of the alternatives in such questions.
• Wherever necessary, neat and properly labeled diagrams should be drawn.

SECTION – A
(Section A has 6 Questions of 2 marks each.)

Question 1.
(A) Name the property of normal cells, the absence of which in cancerous cells leads to the formation of tumours.
(B) The tumors are classified into two types, in which one of the types remains confined to the original location and the other type of tumours invade neighboring tissues. Name the two types of tumours. (2)

Question 2.
(A) Given below is the picture of the plant Rauwolfia vomitoria. Genetic variation is very important in this plant. Why?

(B) A bird named Great Indian Bustard found in Rajasthan is a threatened species. What do you mean by threatened species?
OR
Rahul read an article that hunting and human population growth is major threats to the Bengal tigers. Due to these reasons, it has become an endangered species. What do you mean by endangered species? Give two examples. (2)

Question 3.
(A) Name the gases that are produced in anaerobic sludge digesters.
(B) Which enzyme is used as a clot-buster in patients who have undergone myocardial infarction, to remove clots from blood vessels? (2)

Question 4.
Some individuals experience sudden sneezing and wheezing after reaching a hill station, while these symptoms disappear when they return to the plains. What is such a response called? How does the body produce it? (2)

Question 5.
Egrets and grazing cattle are often seen together as depicted in the picture. What is this type of interaction called? Give a reason for such kind of association.

OR
Illustrate the relationship between biotic potential and environmental resistance. (2)

Question 6.
(A) Define Antibiotics.
(B) Which was the first antibiotic and who discovered it? (2)

SECTION – B
(Section B has 6 Questions of 3 marks each.)

Question 7.
(A) If one of the insect pollinators becomes extinct in an ecosystem. Would it affect the ecosystem? If yes, how?
(B) Seed banks are used to store seeds of different genetic strains. Briefly describe the conservative strategy involved in this.
(C) From the study of the fossil records, it was found that mass extinction of species happened earlier, even before humans appeared. What could have triggered mass extinctions of species in the past? (3)

Question 8.
(A) Why malaria is restricted to the tropics?
(B) Explain why tetanus antitoxin is given to a person injured in an accident with a bleeding wound and not a tetanus vaccine? (3)

Question 9.
Explain briefly how desert ephemeral plants are adapted to withstand hot and dry environments?
OR
(A) How would you classify organisms on the basis of their tolerance to salinity?
(B) Certain animals like fungi, zooplankton and bears have the ability to cope up with the temporary short-lived stressful environmental conditions. How do they do it? (3)

Question 10.
(A) Reema (who works in a genetic engineering firm) was making her friend understand that manipulation of living organisms by the human race is not ethical and should be regulated. Is it true? Why? Name the set of standards which are used to regulate human activities in relation to biological world.
(B) A human protein a-1 antitrypsin is produced in a transgenic animal created by the introduction of the gene which codes for it. For what purpose this protein is used? (3)

Question 11.
(A) In most of the genetic engineering experiments, mice are used as the preferred organism for transgenic production. Why?
(B) RNA interference involves the silencing of specific mRNA that helps to prevent many parasitic infestations. How? (3)

Question 12.
(A) What are the different places where methanogens are found?
(B) How do methanogens help in producing biogas? (3)

SECTION – C
(Section C has a case-based question of 5 marks.)

Question 13.
During o visit to a genetic engineering laboratory, students saw a PCR machine as shown in the picture. The students were amazed os their teacher told them, this small machine can make miLlions of copies of DNA from a very small sample of blood, hair, etc, and that too within few hours. Based on this answer the following questions:

(A) Who deveLoped the technique of PCR?
(B) What ¡s the basis of the technique of PCR?
(C) What ¡s the source of Tag Potymerose enzyme used in PCR?
(D) Name the finaL step of PCR and explain it briefly.

OR
The following picture shows different types of bioreactors that are used in downstream processing.
Based on the below picture answer the following questions:

(A)
(i) Why mechanical agitation is required in stirred tank bioreactors and not in air-lift reactors?
(ii) What do you mean by continuous culture? List one of its advantages.
(B) List any two advantages of stirred tank bioreactors over shake flasks. (5)

Students can access the CBSE Sample Papers for Class 12 Physics 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 Physics Standard Term 2 Set 4 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.
What is an ideal diode and draw the output waveform across R for the input waveform given below? (2)

Answer:
The diode offering zero resistance in forward bias, infinite resistance in reverse bias and having no potential barrier ideal diode. In the first half cycle the diode is forward biased and conducts, shown output is obtained across r, while in the next half-cycle the diode is reverse biased does not conduct, so no output is obtained across R.

Question 2.
In the given fusion reaction, find the mass defect, Q-value and state whether the fusion is energetically favorable? He⁴+He⁴ → Be⁸
m (Be⁸) = 8.0053 u and m(He⁴) = 4.0026 u.
OR
In Young’s double-slit experiment, plot a graph showing the variation of fringe width versus the distance of the screen from the plane of the slits keeping other parameters same. What type of information can we obtain from slope of the curve? (2)
Answer:
He⁴, He⁴ →, Be⁸ + Q
Δm=2 x 4.0026-8.0053 = 8.0052-8.0053
=-0.0001
Q=(2m_He — m_He)c = (2 x 4.0026 – 8.0053) x 931.5 MeV = – 931.5 KeV
Since Q is negative, the fusion is not energetically favorable.
OR
The fringe width in Young’s double-slit experiment is given by
β = \(\frac{\lambda D}{d}\)
D = distance between slit and screen
d = distance between slits

So, β ∝ D
It is linear graph with slope \(\frac{\lambda}{d}\) . So the fringe width to vary Linearly with distance of screen from the slits, the ratio of wavelength to distance between the slits should remain constant. Therefore, it is advised to take wavelengths incident light nearly equal to the width of the slit.
Related Theory:
In Youngs double-slit experiment fringe width is directly proportional to the distance of screen the slit.

Question 3.
Describe a reflecting type telescope and mention its advantages. (2)
Answer:
Cassegrain telescope is a reflecting type telescope.

Telescopes with mirror objectives are called reflecting telescopes. Advantages of taking mirror objectives are:
(A) There is no chromatic aberration in mirrors.
(B) MechanicaL support is much less of a problem since Lens weighs more than mirror of equal. equivalent optical quality.
(C) If a parabolic reflecting surface is chooser, spherical aberration is also removed.

Related Theory
The Largest reflecting telescope in the world ore the pair of rack telescopes in Hawai USA. with reflectors
of 10 meters in diameter.

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

Question 4.
An electron and a photon each have a wavelength of 1.50 nm. Find:
(A) their momentum.
(B) the energy of the photon.
(C) the kinetic energy of the eLectron. (3)
Answer:
(A) For each electron or photon, momentum,
p = \(\frac{h}{\lambda} \)
= \(\frac{6.63 \times 10^{-34}}{1.5 \times 10^{-9}}\)
= 4.42 x 10^-25kg\(\frac{m}{\text { sec }}\)

(B) E = \(\frac{h c}{\lambda}\)
= \(\frac{\left.\left.6.63 \times 10^{-34}\right) 3 \times 10^{8}\right) e V}{\left(1.5 \times 10^{-9}\right)\left(1.6 \times 10^{-19}\right)}\)
= 828.75 eV

(C) E_k = \(\frac{1}{2} \frac{p^{2}}{m}\)
= \(\frac{1\left(4.42 \times 10^{-25}\right)^{2}}{2 \times 9.1 \times 10^{-31} \times 1.6 \times 10^{-19}}\)eV
= 0.671 eV

Question 5.
(A) Prove the behavior of a plane wavefront by using Huygens’s geometrical construction when:
(i) passing through o biconvex lens and
(ii) reflecting by a concave mirror.
(B) ExpLain why the refracted Light have the same frequency as that of the incident Light when monochromatic Light is incident on a surface separating two media.
OR
How can we convert a full-wave rectifier to a half-wave rectifier changing the connection of wires? (3)
Answer:

(B) The frequency and time period of an electromagnetic wave depends only on the source which produces it. The frequency is independent of the medium through which it travels. But the speed and wavelength depend on the medium through which the wave travels. Because of this, the frequency and time period of sound wave does not change due to change in the medium.
OR
If we change the connection of load resistance from point C to point B due to which no current will flow through second diode in second half cycle.

Related theory:
As the connection is altered, only one diode will be functional which Will result in decrease in efficiency
of the diode, and the output voltage will be cut down to exactly half.

Question 6.
(A) What is the significance of negative signs in the expression for the energy?
(B) Draw the energy level diagram showing how the line spectra corresponding to Paschen series occur due to transition between energy levels. (3)
Answer:
(A) Negative sign indicates that revolving electron is bound to the positive nucleus.
(B) For Paschen series, n = 3 and n_i = 4, 5,..
\(\frac{1}{\lambda}=\mathrm{R}\left(\frac{1}{3^{2}}-\frac{1}{n_{i}^{2}}\right)\)
where, n_i= 4, 5 ……………………..
Total energy, E (eV)

Question 7.
(A) What is the nature of the final image in a compound microscope?
(B) You are given two converging lenses of focal lengths 1.25 cm and 5 cm to design a compound microscope.
If it is desired to have a magnification of 30, find out the separation between the objective and the eyepiece? (2)
Answer:
(A) The final image in a compound microscope is inverted with respect to the object. It is
virtual and magnified.
(B) Given, f_e= 5 cm;
f₀= 1.25 cm
and M=-30.
Let L be the tube length (distance between the objective and the eyepiece).
∴ M = \(\frac{-L}{f_{0}}\left(1+\frac{d}{f_{e}}\right)\)
[d is a constant and equals to the normal distance of clear vision of the human eye].

Hence, -30 = \(-\frac{L}{1.25}\left(1+\frac{25}{5}\right)\)
∴ L = \(\frac{30 \times 1.25}{6}\)
= 6.25 cm
Hence, the tube [ength = 6.25 cm.

Caution
Students are often confused about the separation of objective and eyepiece. Tube Length is the separation between the objective and the eyepiece.

Question 8.
(A) How does the fringe width, in Young’s double-slit experiment change when the distance of separation between the slits and screen is doubled?
(B) In Young’s double-slit experiment maximum intensity is lo, find the intensity at a point on the screen where
(i) the phase difference between the two interfering beams is \(\frac{\pi}{3}\)
(ii) the path difference between them is \(\frac{\lambda}{4}\). (3)
Answer:
(A) Fringe width
β = \(\frac{\lambda D}{d}\)
⇒ β ∝ D
The fringe width becomes double when the distance of separation between the slits and screen is doubled.
(B) (A) I= I‘max Cos² \(\left(\frac{\phi}{2}\right)\)
Here I_max is I_o (i.e.. intensity due to independent sources is \(\frac{I_{0}}{4}\) .).

Therefore, at
Φ = \(\frac{\pi}{3}\)
or
\(\frac{\phi}{2}=\frac{\pi}{6}\)
I =I₀cos² \(\left(\frac{\pi}{6}\right)=\frac{3}{4}\) I₀

(B) Phase difference corresponding to the given path difference Δx = \(\frac{\lambda}{4}\) is
Φ = \(\left(\frac{2 \pi}{\lambda}\right)\left(\frac{\lambda}{4}\right)\)
= \(\frac{\pi}{2}\) (Φ = \(\frac{2 \pi}{\lambda} \Delta x\))
\(\frac{\phi}{2}=\frac{\pi}{4}\)
I = I₀ cos²\(\left(\frac{\pi}{4}\right)=\frac{I_{0}}{2}\).

Question 9.
Explain giving reasons for the following:
(A) Photoelectric current in a photocell increases with the increase in the intensity of the incident radiation.
(B) The stopping potential (V₀) varies linearly with the frequency (v) of the incident radiation for a given photosensitive surface with the slope remaining the same for different surfaces.
(C) Maximum kinetic energy of the photoelectrons is independent of the intensity of incident radiation. (3)
Answer:
(A) Since number of photo electrons emitted is directly proportional to the intensity of incident radiation, therefore, as intensity increases the electron-hole pairs also increases.

(B) hv = hv₀ + eV₀
eV₀= h(v – v₀)
⇒ V₀=\(\frac{h}{e}\) (v-v₀)

Further from graph,\(\frac{h}{e}\) slope is which is a constant and it does not depend on v.
KE_max = h(v – v₀)
Hence, it depends on the frequency and not on the intensity of the incident radiation.
(C) As intensity increases, the number of photons increases but the energy remains same.

