CBSE Sample Papers for Class 12 Physics Set 2 with Solutions

Students must start practicing the questions from CBSE Sample Papers for Class 12 Physics with Solutions Set 2 are designed as per the revised syllabus.

CBSE Sample Papers for Class 12 Physics Set 2 with Solutions

Time Allowed: 3 Hours
Maximum Marks: 70 Marks

General Instructions:

There are 35 questions in all. All questions are compulsory.
This question paper has five sections: Section A, Section B, Section C, Section D and Section E. All the sections are compulsory.
Section A contains eighteen MCQ of 1 mark each, Section B contains seven questions of two marks each, Section C contains five questions of three marks each, section D contains three long questions of five marks each and Section E contains two case study-based questions of 4 marks each.
There is no overall choice. However, an internal choice has been provided in section B, C, D and E. You have to attempt only one of the choices in such questions.
Use of calculators is not allowed.

Section – A (18 Marks)

Question 1.
A circular current carrying coil has a radius R. What is the distance from the centre of the coil on the axis, where the magnetic induction will be (\(\frac{1}{8}\)^th of its value at the centre of the coil? [1]
(a) \(\sqrt{2}\)R
(b) R
(c) \(\sqrt{3}\)R
(d) R²
Answer:
(c) \(\sqrt{3}\)R

Explanation:
Magnetic field on the axis of a circular coil,
B₁ = \(\frac{\mu_0 i R^2}{2\left(R^2+z^2\right)^{3 / 2}}\)
Magnetic field at the center of the circular coil:
B₂ = \(\frac{\mu_0 i}{2 R}\)
Given, B₁ = \(\frac{1}{8}\)B₂
\(\frac{\mu_0 R^2}{2\left(R^2+z^2\right)^{3 / 2}}=\frac{1}{8} \times \frac{\mu_0 i}{2 R}\)
8R³ = (R² + z²)^3/2
64 R⁶ = (R² + z²)³
(4R²)³ = (R² + z²)³
4R² = R² + z²
z = \(\sqrt{3}\)R

Question 2.
What is the expression for energy of an electron in then nth orbit of hydrogen atom? [1]
(a) \(-\frac{K Z^2 e^2}{r^2}\)
(b) \(-\frac{K Z e^2}{r}\)
(c) \(\frac{K Z e^2}{r}\)
(d) \(-\frac{K Z e^2}{r^2}\)
Answer:
(b) \(-\frac{K Z e^2}{r}\)

Explanation:
The total energy of the electron in nth orbit is the sum of potential energy and kinetic energy. The expression for potential energy is given by \(\frac{-K Z e^2}{r}\) and for kinetic energy is \(\frac{K Z e^2}{2r}\) where, r is the radius of orbit
r = \(\frac{n^2 h^2}{4 \pi^2 m K e^2}\)
Then, T.E. = K.E. + P.E.
= \(\frac{K Z e^2}{2 r}-\frac{K Z e^2}{r}\)
= \(\frac{-K Z e^2}{r}\)

Related Theory
The negative sign of total energy indicates that an electron is bound to the nucleus due to electrostatic attraction and some amount of work is required to be done to pull it away from the nucleus.

Question 3.
What is the minimum distance between an object and its real image formed by a convex lens? [1]
(a) 4 f
(b) 2 f
(c) 6 f
(d) f
Answer:
(a) 4f

Explanation:
Let I be the real image of the object O and ‘d’ be the distance between the image and object. If the image distance from the lens is x, then the object distance will be (d – x).
So, we can say that,
u = -(d -x)
v = + x
Substituting in the lens formula,
\(\frac{1}{v}-\frac{1}{u}=\frac{1}{f}\)
\(\frac{1}{x}-\frac{1}{-(d-x)}=\frac{1}{f}\)
\(\frac{1}{x}+\frac{1}{(d-x)}=\frac{1}{f}\)
x² – xd – fd = 0
The value of x must be real for a real image i.e. roots of the equation must be real. This is only possible if
d² ≥ 4 fd
d ≥ 4f
Hence, the required minimum distance is 4f.

Question 4.
When a magnet is inserted into a coil, an induced emf is created. The magnitude of the generated emf is unaffected by: [1]
(a) the number of coil twists
(b) the rate at which the magnet is moved
(c) the magnet’s strength
(d) the resistivity of the coil’s wire.
Answer:
(d) the resistivity of the coil’s wire.

Explanation:
emf, ε = \(-\frac{n d \phi}{d t}=\frac{-d(n B A)}{d t}\)
Where, n is number of turns in coil.
So, The emf is affected by the number of coil turns, the strength of the magnet, and the speed with which the magnet is moved. The emf is independent of the coil’s wire resistance.

Question 5.
1 amu is equivalent to: [1]
(a) 931 MeV
(b) 935 MeV
(c) 937 Mev
(d) 930 MeV
Answer:
(a) 931 Me V

Explanation:
According to Einstein’s energy mass equivalence,
E = mc²
From this relation, we can conclude that energy content of an object is mass times the square of the speed of light. The energy equivalent to one atomic mass unit can be calculated as follows:
m = 1 amu = 1.66 × 10^-27 kg
c = 2.998 × 10⁸ ms^-1
Then E = 1.66 × 10^-27 × (2.998 × 10⁸)² J
= \(\frac{1.66 \times 10^{-27} \times\left(2.998 \times 10^8\right)^2}{1.602 \times 10^{-19}} \mathrm{eV}\)
= 931 MeV
∴ 1 amu = 931 MeV.
This is the required relationship between amu and MeV.