Question 10.
Derive the expression for the wavelength of lines of different spectral series in H-atom.
OR
(A) In semiconductors, thermal collisions are responsible for taking a valence electron to the conduction band. Why does the number of conduction electrons not go on increasing with time as thermal collisions continuously take place?
(B) In a p-n junction, the depletion region is 400 nm wide and an electric field of 5 x 10⁵\(\frac{\mathrm{V}}{\mathrm{m}}\) exists in it. m
(i) What will be the height of the potential barrier?
(ii) What should be the minimum kinetic energy of a conduction electron which can diffuse from the n-side to the p-side? (3)
Answer:
10. By Bohr’s III concept
ΔE = E₂ – E₁
\(\frac{h c}{\lambda}\) = E₂ – E₁
\(\frac{1}{\lambda}=\frac{E_{2}-E_{1}}{h c}\) ……………………… (i)
E₂ = – \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \cdot \frac{1}{n_{2}^{2}}\) ………………. (ii)
E₁ = – \(\frac{m e^{4}}{8 \varepsilon_{0}^{2} h^{2}} \cdot \frac{1}{n_{1}^{2}}\) ………………………. (iii)

Putting (ii) and (iii) n equation (i), we get

(A) The presence of electrons in the conduction band saturates the further production of electron-hole pairs.
(B) (i) Height of potentiaL barrier =5 x 10⁵ V/m x 400 x 10^-9
= 0.2 V
(ii) Kinetic energy
=q x V= 1.6 x 10^-19 x 0.2
= 0.2 V

Question 11.
Angular width of central maxima in the Fraunhofer diffraction pattern of a slit is measured. The slit is illuminated by light of wavelength 6000 A. When the slit is illuminated by light of another wavelength, then the angular width decreases by 30%. Calculate the wavelength of this light. The same decreases in the angular width of central maximum is obtained when original apparatus is immersed in a liquid. Find the refractive index of the liquid. (3)
Answer:
An guiar width of central maximum
= \(\frac{2 \lambda}{a}\)
λ₁ = 6000 Å = 60 nm,
θ₁ = \(\frac{2 \lambda_{1}}{a}\)
θ₂ = θ₁ x 0.7
= \(\frac{2 \lambda_{1}}{a}\)

On dividing both the equations, we get
\(\frac{1}{0.7}=\frac{\lambda_{1}}{\lambda_{2}}=\frac{600}{\lambda_{2}}\)
λ₂ = 420 nm
When apparatus is under liquid,

The refractive index of the given medium is 1.42.

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

Question 12.
A changing electric field produces a changing magnetic field and vice-versa which gives rise to a transverse wave known as electromagnetic waves. The time-varying electric field and magnetic field are mutually perpendicular to each other also perpendicular to the direction of propagation. Thus, the electromagnetic waves consist of sinusoidally time-varying electric and magnetic fields acting at right angles to each other as well as at right angles to the direction of propagation.

(A) The amplitudes of electric and magnetic fields in EM waves are E0 and Bo respectively.
They are related as:
(i) E₀ = \(\frac{\mathrm{B}_{0}}{c}\)
(ii) E₀ = cB₀
(iii) cE₀ = B₀
(iv) E₀ = c²B₀

(B) Electric and magnetic field vectors in EM waves are:
(i) perpendicular to each other
(ii) parallel to each other
(iii) 270° to each other
(iv) 180° to each other

(C) A plane electromagnetic wave of frequency 25 MHz travels in free space along the x-direction. At a particular point in space and time E = 6.3 \(\hat{j} \frac{v}{m}\) the B at this point is:

(D) If E and B represent electric and magnetic field vectors of the electromagnetic waves, then the direction of propagation of the electromagnetic waves is that of:
(i) E
(ii) B
(iii) E x B
(iv) B x E

(E) In an electromagnetic wave, the electric and magnetic fields are 100 V/m and 0.265 A/m.
The maximum energy flow is:
(i) 79 \(\frac{W}{m^{2}}\)
(ii) 13.2 \(\frac{W}{m^{2}}\)
(iii) 53.0 \(\frac{W}{m^{2}}\)
(iv) 26.5 \(\frac{W}{m^{2}}\) (5)
Answer:
(A) (b) E₀ = cB₀
Explanation: Electric field and magnetic field vectors in an electromagnetic wave is related as \(\frac{E}{B}\) = c where c s the speed of light. The electromagnetic waves travel with the speed of Light.

(B) (a) Perpendicular to each other Explanation: The angle between electric and magnetic field vectors in EM waves is 900.

(C) (b) 2.1 x 10^-8 \(\hat{k}\)T
Explanation: f’ = 25 MHz = 25 x 10⁶ Hz,
E=6.3 \(\hat{j} \frac{v}{m}\)
E₀ = 6.3
\(\left(\frac{E_{0}}{B_{0}}\right)\) = c

∴ E is along y-direction and wave is along x-direction hence B has to be perpendicular to both x and y axes. E x B should be along the x-direction. B shouLd be along z-direction B = B₀ k
B=2.1 x 10^-8\(\hat{k}\) T

(D) (c) E x B
Explanation: The electric field and the magnetic field of an electromagnetic wave are mutually perpendicular to each other and their direction of propagation is perpendicular to both the electric and magnetic fields. If E and B represent electric and magnetic field vectors of electromagnetic waves respectively, then the direction of propagation of the electromagnetic wave is along the direction of vector E x B.

Caution:
Students are often confused about the vector product of two quantities The vector product B x E gives the negative direction of the propagation of wove. The electric and magnetic fields of an electromagnetic wave are perpendicular to each other and also perpendicular to the direction of wave propagation.
Hence these are transverse waves.
(E) (d) 26.6 \(\frac{W}{m^{2}}\)

Explanation: Here, amplitude of electric field,
E₀ = 100\(\frac{V}{m}\)
Amplitude of magnetic field,
B₀ =0.265 \(\frac{A}{m}\)
The maximum rate of energj flow,
S = E₀ x B0 = 100 x 0.265
= 26.5 \(\frac{W}{m^{2}}\).

Students can access the CBSE Sample Papers for Class 12 Chemistry 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 Chemistry Term 2 Set 7 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.
Study the change of concentration of reactant with respect to time as depicted in the graphical representation below: (2)
(A) Predict the order of the reaction
Answer:
The variation in the concentration (R) vs. time (t) plot shown here represents a zero order reaction, for which the rate of the reaction is proportional to zero power of the concentration of the reactants.

(B) What does the slope represent?

Answer:
For a zero order reaction, rate constant is given as [R] = [R]₀ – kt
So, the slope of the curve for the variation in the concentration (R) vs. time (t) plot is equal to the negative of the rate constant for the reaction.

Question 2.
Arrange the following compounds in increasing order of their property as indicated. to-carbon bond of alkenes? (Any two)
(A) CH₃COCH₃, C₆H₅—CO—C₆H₅, CH₃CHO (Reactivity towards nucleophilic addition reactions)
Answer:
C₆H₅—CO—C₆H₅, < CH₃COCH₃ < CH₃CHO

Explanation: Ketones are less reactive towards nucleophilic addition reactions due to more steric hinderance.

(B)

Answer:

Explanation: More the number of electron withdrawing groups (-Cl), more the acidic character.

(C) C₂H₅OH, CH₃CHO, CH₃COOH (Boiling points) (2)
Answer:
CH₃—CHO < C₂H₅OH, < CH₃COOH

Explanation: Carboxylic acids have higher boiling points then alcohols due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bonds are not broken completely even in the vapour phase.

Question 3.
Answer the following questions.
(A) What feature of their structure makes aldehydes easier to oxidize than Ketones?
Answer:
The H on the carbonyl carbon atom of aldehyde makes it easier to oxidize.

(B) How does the carbon-to-oxygen bond of aldehydes and ketones differ from the carbon-to-carbon bond of alkenes? (2)
Answer:
The carbon-to-oxygen double bond is polar due resonance; the carbon-to-carbon double bond is non-polar.

Related Theory:
There is a significant contribution from the resonance structure which puts a formal negative charge on the oxygen atom and a formal positive charge on the carbon atom and there by increases the polarity of the bond.

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

Question 4.
Draw diagram to show splitting of d – orbital in octahedral crystal field. Explain the two patterns of filling d⁴ in octahedral crystal FIeld. (3)
Answer:

The splitting of the d orbitals in an octahedral field takes place in such a way that dx² – y², dz² experience a rise in energy and form the eg level, while d_xy, d_yz and d_zx experience a fall in energy and form the t_2g level.
For electric configuration of d⁴

(i) When Δ₀ > P Electronic configuration is t⁴2g eg⁰

When Δ₀ < P
Electronic configuration is t³_2g eg¹

Question 5.
A colloidal solution of Agl is prepared by two methods. (3)

(A) What is the charge on Agl colloidal method.
Answer:
Test tube (A) has negative charge whereas test tube (B) has positive on the colloidal particles.

(B) Give the reason for origin of change.
Answer:
According to preferential adsorption theory, test tube (A) I^– is absorbed on precipitate Agl [or Agl / I^– is formed] and in test tube (B). Ag⁺ is absorbed on precipitate Agl [or Agl/Ag⁺ is formed].

Related Theory:
The electric double layer theory describes the interaction between surface of colloidal particles and ions that are present in the fluid in which the colloidal particles are dispersed.

(C) What ¡s zeta potential?
Answer:
Zeta potential is the potential difference between fixed layer (primary layer) and diffused layer (secondary layer) of colloidal particle. It is also called as electrokinetic potential.

Question 6.
A first order reaction takes 10 minutes for 25% decomposition. Calculate t_1/2 for the reaction. (Given: log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021).
OR
Consider the reaction
2A + B → C + D
Following results were obtained in experiments designed to study the rate of reaction:

(A) Write the rate law for the reaction.
(B) Calculate the value of rate constant for the reaction. (3)
Answer:

OR

Question 7.
(A) Describe the mechanism of the addition of Grignard reagent to the carbonyl group of compound to form an adduct which on hydrolysis yields an alcohol.
(B) Write the IUPAC name of the compound.

OR
(A) Illustrate the following name reactions:
(i) Hell-Vothard-Zelinsky reaction.
(ii) Wolff-Kishner reduction reaction
(B) Write the structural formula of 1-phenylpentan-l-one. (3)
Answer:
(A) a. Mechanism:
(i) Nucleophilic addition of Grignard reagent to carbonyl group to form an adduct.

(ii) Hydrolysis of the adduct to alcohol.

Caution:
Students should draw the curved arrows in proper directions and proper charges on the atoms in the mechanism.

(B) 4-Hydroxy-4-methyl pentan-2-one.
OR
(A) (i) Hell-Volhard-Zelinsky reaction – Carboxylic acids having an α-hydrogen are halogenated at the α-position on treatment with chlorine or bromine in the presence of small amount of red phosphorus to give αx-halocarboxylic acids. The reaction is known as Hell-Volhard- Zelinsky reaction.

(ii) Wolff-Kishner reduction reaction – The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with hydrazine followed by heating with potassium hydroxide in a high boiling solvent such as ethylene glycol.

Caution:
Students should write the chemical reaction also otherwise marks would be deducted during evaluation.

(B)

Question 8.
State Reasons:
(A) Aniline is a weaker base than cyclohexylamine. (3)
Answer:
Aniline is a weaker base than cyclohexylamine because the lone pair of electrons on the N-atom is delocalised over the benzene ring in aniline which results in the decrease in electron density on Nitrogen. In cyclohexylamine, the lone pair of the electrons on N-atom is readily available due to absence of π-electrons.

Caution:
Basicity of the amino group means it is unsuitable for reactions with acids (e.g. H₂SO₄ or AlCl₃) such as nitration, sulfonation and Friedel-Crafts alkylation or acylation.

(B) It is difficult to prepare pure amines by ammonolysis of alkyl halides.
Answer:
It is difficult to prepare pure amines by ammonolysis of alkyl halides as the primary amine formed by ammonolysis acts as a nucleophile. It further produces 2° and 3° alkyl amine.

(C) Etectrophilic substitution in aromatic amines takes place more readily than benzene.
Answer:
Arylamines are potentially very reactive towards electrophilic aromatic substitution. This is because -NH₂, -NHR₂ and -NR₂ are very strong activators and are ortho, para-directing.

Question 9.
Observe the graph and answer the questions that follow:

(A) Which transition metaL of 3d series has positive E°(M²⁺/M) value and why?
Answer:
The E°_(M²⁺/M) for copper is positive. This is because high energy is required to transform Cu to Cu²⁺ which is not balanced by its hydration enthalpy.

(B) E° value for the Mn⁺³/Mn⁺² couple is positive (+1.5 V) whereas that of Cr⁺³/Cr⁺² is negative (-0.4 V). Why?
Answer:
The E° value for the Mn³⁺/Mn²⁺ couple is much more positive than that for Cr³⁺/ Cr²⁺ couple. This is because Mn²⁺ ion is particularly stable due to extra stability of its half filled valence electronic configuration (d⁵). Thus Mn³⁺ ion has a very high tendency to gain an electron and form the much more stable Mn²⁺ ion.

(C) E° values are not regular for first row transition elements? (3)
Answer:
The (M²⁺/M) values are not regular which can be explained from the irregular variation of ionisation enthalpies (ΔiH₁ + ΔiH₂) and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question 10.
What is the lanthanoid contraction? What are its causes and consequences? (3)
OR
Explain the following:
(A) Copper (I) ion is not stable in an aqueous solution.
(B) Generally there is an increase in density of elements from titanium (Z = 22) to copper (Z = 29) in the first series of transition elements.
(C) Transition metals in general act as good catalysts.
Answer:
Lanthanoid contractions – The cumulative effect of the regular decrease in size or radii of Lanthanoid with increase in atomic number is called Lanthanoid contraction.

Causes: With an increase in the atomic number, the positive charge on nucleus increases by one unit and one more electron enters same 4f subshell.
The electrons in 4f subshell imperfectly shield each other. Shielding in a 4f subshell is lesser than in d subshell.
With the increase in nuclear charge, the valence shell is pulled slightly towards the nucleus. This causes lanthanide contraction.

Consequences: Due to Lanthanoid contraction.
1. Radii of the members of the third transition series is similar to those of second transition series.
2. It becomes difficult to separate Lanthanoids.
OR
(A) In an aqueous medium, Cu²⁺ is more stable than Cu⁺. This is because although energy is required to remove one electron from Cu⁺ to Cu²⁺, high hydration energy of Cu²⁺ compensates for it. Therefore, C^u+ ion in an aqueous solution is unstable. It disproportionates to give Cu²⁺ and Cu.

(B) The density of elements from titanium to copper increase in the first series of transition elements. This is due to decrease in metallic radius coupled with increase in atomic mass results in a general increase in the density.
Explanation: As the atomic radii decreases moving across from titanium to Cu, So its volume will decrease and density is expected to increase.

(C) They have variable valencies and show multiple oxidation states and transition metals sometime form unstable intermediate compounds and provide a new path with lower activation energy for the reaction. In some cases transition elements provide a suitable surface for the reaction to take place.

Question 11.
An aromatic compound ‘A’ of moLucutar formula C₇H₅O₂ undergoes a series of reactions an shown below. Write the structures of A, B, C, D and E in the following reactions.

OR
How will you convert the following –
(A) Aniline to chtorobenzene
(B) Ethanoic acid to methanamine
(C) Methyl chloride to ethanamine (3)
Answer:

OR

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:
The potential of each electrode is known as electrode potential. Standard electrode potential is the potential when concentration of each species taking part in electrocde reaction is unity and the reaction is taking place at 298 K. By convention, the standard electrode potential of hydrogen (SHE) is 0.0 V. The electrode potential value for each electrode process is a measure of relative tendency of the active species in the process to remain in the oxidised/reduced form. The negative electrode potential means that the redox couple is stronger reducing agent than H⁺/H₂ couple. A positive electrode potential means that the redox couple is a weaker reducing agent than the H⁺/H₂ couple. Metals which have higher positive value of standard reduction potential form the oxides of greater thermal stability.
(A) What is meant by reference electrode?
(B) Platinum is used in the standard hydrogen electrode. Give reason.
(C) Explain the term Standard electrode potential.
(D) Calculate the emf of the following cell at 298 K: Fe(s) | Fe²⁺ (0.001 M) || H⁺ (1M) | H₂(g)_{(1 bar)}, Pt_(s) (Given E°cell = +0.44V)
OR
Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10. (5)
Answer:
(A) Standard Hydrogen electrode: It is a reference electrode against which the electrode potentials of all electrodes are measured.
Explanation: The Standard electrode potential of SHE is zero volt. This SHE can act as anode or cathode, depending upon the half cell that is attached to it.

(B) Platinum is a less reactive metal. So, it doesn’t easily react with other metals. As a result, it provides the surface for the redox reaction.

(C) The potential difference developed between metal electrode and solution of ions of unit molarity (1M) at 1 atm pressure and 25°C (298 K) is called standard electrode potential. It is denoted by E°.

(D) The cell reaction:

Students can access the CBSE Sample Papers for Class 12 Physics 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 Physics Standard Term 2 Set 3 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.
Under what conditions does the phenomenon of total internal reflection take place? Draw a ray diagram showing how a rag of light deviates by 90° after passing through a right-angled isosceles prism.
OR
What are the (A) momentum, (B) de Broglie wavelength of an electron with kinetic energy of 120 eV? (2)
Answer:
The phenomenon of total internal reflection occurs when,
(A) Angle of incidence is equaL or greater than critical angle that is i ≥ C
(B) Angle of incidence is equal or greater than critical angle that is i ≥ C In case of right-angle isosceles triangle, if light rays fall normally on face AB the light ray incident of face AC with angle of incidence greater than the critical angle.

Hence, total Internat reflection will occur with normal to the surface of BC. The phenomenon in which a ray of light travelling at an angle of incidence greater than the critical angle from denser to a rarer medium is totally reflected back into the denser medium is called total internal reflection.
OR
Here, kinetic energy of electron
E = 120 eV
= 120 x 1.6 x 10^-19 J
= 192 x 10^-19J

(A) Momentum of electron,
P = \(\sqrt{2 m_{e} \mathrm{E}} \)
= \(\sqrt{2 \times 9.1 \times 10^{-31} \mathrm{~kg} \times 192 \times 10^{-19} \mathrm{~J}}\)
= 5.9 x 10^-24 Kg ms^-1

(B) de – Brogue wavelength,
P = \(\frac{h}{\lambda}\)
λ = \(\frac{h}{p}\)
= \(\frac{6.63 \times 10^{34} \mathrm{Js}}{5.9 \times 10^{-24} \mathrm{~kg} \mathrm{~ms}^{-1}}\)
= 1.12 x 10^-10
λ = 1.12 Å.

Question 2.
From the relation R = R_o A^1/3 where R_o is a constant and A is the mass number of a nucleus, show that the nuclear matter density is independent of A? (2)
Answer:
If m is the overage mass of a nucleon and R is the nucLear radius, then mass of nucteus = mA,
where A is the mass number of the element Volume of the nucleus
V = \(\frac{4}{3} \pi R^{3}\)
Given R = R₀ A^1/3
⇒ V= \(\frac{4}{3} \pi\left(R_{0} A^{1 / 3}\right)^{3}\)
⇒ V = \(\frac{4}{3} \pi R_{0}^{3} A\)

Density of nuclear matter,
ρ = \(\frac{m \mathrm{~A}}{\mathrm{~V}}\)
⇒ ρ = \(\frac{m \mathrm{~A}}{\frac{4}{3} \pi \mathrm{R}_{0}^{3} \mathrm{~A}}\)
ρ = \(\frac{3 m}{4 \pi R_{0}^{3}}\)
This shows that nuclear density is independent of A.

Question 3.
The given graph shows the V-I characteristics of a semiconductor diode.

(A) Identify the semiconductor diode used.
(B) Draw the circuit diagram to obtain the given characteristics of this device.(2)
Answer:
(A) Since the voltages quoted in the given diagram are much more than actual values and It is usually < 0.6 V. The semiconductor the diode used is Zener diode.
(B) The circuit diagram for the characteristics of Zener diode is shown as: For forward bias Zener diode

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

Question 4.
Write Einstein’s photoelectric equation and mention which important features in photoelectric effect can be explained with the help of this equation. The maximum kinetic energy of the photo-electrons gets doubled when the wavelength of light incident on the surface changes from λ₁ to λ₂. Derive the expressions for the threshold wavelength λ₀ and work function for the metal surface. (3)
Answer:
Einstein’s photoelectric equation is given by
hv =Φ₀ + K_max
where K_max = Maximum kinetic energy of the photoelectron
v = Frequency of the Light radiation
h = Planck’s constant
Φ₀= Work function

We have.
K_max = \(\frac{1}{2} m v^{2}_{\max }\)
where, v_max = Maximum velocity of the emitted photoelectron
m = Mass of the photoelectron
Therefore.
hv =Φ₀= \(\frac{1}{2} m v^{2}_{\max }\)
If v₀ is the thresho[d frequencj, then the work function can be written as
Φ₀ = hv

Hence,
hv = hv₀ + \(\frac{1}{2} m v^{2}_{\max }\)
hv – hv₀ = \(\frac{1}{2} m v^{2}_{\max }\)
\(\frac{1}{2} m v^{2}\) = h (v -v₀)

The above equations explains the following results:
(i) If v < v₀, then the maximum kinetic energy is negative, which is impossible.
Hence, photoelectric emission does not take place for the incident radiation below the threshold frequency. Thus, the photoelectric emission can take place if v > v₀.

(ii) The maximum kinetic energy of emitted photoelectrons is directly proportional to the frequency of the incident radiation. This means that maximum kinetic energy of photoelectrons depends only on the frequency of incident light not on the intensity According to the photoelectric equation.
K₁ = \(\frac{1}{2}\) mv²
= hv – Φ₀
Let the maximum kinetic energy for the wavelength λ₂
K₁ = \(\frac{h c}{\lambda_{1}}-\phi_{0}\) ……………………… (i)
Let the maximum kinetic energy for the waveLength of the incident λ₂ be K₂
K₂ = \(\frac{h c}{\lambda_{2}}\) – Φ₀ …………………. (ii)
(Kmax)2 = (Kmax)1 (given)

Work function is the energy required to eject o photoeLectron from the metal.
Φ₀ = \(\frac{h c}{\lambda_{0}} [latex]
∴ Φ₀ = [latex]\frac{h c\left(2 \lambda_{2}-\lambda_{1}\right)}{\lambda_{1} \lambda_{2}}\).

Question 5.
Draw a ray diagram showing the image formation by a compound microscope. Hence obtain expression for total magnification when the image is formed at infinity. (3)
Answer:
A compound microscope consists of two convex parallel Lenses separated by some distance. The Lens nearer to the object is called the objective lens. The lens through which the final image is viewed is called the eyepiece or eye lens.

The magnification produced by the compound microscope is the product of the magnifications produced by the eyepiece and objective.
M = M_e X M₀

Where M_e and M₀ are the magnifying powers of the eyepiece and objective respectively. If u₀ is the distance of the object from the objective and v₀ is the distance of the image from the objective, then the magnifying power of the objective is:
M₀ = \(\frac{h^{\prime}}{h}=\frac{L}{f_{0}}\)

Where, h, h’ are object and image heights respectively and f₀ is the focal length of the objective and L is the tube length Le., the distance between the second focal point of the objective and the first focal point of the eyepiece.
When the final image is at infinity,
M_e = \(\frac{D}{f_{e}}\)
Magnifying power of compound microscope.
M = M₀ x M_e
= \(\frac{L}{f_{0}} \times \frac{D}{f_{e}}\)

Question 6.
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.
OR
In a Geiger Marsden experiment, calculate the distance of closest approach to the nucleus of Z = 75, when an a-particle of 5 MeV energy impinges on it before it comes momentarily to rest and reverse its direction.
How will the distance of closest approach be affected when the kinetic energy of the a particle is doubled? (3)
Answer:
Given:
λ₁ = 800 nm
=800x 10^-9m
= 600 nm
λ₂ =600 x 10^-9m
D=14m
d = 0.28 mm
= 0.28 x 10^-3m
Suppose n₁th, maximum corresponds to wavelength λ₁ and it coincides with n₂th maximum corresponding to wavelength λ₂.
∴ \(n_{1} \frac{\lambda_{1} D}{d}=n_{2} \frac{\lambda_{2} D}{d}\)
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\(\frac{\lambda_{1} D}{d}\)
= \(\frac{3 \times 800 \times 10^{-9} \times 1.4}{0.28 \times 10^{-3}}\)
X_min = 12 mm
OR
Let r₀ be the centre to centre distance between the α-particle and nucleus when the α-particle is at its stopping point.
Now, E_k = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{r_{0}}\)
r₀ = \(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{E_{K}}\)
Given, E_k = 5 x 10⁶ eV
= 5 x 10⁶ x 1.6 x 10^-19 V
Z = 75
r₀ = \(\frac{9 \times 10^{9} \times 75 \times 2 \times\left(1.6 \times 10^{-19}\right)^{2}}{5 \times 10^{6} \times 1.6 \times 10^{-19}}\)
= \(\frac{3456 \times 10^{9} \times 10^{-38}}{8 \times 10^{-13}}\)
= 432 x 10^-16 m
= 43.2 x 10^-15 m
= 43.2 fm
Since, distance of cLosest approach (r₀) is given
r₀ =\(\frac{1}{4 \pi \varepsilon_{0}} \frac{(2 e)(Z e)}{E_{K}}\)
⇒ r₀ ∝ \(\frac{1}{E_{K}}\)
So, when kinetic energy of α-particle is doubled the distance between closest approach r₀ is halved.

Related Theory:
kinetic energy of’ α-particle is inverseLy proportional the distance of closest approach.