Question 6.
Name the charge corners in conductor: [1]
(a) Holes
(b) Electrons
(c) Both (a) and (b)
(d) None of these
Answer:
(b) Electrons

Explanation:
The electrons responsible for the conduction in case of metals/ conductors are known as free electrons. These are outermost electrons which are quite loosely bound to the atoms and it takes very little energy to kick them off to move around freely.

Related Theory
In case of intrinsic semiconductors, the charge carriers are the electrons and holes that are responsible for conduction. The concentration of these carriers depends upon the temperature and band gap of the material, thus affecting a material’s conductivity.

Question 7.
How many electrons generally are there in valence shell of a semiconductor? [1]
(a) 1
(b) 3
(c) 4
(d) 2
Answer:
(c) 4

Explanation:
The outermost shell for any semiconductor is called as valence shell.

When the electrons are present in this shell, they tend to form covalent bonds with the neighboring atoms. Most conductors have one electron in the valence shell whereas semiconductors have four electrons. As shown in the diagram, each circle represents a silicon atom and the lines represent the shared electrons. Each silicon atom is bonded with four other silicon atoms because four valence electrons in each silicon atom is shared with one neighboring atom.

Related Theory
The different types of compounded semiconductors are Gallium Arsenide (GaAs), Indium Gallium Arsenide (InGaAs) and Aluminium Gallium Indium Phosphide (AlGalnP) etc.

Question 8.
With increase in frequency of an AC supply, the impedance of a series L-C-R circuit: [1]
(a) remains constant.
(b) increases.
(c) decreases.
(d) decreases at first, becomes minimum and then increases.
Answer:
(d) decreases at first, becomes minimum and then increases.

Explanation:
The frequency vs. impedance graph of a series LCR circuit is as follows:

The impedance lowers at first, then becomes minimal, and finally increases as the frequency increases.

Question 9.
The value of current in the given figure is: [1]

(a) 10A
(b) 15A
(c) 4A
(d) 5A
Answer:
(d) 5A

Explanation:
The two batteries shown in figures are opposing each other. The net emf of the circuit,
(e) = 200 – 10
e = 190 V
Resistance, (R) = 38 Ω
Current through the circuit,
I = \(\frac{e}{R}=\frac{190}{38}\) = 5 A

Question 10.
What is the value for self-inductance of a coil if the current changes from 10 A to 2 A in 0.1 second and induced e.m.f. is 3.28 V? [1]
(a) 0.02H
(b) 0.04 H
(c) 0.08 H
(d) 0.01 H
Answer:
(b) 0.04 H

Explanation:
The induced emf in a coil is given by,
e = – \(\frac{L d I}{d t}\)
3.28 = -L \(\frac{(2-10)}{0.1}\)
3.28 = \(\frac{-L \times(-8)}{0.1}\)
L = \(\frac{3.28 \times 0.1}{8}\)
L = 0.04 H

Related Theory:
The significance of negative sign in the expression of induced emf is that the coil opposes its change of flux If the flux of coil is increased, the current induced in such a direction so that it tends to reduce it and vice versa.

Question 11.
An electron is discharged with uniform velocity along the axis of a long solenoid carrying electricity. Which of the following statements is correct? [1]
(a) The electron will move along the axis.
(b) The electron trajectory around the axis will be circular.
(c) The electron will encounter a force at 45° to the axis and hence follow a helical route.
(d) The electron will continue to travel uniformly along the solenoid’s axis.
Answer:
(d) The electron will continue to travel uniformly along the solenoid’s axis.

Explanation:
The magnetic field within the long current carrying solenoid is uniform. As magnetic field and velocity are parallel, the amount of force on the electron of charge (- e) is given by F = – evB sin θ = 0 (θ = 0°). The electron will continue to travel uniformly along the solenoid’s axis.

Question 12.
An electron and a proton have the same de- broglie’s wavelength of 1 nm. The ratio of their momentum is: [1]
(a) 1 : 1
(b) 2 : 1
(c) 1 : 2
(d) 1 : 4
Answer:
(a) 1 : 1

Explanation:
An electron and proton have de- Broglie wavelength of 1 nm.
i.e., λ_e = λ_p = 1 nm = 10^-9 m
The expression for momentum is given by,
p = \(\frac{h}{\lambda}\)
\(\frac{p_e}{p_p}=\frac{h / \lambda_e}{h / \lambda_p}=\frac{\lambda_p}{\lambda_e}=\frac{10^{-9}}{10^{-9}}\) = 1 : 1
So, p_e : p_p = 1 : 1

Question 13.
The kinetic energy of proton to that of electron is, if they both have same wavelength of 1 nm, equal to: [1]
(a) 5.4 × 10^-2
(b) 5.4 × 10^-8
(c) 5.4 × 10^-4
(d) 5.4 × 10^-12
Answer:
(c) 5.4 × 10^-4

Explanation:
The expression for kinetic energy is given by,

Question 14.
Consider a two-slit interference configuration (Figure) in which the distance between the slits is half the distance between the screen and the slits. [1]

The value of D in terms of X so that the first minimum on the screen is D distance away from the centre O is:
(a) \(\frac{\lambda}{4.214}\)
(b) \(\frac{\lambda}{2.472}\)
(c) \(\frac{\lambda}{1.232}\)
(d) \(\frac{\lambda}{3.313}\)
Answer:
(b) \(\frac{\lambda}{2.472}\)