Question 7.
(A) What type of wavefront will emerge from a point source and a distant light source?
(B) Make a labelled diagram showing the wavefronts in reflection from:
(i) Plane mirror
(ii) Curved mirror (3)
Answer:
(A) From o point source, the wavefront is diverging spherical wavefront and from a distant Light source, the wavefront is plane front.
(B) (i) Reflection from plane mirror:

(ii) Reflection from curved mirror

Question 8.
Distinguish on the basis of features between conductors, insulators and semiconductors. (3)
Answer:

Question 9.
Describe the role of processes involved in the formation of p-n junction with the help of a diagram. (3)
Answer:
Two process occur during the formation of p-n junction are diffusion and drift.
(i) Diffusion: In n-type semiconductors, the concentration of electrons is much greater as compared to concentration of holes arid in p-type semiconductors, the concentration of holes is much greater than concentration of electrons. When p-n junction is formed, concentration gradient is set up. The holes diffuse n-side and electrons dislikes the top-side. This motion of charge carriers gives rise to diffusion current across the junction.

(ii) Drift: The drift charge carriers occur due to the electric field. As potential barrier is built, on electric field is directed from n-side to p-side of the junction. This field causes motion of electrons on p-side of the junction to n-side of the junction and motion of holes on n-side to p-side of the junction. This is known as drift current which is opposite direction of diffusion current

Question 10.
(A) Can a charge moving with a constant velocity act as a source of electromagnetic waves?
(B) In a plane EM wave, the electric field oscillates sinusoidally at a frequency of 2 x 10¹⁰Hz and amplitude 48 vm’1.
(i) What is the wavelength of the wave?
(ii) What is the amplitude of the oscillating magnetic field? (3)
Answer:
(A) A charge moving with a constant velocity has zero acceleration and according to the classical theory of electromagnetism only accelerated charge can produce electromagnetic wave. Therefore, a charge moving with a constant velocity cannot produce electromagnetic wave. Frequency of the electromagnetic wave,
v=2.0x 1010Hz
Electric field amplitude, E₀ = 48 Vm-1
Speed of light, c = 3 x 108 m/s
E₀ = 48 \(\frac{\mathrm{V}}{m}\)
and c=3 x 10⁸ms^-1
(i) λ = \(\frac{c}{v}\)
= \(\frac{3 \times 10^{8} \mathrm{~ms}^{-1}}{2 \times 10^{10} \mathrm{~Hz}}\)
= 1.5 x 10^-2 m
(ii) c = \(\frac{E_{0}}{B_{0}}\)
B₀ = \(\frac{E_{0}}{c}\)
= \(\frac{48 \frac{\mathrm{V}}{m}}{3 \times 10^{8} \mathrm{~ms}^{-1}}\)
= 1.6 x 10^-7 T

Question 11.
Give reason for following observations in YDSE.
(A) The resultant intensity at any point on the screen varies between 0 and 4 times the intensity, due to one slit.
(B) A few coloured fringes, around a central white region, are observed on the screen, find the source of monochromatic light is replaced by white light.
(C) Why no interference is seen when two coherent sources are infinitely close to each other or when they are far apart from each other?
OR
Explain the processes of nuclear fission and nuclear fusion by using the plot of binding energy per nucleon (BE/A) versus the mass number A.(3)
Answer:
(A) The resuLtant intensity, at any point on the screen is given by.
I = 4I₀, cos²\(\frac{\phi_{0}}{2}\)
For constructive interference;
Φ₀= O, 2π. 4π and so on
I = O for minimum intensity
For destructive interference:
Φ₀ = π, 3π, 5π and so on
I = 4I₀ for maximum intensity.
Therefore, intensity varies from zero to four times the intensity due to single slit.

(B) The interference patterns due to different colours of white light overtop incoherently.
The central bright fringes for different colours are at the same position. Therefore the central fringe is white and the fringes closest, on either side of central white fringe, are red and farthest wilt appear blue. After a few fringes, no clear fringe pattern is seen.

(C) When two coherent sources are placed infinitely close to each other, the fringe width becomes very large. Even a single fringe width may occupy the entire screen. When the sources are far apart, the fringe width increases. At a very large separation. it becomes too small to be detected. Hence, the interference pattern cannot be detected when the sources are infinitely close to each other or kept far apart from each other.
OR
A plot of binding energy per nucleon moss number is shown in the figure given below

(A) When we move from the heavy nuclei region to the middle region of the plot, we find that there will be again in the overall binding energy and hence results in release of energy. This indicates that energy can be released when a heavy nucleus (A ~ 240) breaks into two roughly equal fragments.
This process is called nuclear fission.

(B) Similarly. when we move from lighter nuclei to heavier nuclei, we again find that there will be again in the overall binding energy and hence release of energy takes place. This indicates that energy can be released when two or more Lighter nuclei fuse together to form a heavy nucleus. This process is called nuclear fusion.

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

Question 12.
Dispersion is the splitting of white light into its constituent colours. The dispersion occurs in prisms but not in glass slabs because of its geometric design. In a slab, the opposite sides are parallel to each other whereas in case of the prism, the sides are not parallel to each other.

In prism, the dispersion occurring at one face and is enhanced at the other end but in glass slabs, dispersion occurring at one end is neutralized by refraction at the other end. When we put two prisms in series but in opposite direction, they will act as a glass slab together.

Based on the above facts, answer the following questions:
(A) Two parallel rays of red and violet colour pass through a glass stab as shown In the figure. Which of the following is correct?

(i) 3 and 4 are parallel
(ii) 4 and 5 are parallel
(iii) 6 and 3 are parallel
(iv) 2 and 5 are parallel.
(B) A convex lens is made up of two types of material Upper half is made by the refractive index n₁ and Lower half is made by the refractive index n₂. Which of the following is true for rays coming from infinity?
(ï) Two images are formed
(ii) One image is formed
(iii) Continuous ¡mage is formed between the focal points of upper and tower tens.
(iv) Image is not formed

(C) Maximum Lateral displacement of ray Light incident on a stab of thickness t is:
(i) \(\frac{t}{2}\)
(ii) \(\frac{t}{3}\)
(iii) \(\frac{t}{4}\) (iv) t
(D) The refractive index of water, glass, and diamond are 1.33, 1.50, and 2.40, respectively. What is the refractive index of the diamond relative to water and of glass relative to diamond, respectively are nearly?
(i) 1.80, 1.6
(ii) 0.5 54, 1.6
(iii) 1.80, 0.625
(iv) 0.554, 0.625

(E) An air bubble in a glass slab with refractive index 1.5 (near-normal incidence) is 5 cm deep when viewed
from one surface and 3 cm deep when viewed from the opposite face. The thickness (in cm) of the slab is:
(i) 8
(ii) 10
(iii) 12
(iv) 16 (5)
Answer:
(A) (d) 2 and 5 are parallel Explanation: When refraction occurs through a parallel glass slab, the emergent ray is parallel to incident ray. From the given diagram
6||2,5||1
Because 1 || 2
⇒ 2 || 5
(B) (a) Two images are formed
Explanation: The given lens is made of two materials of different refractive indices, so it has two focal lengths. Hence, two images are formed.

‘Related Theory
Due to the two materials of different refractive indices lens will form two images (C) (d)t
Explanation: For I = 90⁰ and r=0⁰, Lateral shift is maximum.
Lateral shift for a sLob of t thickness
LS = t x sin\(\frac{(i-r)}{\cos r}\)
LS = t
Hence, the Lateral shift is equaL to thickness t.

Caution:
Students are often confused about the values of i and r when lateral shift is maximum. in this case i = 90° and r= 0°
(D) (c) 1.80, 0.625
Explanation: n_w = 1.33,
n_g = 1.5
and n_d= 2.4
Now, _wn_d = \(\frac{n_{d}}{n_{w}}\)
= \(\frac{2.4}{1.33}\) = 1.80
Further, _dn_g = \(\frac{n_{g}}{n_{d}}\)
= \(\frac{1.50}{2.4}\) = 0.625

(E) (c)12
Explanation: Refractive index
n= \(\frac{\text { Real thickness }}{\text { Apparent”thickness }}\)
ReaL thickness = n x Apparent thickness
= 1.5 x (5 + 3) = 12 cm

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

CBSE Sample Papers for Class 12 Chemistry Term 2 Set 12 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.
Arrange the following compounds in increasing order of their property as indicated – (Any two) (2)
(A) CH₃CHO, C₆H₅CHO, HCHO (reactivity towards nucleophilic addition reaction).
(B) 2, 4-dinitrobenzoic acid, 4-methoxybenzoic acid, 4 nitrobenzoic acid(acidic character).
(C) CH₃CH₂CH₂CH₃, CH₃OCH₂CH₃,
CH₃CH₂CHO, CH₃COCH₃, CH₃CH₂CH2₂OH
(boiling point)

Question 2.
Why do amines act as nucleophiles? Give example of a reaction in which methylamine acts as a nucleophile. (2)

Question 3.
(A) In the expression of rate of reaction in terms of reactants, what is the significance of negative sign?
(B) What is the use of integrated rate equation? (2)

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

Question 4.
(A) 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 and 3d⁴?
(B) What are interstitial compounds? Why are such compounds well known for transition metals? (3)

Question 5.
When Liquid ‘A’ is treated with a freshly prepared ammoniacal silver nitrate solution, it gives bright silver mirror. The liquid forms a white crystalline solid on treatment with sodium hydrogen sulphite. liquid ‘B’ also forms a white crystalline solid with sodium hydrogen sulphite but it does not give test with ammoniacal silver nitrate. Which of the two liquids is aldehyde? Write the chemical equations of these reactions also.
OR
Complete the following equation and write the structures of A, B, C, D, E, and F.

Question 6.
(A) Differentiate between the following with the help of one example of each
(i) Homoleptic and Heterolytic Complexes
(ii) Double salt and a complex
(B) What is the coordination number of Fe in [Fe(EDTA)]³⁺ ? (3)

Question 7.
(A) How does the precipitation of colloidal smoke takes place in Cottrell precipitator?
(B) Observe the following diagram and answer the questions that follow –

(i) Name the property which is shown by the colloidal solution.
(ii) Explain the reason for the property shown by colloidal solution. (3)

Question 8.
Write the structures of the following compounds.
(A) α-Methoxypropionaldehyde
(B) 3-Hydroxybutanal
(C) 2-Hydroxycyclopentanecarbaldehyde (3)

Question 9.
(A) Why does acetylation of -NH₂ group of aniline reduce its activating effect?
(B) What is the product when C₆H₅CH₂NH₂ reacts with HNO₂ ?
(C) What is the structure and IUPAC name of the compound, allyl amine?
OR
How will you convert the following:
(A) Nitrobenzene into aniline
(B) Ethanoic acid into methanamine
(C) Aniline into N-phenylethanamide (3)

Question 10.
For the complex [Fe(en)₂ Cl₂ ], Cl, (en = ethylene diamine), identify:
(A) the oxidation number of iron,
(B) the hybrid orbitals and the shape of the complex,
(C) the magnetic behaviour of the complex,
OR
(A) What type of isomerism is shown by the complex [Co(NH₃ )₆ ][Cr(CN)₆ ]?
(B) Why a solution of [Ni(H₂O)₆ ]²⁺ is green while a solution of [Ni(CN)₄]^2- is colourless? (At. no. of Ni = 28)
(C) Write the IUPAC name of the following complex: [Co(NH₃ )₅ (CO₃ )]Cl. (3)

Question 11.
For the reaction A + B → products, the following initial rates were obtained at various given initial concentrations:

[A] mol / L	[B] mol / L	Initial rate M/s
1) 0.1	0.1	0.05
2) 0.2	0.1	0.10
3) 0.1	0.2	0.05

Determine the half-life period.
OR
The following data were obtained during the first order thermal decomposition of
SO₂Cl₂ at a constant volume:
SO₂Cl_2(g) → SO_2(g) + Cl_2(g)

Experiment	Time/s-1	Total Pressure/atm
1	0	0.4
2	100	0.7

Calculate the rate constant.
(Given: log 4 = 0.6021, log 2 = 0.3010) (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:
ALl chemical reactions involve interaction of atoms and molecules. A large number of atoms/ molecules are present in a few grams of any chemical compound varying with their atomic/ molecular masses. To handle such large number conveniently, the mole concept was introduced. All electrochemical cell reactions are also based on mole concept. For example, a 4.0 molar aqueous solution of NaCI is prepared and 500 mL of this solution is electrolysed. This leads to the evolution of chlorine gas at one of the electrode. The amount of products formed can be calculated by using mole concept.
(A) How many moles of chlorine gas will be evolved from the reaction given in the above passage? .
(B) In electrolysis of aqueous NaCl solution when Pt electrode is taken, then which will the products formed at cathode and anode?
(C) What will be number of moles of electrons exchanged during electrolysis of aqueous solution of NaCl?
(D) Calculate the time to deposit 1.5 g of silver at cathode when a current of 1.5A was passed through the solution of AgNO₃. (Molar mass of Ag = 108 g mol- 1,1 F = 96500 C mol^-1). ”
OR
How many electrons flow through a metallic wire if a current of 0.5 A is passed for 2 hours? (Given: IF = 96,500 C mol^-1) (5)

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

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

Time Allowed: 2 Hours
Maximum Marks: 35

General Instructions:

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

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

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

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

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

Conductivity and molar conductivity of electrolytes increase with increasing temperature.

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

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

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

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

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

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

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

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

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

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

C₆H₅CHO + NaOl → No yellow ppt

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

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

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

= 0.804 A

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

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

Explanation:

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

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

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

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

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

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

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

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

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

OR

Question 9.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Students can access the CBSE Sample Papers for Class 12 Chemistry 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 Chemistry Term 2 Set 2 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) Arrange the following amines in order of decreasing basic strength—Ethylamine, ammonia, tri ethylamine.
Answer:
Triethyl amine (C₂H₅)₃N> Ethyl amine C₂H₅NH₂> Ammonia NH₃
Explanation: More the number of alkyl groups, more the +1 effect so more the basic strength.

(B) Why aniline is acetylated first to prepare mono bromo derivative ? (2)
Answer:
Due to the strong activating effect of amino group, aromatic amines readily undergo electrophilic substitution reactions and it is difficult to stop the reaction the monosubstitution stage. Therefore, the acetyl group attached with aniline deactivates the aniline due to its electron withdrawing effect. Thus acetylation of aniline reduces its activation effect.

Question 2.
If the half-life period of a first-order reaction in A is 2 minutes, how long will it take to reach 25% of its initial concentration? (2)
Answer:

Question 3.
Draw the structure of the alcohol that could be oxidized to each compound. (2)
(i) 2-propanone
(ii) Cyclohexanone
Answer:

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

Question 4.
Using thermodynamic principles, explain: (3)
(A) Entropy has negative value for adsorption.
Answer:
The randomness of the overall process decreases due to the presence of a force of attraction between adsorbent and adsorbate due to which the gas molecules stick to the surface. So, Entropy has negative value for adsorption.

(B) Adsorption is a exothermic process?
Answer:
Adsorption is an exothermic process since surface particles of the adsorbent are unstable and when the adsorbate is adsorbed on the surface, the energy of adsorbent decreases, and this results in the evolution of heat. Therefore, adsorption is always exothermic.

(C) Relation between enthalpy and entropy at equilibrium.
Answer:
When ΔG becomes zero, adsorption equilibrium is established. Since it is an exothermic process, ΔH is negative. Since the adhering of gas molecules to the surface lowers the randomness, the ΔS is negative. So, ΔH and TΔS become equal at equilibrium to make ΔG zero. (ΔG = ΔH – TΔS)

Question 5.
The following table summarizes the oxidation states of the 3d series elements.

Account for the following questions related to the above table:
(A) Transition metals show variable oxidation state.
(B) At the beginning of the series, +3 oxidation state is stable but towards the end +2 oxidation state is stable.
(C) Mn has six different oxidation states from +2 to +7.
OR
Explain the trends in the properties of the members of the First series of transition elements:
(A) E⁰ (M²⁺/M) value for copper is positive (+0.34V)in contrast to the other members of the series.
(B) Cr²⁺ is reducing while Mn²⁺ is oxidising though both have d⁴ configuration.
(C) The oxidising power in the series increases in the order VO₂⁺ < Cr₂ O₇^2- < MnO₄^– (3)
Answer:
(A) Transition elements show variable oxidation states because their valence electrons are in two different sets of orbitals, that is (n-l)d and ns. The energy difference between these orbitals is very less, so both the energy levels can be used for bond formation. Thus, transition elements have variable oxidation states.

(B) The number of oxidation states increases with the number of electrons available, and it decreases as the number of paired electrons increases.
Explanation: In 3d series, first element Sc has only one oxidation state +3; the middle element Mn has six different oxidation states from + 2 to +7 because number of unpaired electrons is maximum but the last element Cu shows +1 and +2 oxidation states only because it has maximum number of paired electrons.

(C) The different oxidation states of manganese (Mn) are +2, +3, +4, +5, +6 and +7. Manganese with 3d⁵, 4s² as the outer electronic configuration, shows oxidation states ranging from +2(3d⁵) to + 7(3d⁰). The +2 oxidation state is very stable due to exactly half-filled 3d-orbital of higher stability.
OR
(A) Copper has high atomisation Δ_aH° and low hydration energy Δ_hydH°. Due to which the E° value is positive.
Explanation: The E_θ(M2+/M) value of a metal depends on the energy changes involved in the following:
1. Sublimation : The energy required for converting one mole of an atom from the solid state to the gaseous state.
M_(x) → M_(g) Δ_s H(Sublimation energy)

2. Ionization : The energy required to take out electrons from one mole of atoms in the gaseous state.
M_(x) → M_(g) Δ_i H(lonization energy)

3. Hydration: The energy released when one mole of ions hydrated.
M_(g) → M²⁺_(ag) Δ_hyd H(Hydration energy)
Now, copper has a high energy of atomization and low hydration energy. Hence, the E^Φ (M²⁺/M) value for copper is positive.

(B) Cr²⁺ is reducing agent as its configuration changes from d⁴ to d³, when it is oxidized to Cr³⁺ . On the other hand, the reduction of Mn³⁺ to Mn²⁺ results in the half-filled (d⁵) configuration which has extra stability hence Mn³⁺ acts as oxidizing agent.

(C) Greater the oxidation state, higher is the oxidising power.
Explanation: Oxidation state of V in VO₂⁺+ is 5, Cr in Cr₂O₇^2- is 6 and Mn in MnO₄^– is 7.

Question 6.
Based on valence bond theory, explain the bonding in the coordination entity [Co(NH₃)₆]³⁺. Also, comment on the geometry and spin of the given entity. (Atomic no. of Co = 27) (3)
Answer:
[Co(NH₃)₆³⁺
Co³⁺ = 3d⁶

In this complexion, the oxidation state of cobalt is +3.
This complex involves the d²sp³ hybridisation. The six pairs of electrons, one from each NH₃ molecule, occupy the six hybrid orbitals. It proves that complex has octahedral geometry.

Question 7.
A metal complex having composition Cr(NH₃)₄Cl₂Br has been isolated in two forms A and B. The form A reacts with AgN03 to give a white precipitate readily soluble in dilute aqueous ammonia, whereas B gives a pale-yellow precipitate soluble in concentrated ammonia
(i) Write the formulae of isomer A and B
(ii) State the hybridisation of Cr in each of them.
OR
Give reason for each of the following:
(A) Co²⁺ is easily oxidised to Co³⁺ in presence of a strong ligand.
(B) CO is a stronger complexing reagent than NH₃.
(C) The molecular shape of Ni(CO)₄ is not the same as that of [Ni(CN)₄]^2- (3)
Answer:
(i) Formula of A → [Cr(NH₃)₄ Br Cl] Cl
Formula of B →[Cr(NH₃)₄ Cl₂] Br
Explanation: ‘A’ and ‘B’ are Ionisation isomers. ‘A’ has Cl as onion so it gives white ppt. with AgNO₃
Whereas ‘B’ has Br as onion so it gives pale yellow ppt. with AgNO₃.
(ii) Hybridisation state of Cr in each of them is d²sp³.
OR
(A) Co²⁺ ions can be easily oxidised to Co³⁺ ions because the crystal field stabilisation energy of Co³⁺ ions with a d⁶ configuration is higher than d⁷ configuration.
In presence of strong ligand, the d- orbitals of metal gets splits up into two set of orbitals that is t_2g and e_g. The half filled or fully filled t_2g is more stable than other configuration. This is because it gives maximum amount of low energy of t_2g orbital with electrons. Lower the energy, more is the stabilisation.

(B) Since CO can form a as well as n bond, whereas NH₃ has lone pair of electrons and can form a bond only. Therefore, CO is a better complexing reagent than NH₃.
Explanation: CO has empty π orbitals which overlap with filled d-orbitals (t_2g orbitals) of transition metals and form π bonds by back bonding. These π interaction increases the value of crystal field stabilisation energy (∆₀).
As NH₃ cannot from π bonds by back bonding, therefore, CO is a stronger ligand than NH₃.

(C) In [Ni(CO)₄], nickel is present in zero oxidation state {Ni = 3d⁸ 4s²} but in [Ni(CN)₄ ]^2-, oxidation state of Nickel is +2. So, Ni²⁺: 3d⁸ 4s⁰.
Explanation: Electronic configuration of:
Ni [Ar]4s², 3d⁸
So Ni (0): [Ar] 4s⁰, 3d¹⁰
CO ligand causes pairing of electrons and shifting of electrons from 4s to 3d.
[Ni(CO)⁴]

[Ni(CO)₄] has sp³ hybridisation and has tetrahedral shape.

[Ni (CN₄)]^2- has dsp² hybridisation and has square planar shape.

Question 8.
The graphical representation of concentration of A Vs time is given for a general A → B,

The following questions:
(a) What is the order of the reaction?
(b) What is the slope of the curve?
(c) State the units for the rate constant. (3)
Answer:
(A) Zero order reaction
(B) slope = – k
(C) Unit of k = mol L^-1 s^-1

Explanation: In case of a zero-order reaction, the rate of reaction is independent of the concentration of the reactant. The concentration [A]t of the reactant at a time t is given by [A]_t = – kt + [A]₀ (y = – mx +c) where [A]₀ is the initial concentration of the reactant and k is a rate constant.
When the concentration of the reactant is plotted against time, a straight line with the slope equal to – k is obtained.

Question 9.
Complete the following reactions:

OR
Write chemical equations for the following reactions:
(A) Propanone is treated with diLute Ba(OH)₂.
(B) Acetophenone is treated with Zn(Hg)/ Conc. HCI
(C) Benzoyl chloride is hydrogenated in presence of Pd/BaSO₄. (3)
Answer:

Explanation: Reaction is Hell-Vollard-Zelinsky reaction in which carboxylic acids react with chlorine or bromine in presence of small amount of red phosphorus to give a-halocarboxylic acids.
OR
(A)

Explanation: Aldol condensation is an organic reaction in which an enolate ion reacts with carboxyl compound to form β-hydroxy aldehyde or β-hydroxy ketone. Hydroxide functions as a base and therefore moves the acidic a-hydrogen producing the reactive enolate ion.

(B)

Explanation: The Clemmensen reduction is a reaction that is used to reduce aldehydes or ketones to alkanes using hydrochloric acid and zinc amalgam.

(C)

Explanation: Rosenmund reduction is the hydrogenation of an acyl chloride to an aldehyde, in presence of catalyst (Pd supported on barium sulphate). Barium sulphate partially poisons Pd metal and prevents over-reduction. Untreated Pd catalyst is too reactive.

Question 10.
(A) Account for the following:
(i) Cl- CH₂COOH is o stronger acid than CH₃COOH.
Answer:
(i) Cl group is an electron withdrawing creating less electron density on oxygen of carboxylic acid making the release of proton easier than acetate ion. Hence, Cl-CH₂COOH is a stronger acid than CH₃COOH.

Explanation: Chlorine due to its -1 effect increases the acidity of carboxylic acids by stabilising the conjugate base due to delocoalisation of the negative charge by resonance effect.

(ii) Carboxylic acids do not give reactions of the carbonyl group.
Answer:
Carboxylic acids do not give characteristic reaction of carbonyl group. This is because the lone pairs on oxygen atom attached to hydrogen atom in the -COOH group are involved in resonance thereby making the carbon atom less electrophilic Hence, carboxylic acids do not give characteristic reaction of carbonyl group.

Explanation:

(B) Give simple chemical tests to distinguish Benzoic acid and Phenol. (3)
Answer:
Phenol and benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test: Phenol reacts with neutral Fecl₃ to form an iron-phenol complex giving violet colouration. But benzoic acid reacts with neutral FeCl₃ to give a buff coloured ppt.

Question 11.
(A) Aniline is a weaker base than cyclohexylamine. Give reason.
(B) Amino group in aniline is ortho and para directing, Then why does aniline on nitration give substantial amount of m-nitroaniline?
(C) There are two test tubes (abetted A and B. Give suitable test to identify which test tube contains aniline and which contains methylamine.
OR
(A) Arrange the following in increasing order of their pK_a values:
(i) C₆H₅NH₂, C₂H₅NH₂, (C₂H₅) ₂NH, NH₃
(B) Identify the compounds A, B, C and D In the following set of reactions

Answer:
(A) In aniline, benzene group is electron withdrawing in nature. Hence, it decreases the availability of electron pair on N-atom and is less basic. On the other hand, in cyclohexylamine, the cyclohexyl group is electron releasing in nature. Hence, this, increases the availability of electrons on N-atom and is more basic.

(B) Nitration is carried out in an acidic medium. In an acidic medium, aniline is protonated to give anilinium ion (which is meta-directing). Due to which aniline on nitration gives a substantial amount of m-nitroaniline.