Explanation:
From figure,
T₁P = T₂O – OP = (D – x)
T₁P = T₁O + OP = (D + x)
Now, S₁P = \(\sqrt{\left(S_1 T_1\right)^2}+\sqrt{\left(T_1 P\right)^2}\)
= \(\sqrt{D^2}+\sqrt{(D-x)^2}\)
S₂P = \(\sqrt{\left(S_2 T_2\right)^2}+\sqrt{\left(T_2 P\right)^2}\)
= \(\sqrt{D^2}+\sqrt{(D-x)^2}\)
Path difference, S₂P – S₁P = \(\frac{\lambda}{2}\)
For first minimum to occur,
\(\sqrt{D^2}+\sqrt{(D-x)^2}-\left(\sqrt{D^2}+(D+x)^2\right)\) – \(\left(\sqrt{D^2}+\sqrt{(D-x)^2}\right)\)
= \(\frac{\lambda}{2}\)
The first minimum falls at a distance D from the center,
That is, x = D
[D₂ + 4D₂]^1/2 – D = \(\frac{\lambda}{2}\)
D (\(\sqrt{5}\) – 1) = \(\frac{\lambda}{2}\)
D(2.236 – 1) = \(\frac{\lambda}{2}\)
D = \(\frac{\lambda}{2.472}\)

Question 15.
When illuminated by reflected light of wavelength 580 nm, a soap coating with a thickness of 0.3 mm looks black. What is the soap solution’s index of refraction if it is known to be between 1.3 and 1.5? [1]
(a) 1.35
(b) 1.40
(c) 1.30
(d) 1.45
Two statements are given-one labelled Assertion (A) and the other labelled Reason (R). Select the correct answer to these questions from the codes (a), (b), (c) and (d) as given below:
(a) Both A and R are true and R is the correct explanation of A.
(b) Both A and R are true and R is NOT the correct explanation of A.
(c) A is true but R is false.
(d) A is false and R is also false.
Answer:
(d) 1.45

Explanation:
Path Difference is 2 µt.
Now for destructive interference it can be
2µt = \(\frac{\lambda}{2}\) 0r \(\frac{3\lambda}{2}\) or \(\frac{5\lambda}{2}\) and so on….
µ = \(\frac{\lambda}{4 t}, \frac{3 \lambda}{4 t}, \frac{5 \lambda}{4 t}\)
= \(\frac{580 \times 10^{-9}}{4 \times 0.3 \times 10^{-6}}, \frac{3 \times 580 \times 10^{-9}}{4 \times 0.3 \times 10^{-6}}\)
= 0.4833, 3 × 0.4833 …
So, µ = 3 × 0.4833 = 1.45

Question 16.
Assertion (A): Dielectric breakdown occurs under the influence of an intense light beam. [1]
Reason (R): Elecromagnetic radiations exert pressure.
Answer:
(b) Both A and R are true and R is NOT the correct explanation of A.

Explanation:
Electromagnetic radiations exert pressure and dielectric breakdown occurs under intense light beam. Hence, both statements are true but reason is not correct explanation of assertion.

Question 17.
Assertion (A): Isotopes of an element can be separated by using a mass spectrometer. [1]
Reason (R): Separation of isotopes is possible because of the difference in electron numbers of isotopes.
Answer:
(c) A is true but R is false.

Explanation: Isotopes have same number of electrons and protons but different neutron number. That is the reason the mass number of isotopes are different and can be separated by using a mass spectrometer.

Question 18.
Assertion (A): A capacitor of suitable capacitance can be used in an A.C circuit in place of the choke coil. [1]
Reason (R): A capacitor blocks D.C and allows A.C only.
Answer:
(b) Both A and R are true and R is NOT the correct explanation of A.

Explanation:
We can use a capacitor of suitable capacitance as a choke coil, because average power consumed in an ideal capacitor is zero. Therefore like a choke coiL, a capacitor can be used to reduce A.C without much power dissipation.

Section – B (14 Marks)

Question 19.
Write the properties of electromagnetic waves. [2]
Answer:

The electromagnetic waves do not require any medium for the propagation and are produced by accelerated charges.
The direction of electric field, magnetic field and propagation of an EM wave are mutually perpendicular to each other.
All the electromagnetic waves travel with the speed of light.
The amplitude ratio of electric field to that of magnetic field is equal to the speed of Light i.e. c = E₀/ B₀
The electromagnetic waves are not deflected by electric and magnetic fields.
The electromagnetic waves obey the principle of superposition and show the properties of reflection, refraction and interference.

Question 20.
Determine the current in the wire for the setup depicted in figure given below. [2]

The resistance of wire PQ is minimal. B, the magnetic field is emanating from the paper. θ is a fixed angle formed by PQ travelling smoothly over two parallel conducting wires separated by d.
Answer:
The motional electric field caused by movement
along CD is, E = vB
E will be perpendicular to both v and B, because
F = q × B
F is the force acting on the free charge particle PQ
So, the motional emf, e = E along PQ × effective length PQ
Electric field along PQ = v × B = v sin θ.B
= vB sin θ.

∈= (vB sinθ) \(\left(\frac{d}{\sin \theta}\right)\)
∈ = vBd
Induced current, I = \(\frac{\epsilon}{R}=\frac{v B d}{R}\)
It is independent of q.