(C) Nitrous acid Test – Treat aniline and methylamine with sodium nitrite with hydrochloric acid at zero degree Celsius.
In case of aniline: stable diazonium salt will be formed which will give phenol on hydrolysis.
In case of methylamine: unstable diazonium salt of aliphatic amine will be formed which will further get decompose into bubbles of nitrogen gas and methanol.
OR
(A) Increasing order of pKa value-
(C₂H₅)₂NH < C₂H₅NH₂ < NH₃ < C₆H₅NH₂
Explanation: Higher the pKa value, less the basic character.Diethylamine is the most basic due to maximum +1 effect (lowest pKa) and aniline is the least basic due to resonance (highest pKa).
(B)

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 – (5)
Electrolysis, process by which electric current is passed through a substance to effect a chemical change. The process is carried out in an electrolytic cell, an apparatus consisting of positive and negative electrodes held apart and dipped into a solution containing positively and negatively charged ions. The substance to be transformed may form the electrode, may constitute the solution, or may be dissolved in the solution. Electric current (i.e., electrons) enters through the negatively charged electrode (cathode); components of the solution travel to this electrode, combine with the electrons, and are transformed (reduced). The products can be neutral elements or new molecules. Components of the solution also travel to the other electrode (anode), give up their electrons, and are transformed (oxidized) to neutral elements or new molecules.

Electrolysis is used extensively in metallurgical processes, such as in extraction (electrowinning) or purification (electrorefining) of metals from oresor compounds and in deposition of metals from solution (electroplating). Metallic sodium and chlorine gas are produced by the electrolysis of molten sodium chloride; electrolysis of an aqueous solution of sodium chloride yields sodium hydroxide and chlorine gas. Hydrogen and oxygen are produced by the electrolysis of water.

Michael Faraday discovered in 1833 that there is always a simple relationship between the amount of substance produced or consumed at an electrode during electrolysis and the quantity of electrical charge Q which passes through the cell.
(A) Name two metals which can be purified using electrolytic refining.
(B) What are the electrolysis products of an aqueous solution of sodium chloride?
(C) How much charge in terms of Faraday is required for the reduction of 1 mol of Cu²⁺ to Cu?
(D) How many moles of mercury will be produced by electrolysing 1.0 M Hg(NO₃)₂ solution with a current of 2.00 A for 3 hours? [Hg(NO₃)₂ = 200.6 g mol^-1].
OR
A steady current of 2 amperes was passed through two electrolytic cells X and Y connected in series containing electrolytes FeSO₄ and ZnSO₄ until 2.8 g of Fe deposited at the cathode of cell X. How long did the current flow?
(Molar mass: Fe = 56 g mol^-1, Zn = 65.3 g mol^-1, 1F = 96500 C mol^-1)
Answer:
(A) Na (Sodium) and K (Potassium)

(B) H₂ gas is liberated at cathode. Cl₂ gas is liberated at anode. NaOH is formed in the solution.
Explanation:

At anode: 2Cl^– → Cl₂ + 2e^–; At cathode: 2H⁺ + 2e^– → H₂

(C) Cu²⁺ + 2e- → Cu
Quantity of charge required for reduction of 1 mol of Cu²⁺ = 2F(F = Faraday)
Explanation: Faraday’s first law of electrolysis – The amount of any substance deposited or liberated at the electrode is directly proportional to the quantity of electricity passing through the electrolyte.

(D) Mass of mercury produced at the cathode,

OR
(D) Current i = 2 A, Mass m of Fe = 2.8 g
Molar Mass M of Fe = 56 g mol^-1, 1F = 96500 c Fe2+ + 2e” -> Fe (Charge required = 2F)

Students can access the CBSE Sample Papers for Class 12 Physics 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 Physics Standard Term 2 Set 1 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 consists 3 questions of 2 marks each.)

Question 1.
In a pure semiconductor crystal of Si, if antimony is added then what type of extrinsic semiconductor is obtained. Draw the energy band diagram of this extrinsic semiconductor so formed. (2)
Answer:
As given in the statement antimony is added to pure Si crystal, then an n-type extrinsic semiconductor would be so obtained Since antimony (Sb) is a pentavalent impurity. Energy level diagram of n-type semiconductor

Related Theory
Each pentavalent impurity atom donates one extra electron: it is known as donor. Most of the current is carried by negatively charged electrons, so, the semiconductors doped with donor type impurities are known as n-type semiconductors.

Question 2.
Consider two different hydrogen atoms. The electron in each atom is in an excited state. Is it possible for the electrons to have different energies but same orbital angular momentum according to the Bohr model? Justify your answer.
OR
Explain how does (A) photoelectric current and (B) kinetic energy of the photoelectrons emitted in a photocell vary if the frequency of incident radiation is doubled, but keeping the intensity same? Show the graphical variation in the above two cases. (2)
Answer:
No
Because according to Bohr’s model,
E_n = \(-\frac{13.6}{n^{2}}\) and electrons having different energies belong to different levels having different values of n.
So, their angular momenta will be different, as ‘
L = mvr = \(\frac{n h}{2 \pi}\)
OR
(A) The increase in the frequency of incident radiation has no effect on photoelectric current. This is because of incident photon of increased energy cannot eject more than one electron from the metal surface. f = photoelectric current f = frequency of incident radiation fo = threshold frequency

The kinetic energy of the photoelectron becomes more than the double of its original energy. As the work function of the metal is fixed, so incident photon of higher frequency and hence higher energy will impart more energy to the photo electrons.

Question 3.
Name the device which converts the change in intensity of illumination to change in electric current flowing through it. Plot l-V characteristics of this device for different intensities. State any two applications of this device.(2)
Answer:
Photodiodes are used to detect optical signals of different intensities by changing current flowing through them.

I-V Characteristics of a photodiode Applications of photodiodes:
1. In detection of optical signals.
2. In demodulation of optical signals.
3. In light operated switches.
4. In speed reading of computer punched cards.
5. In electronic counters
(any two out of these or any other relevant application)

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

Question 4.
Derive an expression for the frequency of radiation emitted when a hydrogen atom de-excites from LeveL n to Level (n – 1). Also show that for Large values of n, this frequency equals to classical frequency of revolution of an eLectron. (3)
Answer:
From Bohr’s theory, the frequency f of the radiation emitted when an electron de – excites from Level n₂ to Level, n₁ is given as,
f = \(\frac{2 \pi^{2} m k^{2} z^{2} e^{4}}{h^{3}}\left[\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right]\)
Given n₁ = n – 1 = n . derivation of it
f = \(\frac{2 \pi^{2} m k^{2} z^{2} e^{4}}{h^{3}} \frac{(2 n-1)}{(n-1)^{2} n^{2}}\)
For large n. 2n- 1 = 2n,n- 1 =n and z= 1
Thus, f = \(\frac{4 \pi^{2} m k^{2} e^{4}}{n^{3} h^{3}}\)
which is same as orbital. frequency of etectron in nth orbit.
f = \(\frac{v}{2 \pi r}=\frac{4 \pi^{2} m k^{2} e^{4}}{n^{3} h^{3}}\)

Question 5.
Explain with a proper diagram how an ac signal can be converted into dc ( pulsating) signal with output frequency as double than the input frequency using pnjunction diode. Give its input and output waveforms. (3)
Answer:
A junction diode aLlows current to pass only when it is forward biased. So, Wan alternating voltage is applied across a diode the current flows only in that part of the cycle when the diode is forward biased. This property is used to rectiy alternating voltages and the circuit used for this purpose is called a rectifier.

Explanation: A full wave rectifier converts ac signal into dc signaL During positive half of the cycle, diode 1 is forward biased and diode 2 is reverse biased. During negative half of the cycle, diode 1 is forward biased. Output voltage is obtained during both half cycles. This process is known as rectification.

Caution:
Output frequency double of input frequency is not possible in half wave rectifier. In half wove rectifier, there will be no output voltage, hence, diode will conduct during positive half cycle only.

Question 6.
How Long can an electric Lamp of 100 W be kept glowing by fusion of 2 kg of deuterium? Take the fusion reaction as ₁¹H+₁²H → ₂³He+n +3.27 MeV (3)
Answer:
Number of atoms present in 2 g of deuterium = 6 x10²³
Number of atoms present in 20 Kg of deuterium = 6 x 10²⁶
Energy released in fusion of 2 deuterium atoms = 3.27 MeV
Energy reLeased in fusion of 2.0 Kg of deuterium atoms

= \( \frac{3.27}{2}\) x 6 x 1026 MeV
= 9.81 x. 10²⁶ MeV
= 15.696 x 10¹³J
Energy consumed by bulb per sec = 100 J
Time for which bulb will glow
= \(\frac{15.696 \times 10^{13}}{100}\)
s = 4.97 x 10⁴ year

Question 7.
Define wavefront. Draw the shape of refracted wavefront when the plane incident wave undergoes refraction from optically denser medium to rarer medium. Hence prove Snell’s law of refraction. 3
Answer:
A locus of points, which oscillate in phase is called a wavefront. A wavefront is defined as a surface of constant phase.

Explanation: A wavefront is defined as the continuous locus of aLL such particles of the
medium which are vibrating in the same phase at any instant.

sin∠BAC=sini= \(\frac{\mathrm{BC}}{\mathrm{AC}}\)
From the right ΔAEC
sin ∠ECA =sin r = \(\frac{\mathrm{AE}}{\mathrm{AC}}\)
\(\frac{\sin i}{\sin r}=\frac{\mathrm{BC}}{\mathrm{AE}}=\frac{v_{1} \tau}{v_{2} \tau}\)
This is the Snells tow of refraction.

Question 8.
(A) Draw a ray diagram of compound microscope for the final image formed at least distance of distinct vision?
(B) An angular magnification of 30X is desired using an objective of focal length 1.25 cm and an eye piece of focal length 5 cm. How will you set up the compound microscope for the finaln image formed at least distance of distinct vision?
OR
(A) Draw a ray diagram of Astronomical Telescope for the final image formed at infinity.
(B) A small telescope has an objective lens of focal length 140 cm and an eyepiece of focal length 5.0 cm. Find the magnifying power of the telescope for viewing distant objects when
(i) the telescope is in normal adjustment,
(ii) the final image is formed at the least distance of distinct vision. (3)
Answer:
(A) Diagram of Compound Microscope for the final image formed at D:

(B) m₀ = 30, i₀ = 1.25 cm, f_e = 5 cm
when image is formed at least distance of aistinct vision,
D = 25 cm
Angular magnification of eyepiece
m_e = \(\left(1+\frac{D}{f_{e}}\right)=1+\frac{25}{5} \) = 6

Total Angular magnification, m = m₀m_e ⇒
m₀ = \(\frac{m}{m_{e}}=\frac{30}{6}\) = 5
As the objective lens forms the real image,
m₀= \(\frac{v_{0}}{u_{0}}\) – 5 ⇒ v₀ = -5u₀
using Lens equation, u_e = \(\frac{25}{6}\)

Thus, object is to be placed at 1.5 cm from the objective and separation between the two lenses should be
L = v₀ + |U_e| = 11.67 cm
OR
(A) Ray diagram of astronomical telescope when image is formed at infinity.

(B) (i) In normal adjustment: Magnifying power.
m=f₀/f_e=(140/5)= 28
(ii) When the final image is formed at the east distance of distinct vision (25 cm):
m = \(\frac{f_{o}}{f_{e}}\left(1+\frac{f_{e}}{D}\right)(28 \times 1.2)\) = 33.6

Question 9.
Light of waveLength 2000 Å falls on a metal surface of work function 4.2 eV.
(A) What is the kinetic energy (in eV) of the fastest electrons emitted from the surface?
(B) What will be the change in the energy of the emitted eLectrons if the intensity of Light with same waveLength is doubled?
(C) If the same Light faits on another surface of work function 6.5 eV, what will be the energy of emitted electrons? (3)
Answer:
λ = 2000 Å = (2000 x 10^-10)m
W₀ = 4.2 eV
h = 6.63 x 10^-34 Js

(A) Using Einstein’s photoelectric equation
K. E. = (6.2 – 4.2) eV = 2.0 eV

(B) The energy of the emitted electrons does not depend upon intensity of incident Light; hence the energy remains unchanged.

(C) For this surface, electrons will not be emitted as the energy of incident light (6.2 eV) is less than the work function (6.5 eV) of the surface.

Question 10.
The focal length of a convex lens made of glass of refractive index (1.5) is 20 cm. What will be its new focal length when placed in a medium of refractive index 1.25 ? Is focal length positive or negative? What does it signify? (3)
Answer:
Given a_μg = 1.5
Focal length of the given convex lens when it is placed in air is
f= + 20 cm
Refractive index of the given medium with respect to air is
a_μm = 1-25
New focal length of the given convex lens when placed in a medium is f’

New focal length is positive.
The significance of the positive sign of the focal length is that given convex lens is still converging in the given medium.

Question 11.
(A) Name the e.m. waves which are suitable for radar systems used In aircraft navigation. Write the range of frequency of these waves.
(B) If the Earth did not have atmosphere, would its average surface temperature be higher or lower than what it is now? Explain.
(C) An e.m. wave exerts pressure on the surface on which it is incident. Justify.
OR
(A) If the slits in Young’s double-slit experiment are identical, then intensity at any point on the screen may vary between zero and four times to the intensity due to single slit”. Justify the above statement through a relevant mathematical expression.
(B) Draw the intensity distribution as function of phase angle when diffraction of light takes place through coherently illuminated single slit. 3
Answer:
(A) Microwaves are suitable for the radar system used in aircraft navigation. Range of frequency of microwaves is 108 Hz to 1011 Hz.

(B) If the Earth did not have atmosphere, then there would be absence of greenhouse effect of the atmosphere. Due to this reason, the temperature of the earth would be lower than what it is now.

(C) An e.m. wave exerts pressure on the surface on which it is incident. Justify An e.m. wave carries a linear momentum with it and the linear momentum carried by a portion of wave having energy U, which is given by p = U/c, where U = potential energy and c = speed of light.
Thus, if the wave incident on a material surface is completely absorbed, it delivers potential energy U and momentum p = Uc to the surface.
If the wave is totally reflected, the momentum delivered is p = 2U/c because the momentum of the wave changes from p to -p.
Therefore, it follows that an e.m. waves incident on a surface exert a force and hence a pressure on the surface.
OR
(A) The total intensity at o point where the phase difference is Φ. is given by I = I₁ + I₂ + 2\( \sqrt{1_{1} I_{2}}\) Φ
Here I₁ and I₂ are the intensities of two individual sources which are equal.
When Φ is 0,I = 4I₁
When Φ is 90°, I = 0
Thus intensity on the screen varies between 4I₂ and 0.
(B) Intensity distribution as function of phase angle, when diffraction of light takes place through coherently illuminated single slit

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

Question 12.
CASE STUDY: MIRAGE IN DESERTS

To a distant observer, the light appears to be coming from somewhere below the ground. The observer naturally assumes that light is being reflected from the ground, say, by a pool of water near the tall object.
Such inverted images of distant tall objects cause an optical illusion to the observer. This phenomenon is called mirage.
This type of mirage is especially common in hot deserts. Based on the above facts, answer the following questions:
(A) Which of the following phenomena is prominently involved in the formation of mirage in deserts?
(i) Refraction, Total internal Reflection
(ii) Dispersion and Refraction
(iii) Dispersion and scattering of light
(iv) Total internal Reflection and diffraction.

(B) A diver at a depth 12 m inside water
\(\left(a_{\mu \omega}=\frac{4}{3}\right)\) sees the sky in a cone of semi-vertical angle

• sin^-1\(\frac{4}{3}\)
• tan^-1 \(\frac{4}{3}\)
• sin^-1 \(\frac{4}{3}\)
• 90°

(C) In an optical fibre, if n₁ and n₂ are the refractive indices of the core and cladding, then which among the following, would be a correct equation?
(i) n₁ < n₂
(ii) n₁ = n₂
(iii) n₁ < n₂
(iv) n₁ > n₂

(D) A diamond is immersed in such a liquid which has its refractive index with respect to air as greater than the refractive index of water with respect to air. Then the critical angle of diamond- liquid interface as compared to critical angle of diamond -water interface will:
(i) depend on the nature of the liquid only
(ii) decrease
(iii) remain the same
(iv) increase.

(E) The following figure shows a cross-section of a ‘light pipe’ made of a glass fiber of refractive index 1.68.
The outer covering of the pipe is made of a material of refractive index 1.44.
What is the range of the angles of the incident rays with the axis of the pipe for the following phenomena to occur. (5)

Answer:
(A)
(i) Refraction, Total internal reflection
Explanation: When Light passes from cold air to hot air, tight tends to bend from its path which is known as refraction. As the tight is refracted, it reaches to o point where it forms a 90° angle. Hence refraction and total internal reflection are involved in formation of mirages.

(B) (iii) sin^-1 \(\left(\frac{3}{4}\right)\)

(C) (iv) n₁ > n₂
The refractive index of the core should be greater than the refractive index of the dadding.

(D) (iv) increase
\(\mathrm{I}_{\mu d}=\frac{1}{\sin c}=\frac{\mu_{d}}{\mu_{1}}\)
μ₁ = μ_ω
Thus C>C’

(E) (ii) 0<i<60°, ^lμ₂ = \(\frac{1}{\sin C^{\prime}}\)
SinC’= \(\frac{1.44}{1.68}\) = 0.8571
⇒ C = 59°
Total, internat reflection wilL occur if the angle i > i’c,
i.e., if i’> 59° or when r < r_max where r_max = 90° – 59°= 31°.
Using Snell’s Law.

\(\frac{\sin i_{\max }}{\sin r_{\max }}\) = 1.68
sin i_max = 1.68 x sin r_max
= 1.68 x sin 31° = 1.68 x 0.5 150 = 0.8662
Thus all incident rays which make angles in the range 0 < i < 60° with the axis of the pipe wilt suffer total internal reftections in the pipe.

Students can access the CBSE Sample Papers for Class 12 Maths with Solutions and marking scheme Term 2 Set 9 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Maths Term 2 Set 9 with Solutions

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper contains three sections-A. B and C. Each part is compulsory.
• Section-A has 6 short answer type (SA1) questions of 2 marks each.
• Section-B has 4 short answer type (SA2) questions of 3 marks each.
• Section-C has 4 long answer type questions (LA) of 4 marks each.
• There is an Internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A
(Section – A has 6 short answer type (SA-1) questions of 2 marks each.)

Question 1.
Evaluate (2)

Answer:

Question 2.
Show that the solution of differential equation x dy – y dx = 0 represents a straight line.
OR
Write the general solution of the differential equation cos x sin y dx + sin x cos y dy = 0. (2)
Answer:
We have,
x dy – y dx = 0
or x dy = y dx
⇒ \(\frac{d y}{y}\) = \(\int \frac{d x}{x}\)
Integrating both sides, we get
⇒ \(\int \frac{d y}{y}\) = \(\int \frac{d x}{x}\)
⇒ log y = log x + log C,
where log C is constant of integration.
⇒ log y = log Cx
⇒ y = Cx
which is an equation of a straight line passing through the origin.
Hence proved.
OR
We have,
cos x sin y dx + sin x cos y dy = 0
or, cos x sin y dx = – sin x cos y dy
⇒ \(\frac{\cos x}{\sin x}\)dx = \(\frac{\cos y}{\sin y}\)dy
Integrating both sides, we get
\(\int \frac{\cos x}{\sin x}\) dx = –\(\int \frac{\cos y}{\sin y}\)dy
⇒ log |sin x| = – log |sin y| + log c [∵ \(\int \frac{f^{\prime}(x)}{f(x)}\)dx = log|f(x)| + C]
⇒ log (sin x. sin y) = log C
or, sin x sin y = C

Question 3.
For any two vectors \(\vec{a}\) and \(\vec{b}\), show that |\(\vec{a}\) + \(\vec{b}\)| ≤ |\(\vec{a}\)| + |\(\vec{b}\)|. (2)
Answer:
We have,

Question 4.
If |\(\vec{a}\)| = 10, |\(\vec{b}\)| = 2 and \(\vec{a} \cdot \vec{b}\) = 12, then find the value of \(|\vec{a} \times \vec{b}|\). (2)
Answer:

Question 5.
A man is known to speak truth 3 out of 4 times. He throws a die and reports that it is six. Find the probability it is actually a six. (2)
Answer:
Let E be the event that the man reports that six has occurred on throwing a die.
Also, let E₁ be the event “six has occurred” and E₂ be the event “six does not occurred”.
∴ P(E₁) = \(\frac{1}{6}\) and P(E₂) = \(\frac{5}{6}\)
Also,
P\(\left(\frac{E}{E_{1}}\right)\) = Probability that the man does not speaks truth = \(\frac{1}{4}\)
Now, using Bayes’ theorem,
Required probability = P\(\left(\frac{E_{1}}{E}\right)\)

Question 6.
In answering a question on a multiple choice test, a student either knows the answer or guesses. Let \(\frac{3}{4}\) be the probability that he knows the answer and \(\frac{1}{4}\) be the probability that he guesses. Assuming that a student who guesses at the answer will be correct with probability \(\frac{1}{4}\). What is the probability that the student knows the answer given that he answered it correctly? (2)
Answer:
Let E₁ be the event “the student knows the answer” and E₂ be the event “guesses the answer”.
Also, let A be the event “answer is correct”

Caution:
P\(\left(\frac{\mathrm{A}}{E_{1}}\right)\) = 1, as be knows the answer so his answer wili definitely be correct.

SECTION – B
(Section – B has 4 short answer type (SA-2) questions of 3 marks each.)

Question 7.
Evaluate \(\int \frac{x}{(x+2)(3-2 x)}\) dx.
OR
Evaluate (3)

Answer:

OR

Put x = a cos² t + b sin² t
⇒ dx = a(2 cos t) (- sin t) dt + b(2 sin t cos t) dt
= 2(b – a) sin t cos t dt
Also, x – a = (b – a) sin² t
and b – x = (b – a) cos² t
when x = b, t = \(\frac{\pi}{2}\); and when x = a, t = 0

Question 8.
Find the equation of the plane passing through the points (2, 1, – 3), (- 3, – 2, 1) and (2, 4, -1). (3)
Answer:
We know that the equation of the plane passing through three points (x,₁, y₁, z₁), (x₂, y₂, z₂) and (x₃, y₃, z₃) is given by:
\(\left|\begin{array}{ccc}
x-x_{1} & y-y_{1} & z-z_{1} \\
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
x_{3}-x_{1} & y_{3}-y_{1} & z_{3}-z_{1}
\end{array}\right|\) = 0 …….. (i)
Here,
(x₁, y₁, z₁) = (2, 1, – 3)
(x₂, y₂, z₂) = (- 3, – 2, 1)
and (x₃, y₃, z₃) = (2, 4, – 1)
Substituting these values in (i), we get

(x- 2) (- 6 – 12) – (y – 1) (- 10 – 0) + (z + 3) (- 15 + 0) = 0
⇒ 18 (x – 2) + 10(y – 1) – 15(z + 3) = 0
⇒ 18x + 10y – 15z – 19 = 0
Thus, the required equation of the p[ane is:
⇒ 18x + 10y – 15z – 19 = 0

Question 9.
Sketch the bounded region and find the area of the region bounded by the curve y = |x|, the x-axis and the ordinates x = -1 and x = 1.
OR
Find the area of the region bounded by the circle x² + y² = 4, the line x = √3y and the x-axis in the first quadrant. (3)
Answer:
Required area = ar(∆OAB) + ar(∆OCD)


OR
x² + y² = 4 represents a circle with centre at the origin and radius √4 i.e., 2 units; and the line x = √3y passes through the origin.

The line intersects the circle at point B (√3, 1) in the first quadrant.
∴ Required area = ar (∆OAB) + ar(region CAB)

Question 10.
If AC and BD are the diagonals of a quadrilateral ABCD, then show that its vector area is given by \(\frac{1}{2}(\overrightarrow{A C} \times \overrightarrow{B D})\). (3)
Answer:
Vector area of quadrilateral ABCD = Vector area of ∆ABC + Vector area of ∆ACD

Using triangle law of vector addition in ∆ABD]
Hence proved.

Section – C
(Section – C has 4 Long answer type questions (LA) of 4 marks each.)

Question 11.
Evaluate \(\int x \sin ^{-1}\left\{\frac{1}{2} \sqrt{\frac{2 a-x}{a}}\right\}\) (4)
Answer:

Question 12.
Consider an experiment of throwing two dice. Let E denote the event “odd number on the first die”, F denote the event” odd number on the second die” and G denote the event “sum of numbers on two dice is odd”. Show that E, F and G are pairwise independent events, but not mutually independent. (4)
Answer:
We have,
E = {(1, 1), (1. 2), (1, 3), (1, 4), (1, 5) (1, 6), (3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (5,1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)}
F = [(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5)}
and G = {(1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (3, 6), (4,1), (4, 3), (4, 5), (5, 2), (5, 4), (5, 6), (6, 1), (6, 3), (6, 5)}
Further, E ∩ F = {(1,1), (1, 3), (1, 5), (3, 1), (3, 3), (3, 5), (5,1), (5, 3), (5, 5)
E ∩G = [(1, 2), (1, 4), (1, 6), (3, 2), (3, 4), (3, 6), (5, 2), (5, 4), (5, 6)]
And F ∩ G = {(2, 1), (2, 3), (2, 5), (4, 1), (4, 3), (4, 5), (6, 1), (6, 3), (6, 5)}

⇒ E and G are independent.
Thus, E, F and G are pairwise independent events.
Hence proved.
Now,
E ∩ F ∩ G = Φ
P(E ∩ F ∩ G) = O
Also,
P(E) × P(F) × P(G) = \(\frac{1}{2} \times \frac{1}{2} \times \frac{1}{2}\)
= \(\frac{1}{8}\) ≠ P(E ri Fri G)
Thus, E, F and G are not mutuatlly independent events.
Hence proved.