Question 21.
How can you classify semiconductors on the basis of its chemical composition? [2]
Answer:
On the basis of chemical composition, the semiconductors can be divided into following categories:

Elemental semiconductors: Si and Ge
Compound semiconductors:
1. Inorganic: CdS, CdSe, InP and GaAs etc.
2. Organic polymers: Polypyrrole, polyaniline, polythiophene etc.

Question 22.
Derive the expression for the critical angle in terms of the speed of light in the respective media when a ray of light undergoes total reflection while travelling from a denser to a rarer medium.
OR
A ray of light passing through an equilateral triangular glass prism from air undergoes minimum deviation when angle of incidence is \(\frac{3^{\text {th }}}{4}\) of the angle of prism. Calculate the speed of light in the prism. [2]
Answer:
Let the speed of light in vacuum = c
speed of light in glass = v
Using snell’s law for glass air interface,
µ_g sin i = µ_a sin r
\(\frac{\sin i}{\sin r}=\frac{\mu_a}{\mu_g}\) = ^gµ_a = \(\frac{v}{c}\)
Ifa ray of light travelling from a denser to rarer medium undergoes total internal reflection, then
i = i_c and r = 90°
\(\frac{\sin i_c}{\sin 90^{\circ}}=\frac{v}{c}\)
\(\frac{\sin i_c}{(1)}=\frac{v}{c}\)
sin i_c = \(\frac{v}{c}\)
i_c = sin^-1(\(\frac{v}{c}\))
This is the required expression.

Given A = 60°,
i = \(\frac{3}{4}\) A = \(\frac{3}{4}\) × 60° = 45°
For minimum deviation
i₁ = i₂ = i
and r₁ = r₂ = \(\frac{A}{2}\) = 30°
∴ Refractive index of prism,
n = \(\frac{\sin i}{\sin r}\)
= \(\frac{\sin 45^{\circ}}{\sin 30^{\circ}}=\frac{1 / \sqrt{2}}{1 / 2}\)
= \(\sqrt{2}\) = 1.41
Speed of light in prism.
v = \(\frac{c}{n}=\frac{3 \times 10^8}{1.41}\)
= 2.13 × 10⁸ ms^-1

Question 23.
Determine the focal length of a biconvex lens if its surface radii are 60 cm and 15 cm, respectively, and the index of refraction of the lens glass is 1.5. [2]
OR
What is the lateral shift of the emerging ray when a ray travels through a glass slab of thickness t at an angle i with an angle of refraction r?

Answer:

OR
Using Snell’s law, for ray 1 and its refraction, sin i = µ sin r and sin e = µ sin r

⇒ ∠i = ∠e
The emergent ray is parallel to the incident ray,
Further,
AB = \(\frac{\mathrm{AN}}{\cos r}=\frac{t}{\cos r}\)
and BC = AB sin (i – r)
⇒ lateral shift, (d) = \(\frac{t}{\cos r}\) sin(i – r).

Question 24.
Explain the process of production of electromagnetic waves. [2]
Answer:
When electric charges are accelerated, electromagnetic waves are created. This enables the generation of electromagnetic waves by passing an alternating current through a wire, which serves as an antenna. The frequency of the waves produced in this manner is equal to the frequency of alternating current.

Question 25.
Which will possess more energy if: [2]
(A) an electron and a proton and
(B) an electron and photon having same wavelength?
Answer:
(A) The total relativistic energy if a particle is given by,
E = \(\sqrt{\left(m_0 c^2\right)^2+(p c)^2}\)
= \(\sqrt{m_0^2 c^4+p^2 c^2}\)
The wavelength is same for both electron and proton. Thus, momentum will also be same i.e. p = \(\frac{h}{\lambda}\) and p²c² value will also be A. same for both.
As far as the rest mass (m₀) is concerned, it is greater for proton than that of electron. So, energy of proton is more than that of electron of same wavelength.

(B) The expression for momentum is given by,
p = \(\frac{h}{\lambda}\)
The wavelength of both electron and photon is same, so they have equal momentum. The relativistic energy is given by
E = \(\sqrt{m_0^2 c^4+p^2 c^2}\)
The rest mass of photon (m₀) is zero and rest mass of electron is positive. Thus, the energy of electron will be more than that of photon.

Section – C (15 Marks)

Question 26.
As shown in figure, the rails are connected by a capacitor of capacitance C. A wire of mass m and length l can slide on a pair of parallel, smooth, horizontal rails placed in a vertical magnetic field B. If a constant force F acts on the wire, calculate the acceleration of the wire if electric resistance of rails and wire is zero. [3]

Answer:
Let the velocity of wire at any instant be v. The value of motional emf induced in the wire is given by e = Blv.
The electrical resistance of the rails and wire is zero, charge on the capacitor is,
q= Ce
= CBlv
Current, I = \(\frac{d q}{d t}=\frac{d}{d t}\)(CBlv)
= CBl\(\left(\frac{d v}{d t}\right)\)
I = CBla
The magnetic force exerted by current on the wire is given by,
F’ = ILB
= (CBla) lB
F’ = CB²l²a. towards left.
The vaLue of net fibre on wire will be.
F_net = F – F’
ma = F – CB²l²a
ma + CB²l²a = F
a (m + CB²l²) = F
a = \(\frac{F}{m+C B^2 l^2}\)
This is the required expression for acceleration.

Question 27.
Derive an expression for the focal length of the combination of two lens when they are separated by a certain distance. [3]
OR
(A) A thin double convex lens of focal length f as shown in fig (i) is broken down into two equal halves at the axis. The two halves are combined as shown in fig. (ii) and (iii). What is the focal length of the combination shown in figure (ii) and (iii)?

(B) Show that the power of a lens in inversely dependent upon the focal length of the lens.
Answer:
Consider two Lenses L₁ and L₂ of focal lengths f₁ and f₂ respectively. The two lenses are placed as shown in figure.

The deviation produced by L₁ is,
δ₁ ≈ tan δ₁ = \(\frac{h_1}{f_1}\)
The deviation produced by second lens L₂ s
δ₁ ≈ tan δ₁ = \(\frac{h_2}{f_2}\)
δ is the final deviation produced by the combination of lens. A single thin lens placed at C will produce the same deviation the focal length of the combination, then.

Thus,

This is the required value of the equivalent focal, length.

(A) If f₁ and f₂ are the focal lengths of two halves of the lens.
f₁ = f and f₂ = -f
The focal length of the combination of lenses is given by,

In this case, the focal lengths of both the lenses L₁ and L₂ are same.
i.e f₁ = f₂ = f
The effective focal Length.
\(\frac{1}{f_{e q}}=\frac{1}{f_1}+\frac{1}{f_2}\)
\(\frac{1}{f_{\text {eq }}}=\frac{1}{f}+\frac{1}{f}\)
\(\frac{1}{f_{e q}}=\frac{2}{f}\)
f_eq = \(\frac{f}{2}\)

(B) Power of a lens is defined as the tangent of the angle by which convergence or divergence takes place for a beam of light falling at unit distance from optical centre.

Question 28.
What is the ratio of electric fields at the surfaces of the two charged conducting spheres having radii ‘a’ and V connected to each other by a wire. Explain why charge density on the sharp ends of a conductor is more than that on its flatter portions? [3]
Answer:
When two charged conducting spheres are connected to each other by a wire. the charges will continue to flow between two spheres till their potentials become equaL Thus, charges on two spheres would be,

The ratio of electric fields at the surface of two spheres will be,

Thus, we can conclude that surface charge densities are inversely proportional to the radii of spheres. A flat portion can be taken as a sphere of very large radius whereas a pointed portion will have a small radius. The value of surface charge density is much higher on sharp and pointed surfaces as compared to flat ones.

Question 29.
Explain in brief two processes which are responsible for the breakdown of a junction diode. [3]
OR
A potential barrier of 0.5V exists across a p-n junction.
(A) If the depletion region is 5 × 10^-7 m wide. What is the average intensity of the electric field in this region?
(B) An electron with speed 5 × 10⁵ m/s approaches the p-n junction from the n-side with what speed will it enter the p-side?
Answer:
The two processes responsible for the breakdown of a p-n junction diode are as follows:
(1) Zener breakdown: This type of breakdown occurs in such a diode that is quite heavily doped. Due to the dopant density being high, the barrier field is quite high and width of junction layer is quite small. The depletion layer and energy bands get charged when a large reverse biased voltage is applied.

The high electric field set up in the region of depletion layer, strips off many electrons from the valence band which starts moving towards n-side of the diode. A reverse biased voltage is set up beyond the limit that gives rise to the phenomenon of internal field emission. The breakdown of diode, giving rise to a very high value of reverse current, due to band-to-band tunneling is known as Zener breakdown.

(2) Avalanche breakdown : This type of breakdown occurs in a diode which is having wide depletion layer and has low level of doping. The minority charge carriers get highly accelerated when reverse biased voltage is quite high, leading to the increase in the kinetic energy.

The newly generated electron hole pairs cause the ionisation due to the acceleration. A very large number of charge carriers are produced due to a number of collisions taking place, leading to a large value of reverse current. Such a process is known as avalanche breakdown in a diode.

(A) Value of potential barrier (V) = 0.50 volts
Width of depletion region (d) = 5 × 10^-7 m
Electric field,
E = \(\frac{V}{d}=\frac{0.5}{5 \times 10^{-7}}\)
= 0.1 × 10⁷
E = 10⁶ V/m

(B) Let, v₁ = speed with which an electron enters depletion region
= 5 × 10⁵ m/s
v₂ = speed of electron while entering p-region.
The potential energy of an electron increases by eV joule.
According to Law of conservation;
\(\frac{1}{2} m v_2^2+e V=\frac{1}{2} m v_1^2\)
\(\frac{1}{2} m v_2^2=\frac{1}{2} m v_1^2\) – eV
= \(\frac{1}{2}\) × 9.1 × 10^-31 × (5 × 10⁵)² – 1.6 × 10^-19 × 0.5
= \(\frac{1}{2}\) × 9.1 × 10^-31 × 25 × 10¹⁰ – 0.8 × 10^-19
= 4.55 × 10^-21 × 25 – 0.8 × 10^-19
= 1.13 × 10^-19 – 0.8 × 10^-19
\(\frac{1}{2}\)mv₂² = 0.33 × 10^-19J
\(\frac{1}{2}\) × 9.1 × 10^-31 × v₂² = 0.33 × 10^-19
v₂² = \(\frac{2 \times 0.33 \times 10^{-19}}{9.1 \times .10^{-31}}\)
v₂² = \(\frac{0.66 \times 10^{-19}}{9.1 \times 10^{-31}}\)
= 0.0725 × 10^{-19 + 31}
= 0.0725 × 10¹²
v₂² = 7.25 × 10¹⁰
v₂² = 2.69 × 10⁵ m/sec
This is the required velocity.