Question 13.
Show that the lines \(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) are coplanar. Also,find the point of intersection. (4)
Answer:
Given lines are
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\)
and \(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\)
If two lines are coplanar, then
\(\left|\begin{array}{ccc}
x_{2}-x_{1} & y_{2}-y_{1} & z_{2}-z_{1} \\
a_{1} & b_{1} & c_{1} \\
a_{2} & b_{2} & c_{2}
\end{array}\right|\) = 0
Here,
(x₁, b₁, c₁) = (1, 2, 3)
(a₁, b₁, c₁) = (2, 3, 4)
(x₂, y₂, z₂) = (2, 3,4)
(a₂, b₂, c₂) = (3, 4, 5)

= 1(15 – 16) – 1(10 – 12) + 1(8 – 9)
= – 1 + 2 – 1 = 0
Hence, the given lines are coplanar.
Hence proved.
Now, any general point on the line
\(\frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}\) = λ (say) is (2λ + 1, 3λ + 2, 4λ + 3)
Also, any general point on the line
\(\frac{x-2}{3}=\frac{y-3}{4}=\frac{z-4}{5}\) = µ(say) is (3µ + 2, 4µ + 3, 5µ + 4).
Since, the given two lines are coplanar, so they must be intersecting.
∴ 2λ + 1 = 3µ + 2;
3λ + 2 = 4µ + 3;
4λ + 3 = 5µ + 4
On solving these three equations, we get λ = – 1, µ = – 1.
Hence, the point of intersection of the given two lines is (- 1, – 1, – 1).

Related Theory:
Intersecting Lines are always coplanar.

CASE-BASED/DATA-BASED

Question 14.
The rate at which a body cools is proportional to the difference between the temperature of the body and that of the surrounding air. A body in air at 25°C will get cool from 100°C to 75°C in one minute.

Based on the above information, answer the following two questions:
(A) Find its temperature at the end of 3 minutes.
(B) After how many minutes, its temperature will be 58°C?
OR
The population of Swastik’s village increases continuously at the rate proportional to the number of its inhabitants present at any time. The population of the village was 20,000 in 1999 and 25,000 in the year 2004.

Based on the above information, answer the following two questions:
(A) What was the population of the Swastik’s village in the year 2009? (2)
(B) What will be the population of the village after 15 years from the year 1999? (2)
Answer:
Let the temperature of the body be T°C.
Then, according to the question, dT ,
\(\frac{d T}{d t}\) = k(T – 25)
⇒ \(\frac{d T}{\mathrm{~T}-25}\) = k dt
Integrating both sides, we get
log (T – 25) = kt + log C
⇒ log \(\frac{C}{\mathrm{~T}-25}\) = kt
or T – 25 = Ce^kt ………….. (i)
when t = 0, T = 100
Putting these values in (i), we get
100 – 25 = Ce⁰
⇒ C = 75
∴ T – 25 = 75e^kt …………….. (ii)
When t = 1, T = 75
Thus, from (ii), we get
T – 25 = \(\left(\frac{2}{3}\right)^{2}\)
or T = 25 + 75\(\left(\frac{2}{3}\right)^{t}\) ……… (iii)

(A) Putting t = 3, in eq. (iii), we get
T = 25 + 75\(\left(\frac{2}{3}\right)^{3}\)
= 25 + 75 × \(\frac{8}{27}\)
= 47.22°C

(B) Putting T = 58°C, in eq. (iii), we get
58 = 25 + 75\(\left(\frac{2}{3}\right)^{t}\)
⇒ \(\left(\frac{2}{3}\right)^{t}=\frac{33}{75}\)
or \(\left(\frac{2}{3}\right)^{t}=\frac{11}{25}\)
⇒ t = 2 minutes
OR
Let y be the populations of the Swastik’s village at any time t.
Then, according to the question, dy
\(\frac{d y}{d t}\)
where, k is an arbitrary constant.
⇒ \(\frac{d y}{y}\) = kdt
Integrating both sides, we get
log y = kt + C …(i)
Let the year 1999 be the initial year.
So, at t = 0, y = 20,000
putting these values in (i), we get
log 20000 = k(0) + C
⇒ C = log 20,000
∴ (i) becomes
log y = kt + log 20,000 …(ii)
So, in 2004, t = 5 and y = 25,000.
Putting these values in (ii), we get
log 25,000 = 5k + log 20,000
⇒ k = \(\frac{1}{5}\)log\(\left(\frac{5}{4}\right)\)
Hence, by (ii), we have

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

CBSE Sample Papers for Class 12 Physics Standard Term 2 Set 11 for Practice

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.
What kind of fringes will be observed if we replace monochromatic light by white light?
OR
In a single-slit diffraction experiment, the width of the slit is made double the original width. How does this affect the size and intensity of the central diffraction band? (2)

Question 2.
How does the process of conduction takes place in n-type semiconductors and p-type semiconductors? (2)

Question 3.
How would the ionization energy change when electron in hydrogen atom is replaced by a particle of mass 100 times that of the electron but having the same charge? (2)

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

Question 4.
A proton and an alpha particle are accelerated through the same potential? Which one of the two has:
(A) greater value of de-Broglie wavelength associated with it and
(B) less kinetic energy.
Give reasons to justify your answer. (3)

Question 5.
A convex lens made up of glass of refractive index 1.5 is dipped, in turn, in:
(A) a medium of refractive index 1.65,
(B) a medium of refractive index 1.33.
(i) Will it behave as a converging or a diverging lens in the two cases?
(ii) How will its focal length change in the two media? (3)

Question 6.
(A) Draw a labelled ray diagram showing the formation of a final image by a compound microscope at least distance of distinct vision.
(B) The total magnification produced by a compound microscope is 20. The magnification produced by the eye piece is 5. The microscope is focused on a certain object. The distance between the objective and eye-piece is observed to be 14 cm. If least distance of distinct vision is 20 cm, calculate the focal length of the objective and the eye-piece. (3)

Question 7.
A 12.3 eV electron beam is used to bombard gaseous hydrogen at room temperature. Up to which energy level the hydrogen atoms would be excited? Calculate the wavelength of the second member of Lyman series and second member of Balmer series. (3)

Question 8.
Induced electric field due to changing magnetic flux are more readily observed than induced magnetic field due to changing electric field. Why?
OR
State the laws of photoelectric emission. (3)

Question 9.
Why are Si and GaAs are preferred for making solar cells? Briefly explain working principle of a solar cell with the help of a necessary circuit diagram. (3)

Question 10.
What is a rectifier? Draw a labelled diagram
of Full-wave rectifier and its input and out¬put waveform. (3)

Question 11.
A composite prism ABC is made of two identical right-angled prisms ABD and ADC made of different materials of refractive indices √3 andm respectively. A ray of light is incident on face AB of the prism at 60° and the emergent ray grazes along the face AC. Find the value of n.

Consider two coherent sources S1 and S2 producing monochromatic waves to produce interference pattern. Let the displacement of the wave produced by S1 be given by Y1 = a cos ωt and the displacement by S2 be Y2 = a cos (ωt + Φ). Find out the expression for the amplitude of the resultant displacement at a point and show that the intensity at
that point will be I = 4a2 cos2 \(\frac{\phi}{2}\). Hence
establish the conditions for constructive and destructive interference. (3)

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

Question 12.
There wilt be a gain in the overall binding energy when we move from the heavy nuclei region to the middle region of the plot, hence release of energy when a heavy nucleus (A = 240) breaks into two roughly equal fragments, energy can be released. This process is called nuclear fission. Similarly, there will be gain in

(A) The binding energy per nucleon for the parent nucleus is E₁ and that for the daughter nuclei is E₂. Then A nucleus of mass M + Am is at rest and decays into two daughter nuclei of equal mass M/2 each. Speed of light is c.
(i) E1 = 2E2
(ii) E2 = 2E1
(iii) E1 > E2
(iv) E2 > E1
the overall binding energy, when we move from lighter nuclei to heavier nuclei, hence release of energy. When two or more lighter nuclei fuse together to form a heavy nucleus energy can be released. This process is called Nuclear Fusion. This is the energy source of sun.

(B) Solar energy is mainly caused due to:
(i) burning of hydrogen in the oxygen
(ii) fission of uranium present in the sun
(iii) fusion of protons during synthesis of heavier elements
(iv) gravitational contraction

(C) 200 MeV of energy may be obtained per fission of ²³⁵U. A reactor is generating
1000 KW of power. The rate of nuclear fission in the reactor is:
(i) 1000
(ii) 2 × 10⁸
(iii) 3.125 × 10¹⁶
(iv) 931
(D) Which type of reaction produces the most harmful radiation?
(i) Fusion reaction
(ii) Chemical reaction
(iii) Fission reaction
(iv) Photo-Chemical reaction

(E) The average energy released per fission of \(\begin{gathered}
235 \\
92
\end{gathered}\)U is: ,
(i) 200 eV 00
(ii) 200 MeV
(iii) 400 meV
(Vi) 200 GeV (5)

Students can access the CBSE Sample Papers for Class 12 Maths 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 Maths Term 2 Set 11 for Practice

Time Allowed: 2 Hours
Maximum Marks: 40

General Instructions:

• This question paper contains three sections-A. B and C. Each part is compulsory.
• Section-A has 6 short answer type (SA1) questions of 2 marks each.
• Section-B has 4 short answer type (SA2) questions of 3 marks each.
• Section-C has 4 long answer type questions (LA) of 4 marks each.
• There is an Internal choice in some of the questions.
• Q14 is a case-based problem having 2 sub parts of 2 marks each.

Section – A
(Section – A has 6 short answer type (SA-1) questions of 2 marks each.)

Question 1.

Question 2.
Solve the differential equation
\(\frac{d y}{d x}\) + 2x = e^3x
OR
Find the particular solution of the differential equation cot\(\left(\frac{d y}{d x}\right)\) = a (- 1 ≤ a ≤ 1), given that

Question 3.
If \(\vec{b} \times \vec{c}=\vec{c} \times \vec{a} \neq \overrightarrow{0}\), then show that \(\vec{a}+\vec{b}=k \vec{c}\) = kc, where k is a scalar. (2)

Question 4.
Show that \(|\vec{a}| \vec{b}+|\vec{b}| \vec{a}\) is perpendicular \(|\vec{a}| \vec{b}-|\vec{b}| \vec{a}\), for any two non-zero vectors \(\vec{a}\) and \(\vec{b}\) (2)

Question 5.
Let X represents the difference between the number of heads and the number of tails obtained when a coin is tossed 6 times. What are the possible values of X? (2)

Question 6.
Find the probability of the occurrence of a number greater than 2 in a throw of a die1 if it is know that only even numbers can occur. (2)

SECTION – B
(Section – B has 4 short answer type (SA-2) questions of 3 marks each.)

Question 7.
Find the sine of the angle between the vectors
\(\vec{a}\) = 3î + ĵ + 2k̂ and \(\vec{b}\) = 2î – 2ĵ + 4k̂
OR
Using vectors, show that the angle in a semi-circle is a right angle. (3)

Question 8.
Evaluate \(\int \frac{2 x+1}{\sqrt{x^{2}+4 x+3}}\)dx. (3)

Question 9.
The lines \(\frac{x-2}{1}=\frac{y-3}{1}=\frac{z-4}{-k}\) and \(\frac{x-1}{k}=\frac{y-4}{2}=\frac{z-5}{1}\) are coplanar. Find k.
OR
Find the cartesian equation of a plane that passes through the point (3, – 6, 4) and normal to the line joining the points (2, – 1, 5) and (3, 4, – 1). (3)

Question 10.
Find the particular solution of the differential equation (1 + x²)\(\frac{d y}{d x}\) + 2xy = \(\frac{1}{1+x^{2}}\), given that y = 0 when x = 1. (3)

Section – C
(Section – C has 4 long answer type questions (LA) of 4 marks each.)

Question 11.
Prove that the lines x = ay + b = cz + d and x = αy + β = γz + δ are coplanar, if (γ – c) (αβ – bα) – (α – a) (cδ – dγ) = 0. (4)

Question 12.
Let X denotes the number of colleges where you will apply after your result and P(X = x) denotes your probability of getting admission in x number of colleges. It is given that

where k is a positive constant. Find the value of k. Also, find the probability that you will get admission in (A) exactly one college, (B) at most 2 colleges, (C) at least 2 colleges.
OR
Three persons A, B, C throw a die in succession till one gets a six and wins the game. Find their respective probabilities of winning, if A begins the game. (4)

Question 13.
Evaluate

CASE-BASED/DATA-BASED

Question 14.
Three friends, Anmol, Sanjay, and Sammer live in a same society. Location of their houses in the society is represented by the points A(-l, 0), B(l, 3) and C(3, 2) as shown in the figure.

Based on the above information, answer the following two questions:
(A) Find the equation of the lines AB, BC and CA. (2)
(B) Find the area of the region of ABCDA. (2)