Question 30.
Explain the variation of resistivity of metals with the change in temperature with the help of suitable graphs. [3]
Answer:
Resistivity can be defined as the intrinsic property of a material that shows the opposition to the flow of current. It is the resistance offered by a conductor having unit length and unit area of cross section. So, we can say that it does not depend upon the length and area of cross section of a material.

The SI unit of resistivity is ohm meter The expression for resistivity is given by as,
ρ = R\(\frac{\mathrm{A}}{l}\)
where, R is the resistance in ohms, A is the area of cross section in square meters and l is the length in meters.
The resistivity of materials depend on the temperature as,
ρ_t = ρ₀ [1 + α (T- T₀)]
where, ρ₀ = resistivity at a standard temperature
ρ_t = resistivity at t °C
T₀ = reference temperature
α = temperature coeffcient of resistivity In the presence of potential difference, there is the movement of current in the form of free electrons. In case of conductors, the valence band and the conduction band overlap each other. The valence electrons are loosely bound to the nucleus in conductors. The ionization energy of the metals is quite low and the electrons are lost, quite easily.

Due to this reason, the movement of electron inside the metal happens at normal temperature. With the increase in temperature, the atoms starts to vibrate with higher amplitude.

As a result of this, the collisions are increase by the frequent collisions between the free electrons and the other electrons. Due to the number of collisions taking place, the drift velocity of the electrons decreases. It leads to the increase in the resistivity of the metals and current flow in the metal is decreased.

The value a is positive for metals or conductors. The resistivity increases Linearly with increase in temperature for a range of 500K for e.g. silver, copper, gold etc.

Temperature dependence on resistivity for metals.

Section – D (15 Marks)

Question 31.
(A) Stating Gauss theorem, derive an expression for electric field due to a long straight thin uniformly charged wire, having linear charge density λ, at a point which is lying at a distance V from the wire. [5]
(B) Two identical metallic spheres A and B each carrying a charge ‘2q’ repel each other with force ‘F’. A third metallic sphere C of same size, carrying charge ‘q’ is made to touch the spheres A and B alternately and the moved away. By what factor does the force acting between the two spheres change?
OR
(A) Write two important properties of equipotential surfaces.
(B) Derive an expression for the electric field at point lying on the axis of uniformly charged ring having linear charge density ‘λ’. Hence, prove that the ring behaves as a point charge for large distances (x >> r).
Answer:
(A) Gauss theorem states that the net electric flux through an closed surface is equal to \(\frac{1}{\varepsilon_0}\) times the charge within that closed
surface.
\(\oint \vec{E} \cdot \overrightarrow{d S}=\frac{q_{i n}}{\varepsilon_0}\)
Consider a gaussian surface of radius r and length T.
The net charge enclosed inside the Gaussian surface is q_in = λl
The electric field will be radially outwards.
According to Gauss law,
\(\oint \vec{E} \cdot d \vec{S}=\frac{q_{\text {in }}}{\varepsilon_0}\)
\(\oint_1 \vec{E} \cdot d \vec{S}+\oint_2 \vec{E} \cdot d \vec{S}+\oint_3 \vec{E} \cdot d \vec{S}=\frac{\lambda l}{\varepsilon_0}\)
The electric flux through surfaces 1 and 3 will. be zero because the electric field and area vector are perpendicular to each other.
\(\oint_1 \vec{E} \cdot d \vec{S}=0 \text { and } \oint_3 \vec{E} \cdot d \vec{S}\) = 0
Thus,
\(\oint_2 \vec{E} \cdot \overrightarrow{d S}=\frac{\lambda}{\varepsilon_0}\)
\(E \times 2 \pi r l=\frac{\lambda}{\varepsilon_0}\)
E = \(\frac{\lambda}{2 \pi \varepsilon_0 r}\)

(B) When two charges A and B are placed ‘d’ distance apart.
F = \(\frac{K q_1 q_2}{d^2}\)
F = \(\frac{K(2 q)(2 q)}{d^2}=\frac{K 4 q^2}{d^2}\) …………………. (i)
When a third metallic sphere C of same size, carrying a charge q. comes in contact with A, each sphere gets a charge of \(\frac{3 q}{2}\)

= \(\frac{21}{32}\)
F’ = \(\frac{21}{32}\) F
Thus, the force changes by a factor of \(\frac{21}{32}\).

(A) The two important characteristics of equipotential surfaces:

The electric potential remains constant at all the points on the equipotential surfaces.
No work is required to move a charge on an equipotential surface.

(B) Consider a thin circular ring of radius V with charge density X A point P is taken at a distance x from the centre of the ring.
A small element dq having length dx is considered.
λ = \(\frac{d q}{d x}\)
dq = λ dx
Electric field due to this charge element at a point P is given by,
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{d q}{P Q^2}\)
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{d q}{r^2+x^2}\)
dE = \(\frac{1}{4 \pi \varepsilon_0} \frac{\lambda d x}{\left(r^2+x^2\right)}\)
The electric field at the point P has two components i.e. one in vertical direction and second in horizontal direction.

dE cos θ along the horizontal direction
dE sin θ along the vertical direction.

The vertical components will cancel out the effect on each other due to the presence of a diametrically opposite elements. So, the horizontal components will be added up to find electric field at point P.

where Q is total charge on the ring. E_x is the value of total electric field at point P.
For larger distances;

x² + r² ≈ x²
E_x = \(\frac{x Q}{4 \pi \varepsilon_0\left(x^2\right)^{3 / 2}}\)
= \(\frac{x Q}{4 \pi \varepsilon_0 x^3}\)
E_x = \(\frac{Q}{4 \pi \varepsilon_0 x^2}\)
Thus, the ring behaves as a point charge for x >>> r since E ∝ \(\frac{1}{r^2}\).

Question 32.
(A) What do you mean by a wavefront? What are spherical, cylindrical and plane wavefronts? Show with the help of diagrams. [5]
(B) State the assumptions on which the Huygens’ principle is based.
OR
(A) What do you meant by coherent and incoherent sources of light?
(B) In Young’s double slit experiment using monochromatic light of wavelength K, the intensity of light at a point on the screen where path difference is λ, is K units. Find out the intensity of light at a point, where path difference is \(\frac{\lambda}{3}\).
Answer:
(A) A wavefront is defined as the continuous locus of all such particles of the medium that are vibrating in the same phase at any instant
The different types of wavefronts are as follows:
(1) Spherical wavefront: The wavefronts are spherical in shape when the waves are travelling in all the directions from a point source. All such points which are equidistant from the point source, Lie on a sphere of common radius and the wave wilt reach the next circle of larger radius at the same time at all points.

(2) Cylindrical wavefront: The wavefront is cylindrical in shape when the source of light is linear in shape. This is the reason due to which all such points which are equidistant from the linear source will be a cylinder.

(3) Plane wavefront: The curvature decreases when the spherical or cylindrical wavefront advances. A small portion of such wavefronts will be planar ones at a larger distances.

(B) According to Huygens’ principle, each point on the wavefront is a source of secondary waves, which adds up to give a wavefront at any later time. The principle is based on the following assumptions:

Every point on the given wavefront called ‘primary wavefront’ acts as a fresh source of new disturbance, called ‘secondary wavelets’ that travel in all directions with the velocity of Light in the medium.
A surface touching these secondary wavelets move tangentially in the forward direction at any instant gives a new wavefront at that instant. This is the secondary wave front.

The wavefronts will always move in the forward direction. All the secondary sources emit wavelets. The tangent drawn to all the wavelets is the new position of the waveform.

(A) Two sources of light which continuously emit light waves of same frequency and wavelength with a zero or constant phase difference between them, are called as coherent sources.

Two sources of light which do not emit light of a constant phase difference between them are known as incoherent sources.

(B) Intensity of given light = K units
when the path difference is λ, the phase difference is 2π.
Path diff. λ, → Phase difference 2π
Phase diff. 1 → Phase difference \(\frac{2 \pi}{\lambda}\)
Path diff, \(\frac{\lambda}{3}\) → Phase difference \(\frac{2 \pi}{\lambda} \times \frac{\lambda}{3}\)
∆P = \(\frac{2 \pi}{3}\)
The expression for resultant intensity is given by,
I = I₁ + I₂ + 2\(\sqrt{I_1 I_2}\) cos Φ
Let I₁ = I₂ = I₀
So, I = I₀ + I₀ + 2 \(\sqrt{I_0 \times I_0}\) cos Φ
I = 2I₀ + 2I₀ cos 2π
= 2I₀ + 2I₀
I = 4I₀ ………………….. (1)
The intensity is 4I₀ when the path difference is λ.
When path difference is \(\frac{\lambda}{3}\), the resultant intensity,
I = I₀ + I₀ + 2 \(\sqrt{I_0 \times I_0}\) cos \(\frac{2 \pi}{3}\)
I = 2I₀ + 2I₀ cos \(\frac{2 \pi}{3}\)
I = 2I₀ + 2I₀ (\(\frac{-1}{2}\))
= 2I₀ – I₀
I = I₀ ………………………. (2)
Thus, the required intensity is I₀.
Divide (2) by (1),
\(\frac{I^{\prime}}{I}=\frac{I_0}{4I_0}\)
\(\frac{I^{\prime}}{I}=\frac{1}{4}\)
I’ = \(\frac{I}{4}\)
I’ = \(\frac{K}{4}\)
This is the required result.

Question 33.
Using phasor diagram, derive an expression for impedance of a series LCR-circuit. [5]
OR
(A) How can you prove mathematically that average value of alternating current over one complete cycle is zero?
(B) What is meant by root mean square value of current? Derive its relationship with the peak value of alternating current.
Answer:
Let I be the current in the series circuit,

The voltage across the resistor will be in phase with current. So, phasors \(\overrightarrow{V_R}\) and \(\vec{I}\) are having same direction. The magnitude of \(\vec{V}_L\) is,
V_R = I₀R
The voltage across the inductor is ahead of current with a phase difference of \(\frac{\pi}{2}\) radian. The magnitude of \(\vec{V}_L\) is,
V_L = I₀X_L
The voltage across the capacitor is behind the current with a phase difference of \(\frac{\pi}{2}\) radian the magnitude of \(\vec{V}_C\) is V_C = I₀X_C

As it is clear from the figures that \(\vec{V}_L\) and \(\vec{V}_C\) are in opposite directions. So, the resultant of these two can be taken as
(\(\vec{V}_L\) – \(\vec{V}_C\)) considering \(\vec{V}_L\) > \(\vec{V}_C\).
Using Pythagorus theorem,
e₀² = (V_R)² + (V_L – V_C)²
e₀² = (I₀R)² + (I₀X_L – I₀X_C)²
= (I₀R)² + I₀²(X_L – X_C)²
= I₀²R² + I₀²(X_L – X_C)²
e₀² = I₀²(R² + (X_L – X_C)²)
I₀² = \(\frac{e_0^2}{R^2+\left(X_L-X_C\right)^2}\)
I₀ = \(\frac{e_0}{\sqrt{R^2+\left(X_L-X_C\right)^2}}\)
I₀ = \(\frac{e_0}{Z}\)
Thus, the quantity, Z = \(\sqrt{R^2+\left(X_L-X_C\right)^2}\) is the effective resistance of the LCR series circuit which opposes the flow of current through it. This quantity is known as the impedence of the circuit.

(A) The alternating current at any instant is given by, I₀ = I₀ sin ωt
The amount of charge flowing in small time dt is, dq = I dt
dq = I₀ sin ωt dt
The totaL charge that flows through the circuit in one complete cjc1e is,

= 0
The average value of alterating current over one complete cycle,
I_av = \(\frac{q}{T}\) = 0

(B) The alternating current at any instant is
given by,I = I₀ sin ωt
The amount of charge flowing in small time dt is
dq = Idt
dq = I₀ sin ωt dt
The total charge that flows through the circuit in one complete cycle is,

The average value of c.c. over one complete cycle of a.c..
I_av = \(\frac{q}{T}\) = 0.
Thus, the average value of a.c. over a complete cycle of a.c. is zero.

Section – E (8 Marks)

Question 34.

The energy required to fire an automobile spark plug is derived from magnetic-field energy stored in the ignition coil. Magnetic-field energy plays an important role in the ignition systems of gasoline powered automobiles. A primary coil of about 250 turns is connected to the car’s battery and produces a strong magnetic field.

This coil is surrounded by a secondary coil with some 25,000 turns of very fine wire. When it is time for a spark plug to fire, the current to the primary coil is interrupted the magnetic field quickly drops to zero, and an emf oftens of thousands of volts is induced in the secondary coil.

The energy stored in the magnetic field thus goes into a powerful pulse of current that travels through the secondary coil to the spark plug, generating the spark that ignites the fuel-air mixture in the engine’s cylinders

(A) A transformer is used to light a 100 Wand 110V lamp from a 220V mains. If the main current is 0.5A, what is the approximate effiiciency of the transformer? [1]
(B) A coil of 100 turns carries a current of 5 mA and creates a magnetic flux of 10^-5 weber. Calculate the inductance. [1]
(C) If number of turns in primary and secondary coils is increased to two times each, then determine the mutual inductance.
OR
If two coils of self-inductance L₁ and L₂ are coupled together, then what will be their mutual inductance? [2]
Answer:
(A) Output power for transformer,
(P) = 100 W
Voltage across primary coil,
(V_p) = 220 V
Current in the primary,
(I_p) = 0.5 A
The efficiency of a transformer is given by,
η = \(\frac{\text { output power }}{\text { input power }}\) × 100
= \(=\frac{P}{V_p I_p}\) × 100
= \(\frac{100}{220 \times 0.5}\) × 100
= 90%

Related Theory
The losses are the sum of copper losses in the windings, the iron loss, dielectric loss and humming loss. The efficiency of transformers is greatly affected by the different types of losses. The dielectric losses takes place inside the transformer which can be neglected for low voltage transformers.

(B) ∵ L = \(\frac{\phi}{NI}=\frac{10^{-5}}{100 \times 5 \times 10^{-3}}\)
= \(\frac{10^{-4}}{5} H=\frac{10^{-1}}{5} m H\)
= 0.02 mH

(C) Mutual inductance is directly proportional to the number of turns in primary as well as number of turns in secondary. Hence, mutual inductance becomes 4 times.

If two coils of self inductance Li and L2 are coupled together, their mutual inductance becomes,
M = k\(\sqrt{L_1 L_2}\)
Where, k = coupling constant whose value lies between 0 and 1.

Question 35.

The Hubbie Space Telescope is a space telescope that was launched into low Earth orbit in 1990 and remains in operation. It was not the first space telescope, but it is one of the largest and most versatile, well both known as a vital research tool and as a public relations boon for astronomy.

When light strikes the concave primary mirror of the Hubble Space Telescope, it is reflected to the convex secondary mirror, then back through a hole in the center of the primary mirror. There, the light comes to the focal point, and passes to one of Hubble’s instruments. Telescopes of this design are called Cassegrain telescopes, after the person who designed the first one.

(A) A terrestrial telescope has an objective of focal length 180 cm and an eye piece of focal length 5.0 cm. What will be the magnifying power of the telescope? [1]
(B) In the above problem, if focal length of the erecting lens is 3.5 cm then calculate the separation between the objective and eye piece in normal adjustment. [1]
(C) The magnifying power of a telescope in normal adjustment is 24, When the length of the telescope tube is 1 metre. What is the focal length of the eye lens?
OR
Draw a ray diagram of an astronomical telescope in the normal adjustment position. [2]
Answer:
(A) Magnifying power is ratio of focal length of objective and the focal length of eye lens,
M = \(\frac{f_0}{-f_e}=\frac{180}{5}\) = -36

(B) Length of telescopic tube is sum of focal length of objective plus four times the focal length of erecting lens plus focal length of eye lens.
f₀ + 4f + f_e = 180 + 4 × 3.5 + 5 = 199

Related Theory
The image formed by objective tens Lies at 2f of erecting Lens and it forms image at 2f on other side and erects it.

(C) M = \(\frac{f_o}{f_e}\) = 24
f₀ = 24 f_e
Length of telescope tube
= 1 m = 100 cm
So 24 f_e + f_e = 100 cm
25 f_e = 100
f_e = 4 cm.