Physics Unsolved Sample Paper Solutions CBSE Class 12

Section-A

The following questions are multiple-choice questions with one correct answer. Each question carries 1 mark. There is no internal choice in this section.

  1. (d) 4

Explanation: Electric field inside the sphere will be zero.

2. (b) (σ/ϵ0)

Explanation: As we know that,

E = (σ/ϵ0)

3. (b) decreases from A to B

Explanation: As area increases, vd decreases.

4. The coil of a moving coil galvanometer has 1000 turns each of area 3 × 10–4 m2. Its suspension fibre hasrestoring torque of 2 × 10–7 Nm per degree and the radial magnetic field of induction 0.08 Wb/m2. If a currentof 10 μA is passed through it, then the deflection produced is:

(a) 0.8°

(b) 0.6°

(c) 1.2°

(d) 1.0°

Ans. (c) 1.2°

Explanation: We know that,

NIAB = , where k is restoring torque

∴  α =(NIAB/k)

∴  α =(1000×10-5×3×10-4×8×10-2/2×10-7)

= 1.2°

5. (b) 3.2 × 10−13 N

Explanation: Force acting on a charge q moving with velocity v in magnetic field of intensity B is given by,

F = qvB sin θ

= qvB     (∵ θ = 90° ∴ sin θ = 1)

Given that,  q = 2 × 1.6 × 10–19 (α-particle)

v = 106 m/s

B = 1 T

on substituting these values, we get

F  = 2 × 1.6 × 10–19 × 106 × 1
= 3.2 × 10–13 N

6. (a) two times

Explanation: Magnetic flux density at the centre of a circular coil is given by,

B =(μ02Ni/2R)

or  B ∝(Ni/R)

7. (a) self-induction

Explanation: Self-induction is the property of current carrying coil that opposes the change of current  flowing through it.

8. (d) Radio waves, Infra-red, Visible light, X-rays

Explanation: λRadio waves > λInfra-red > λVisible Light > λ X-rays

The e.m. spectrum is generally divided into seven regions, in order of decreasing wavelength and increasing energy and frequency. The common designations are : radio waves, microwaves, infrared (IR), visible light, ultraviolet (UV), X-rays and gamma rays.

9. (d) not be seen at all

Explanation: For μ = 1.6, the critical angle, μ = 1/sin C, we have C = 38.7°, when viewed from AD, as long as angle of incidence on AD of the ray emanating from pin is greater than the critical angle , the light suffers from total internal reflection and cannot be seen through AD.

10. (a) 159.2 Ω

Explanation: We know that

XC =(1/ωC)

=(1/2πfC)

=(1/2× 3.14×100×10×-5)

= 159.2 Ω

11. (b) E1/2

Explanation:

E = hv2 =(hc/E) or λ2 =(p2/2m)

∴ λ1 =(h/p)=h/(√2mE)

λ1 : λ2 =(h/√2mE):(hc/E)=(1/c√2m)√E

∴ λ1 : λ2 ∝√E

12. (c) 4 : 1

Explanation: As r ∝ n2, for ground state n = 1

For first excited state n = 2,

So, (r2/r1)=(n22/n21)=((n2/n1))2=(2/1)2=(4/1)

13. (a) – 1.46 cm

Explanation:

me =(D/fe)=(25/5)=5
m = mo × me
mo =(m/me)=(-30/5)=-6
⇒   mo=(vo/uo)⇒6=(v0/u0)
So, vo = – 6uo
Using  (1/fe)(1/v0)(1/u0)
⇒ (1 /1.25)=(1/-6u0)-(1/u0)=(-1-6/6u0)=(-7/6u0)
⇒ uo =(-7×1.25/6)=-1.46cm

14. (a) 0.5

Explanation: Hence, area of coil

A = 0.1 m × 0.05 m

= 5 × 10–3 m2

N = 100

Initial flux linked with the coil,

ϕ1 = BA cos θ
= 0.1 × 5 × 10–3 cos 0°
= 25 × 10–5 Wb
= 2.5 × 10–4 Wb

The magnitude of induced emf in the coil is,

e =(N|Δϕ|/Δt)

=(N|ϕ21|/t)
=(100|2.5×10-4-5×10-4|/0.05)
=(100×2.5×10-4/0 05)V

= 0.5 V

15. (a) B ⊥ v

diagram15
diagram16

16. (b) Both A and R are true and R is not the correct explanation of A

Explanation: The energy gap is greater in silicon than in germanium because the minimum energy required to break a covalent bond is higher in Si (1.1 eV) than Ge (0.72 eV).

17. (b) Both A and R are true and R is not the correct explanation of A

Explanation: Combination of lenses helps to obtain desired magnification. It also enhances sharpness of the image. Since, the image formed by the first lens becomes the object for the second, the total magnification of the combination is a product of magnification of individual lenses.

18. (b) Both A and R are true and R is not the correct explanation of A

Explanation: For frequency greater than threshold frequency, photoelectric current increases with increase in intensity of incident light. Number of photoelectrons emitted from surface is independent of frequency of incident light.

Section-B

19. (i) Electromagnetic waves are produced by accelerating or oscillating charges and do not need any material medium for propagation. The source of energy of these waves are energy of oscillating charged particles.
Mathematical expressions for electric and magnetic fields of an electromagnetic wave propagating along the z-axis are :

diagram17

(ii) Characterstics of electromagnetic waves are:

(a) The direction of E, B and direction of propagation of waves are mutually perpendicular to one another.

(b) The amplitude ratio of E and B gives velocity of light.

20. Energy stored before introduction of dielectric,

diagram18
diagram19

21. According to Bohr’s postulate,

mvr =(nh/2π)

i.e.,     2πr =(nh/mv)

or  (h/mv)=(2πr/n)

Since, (h/mv)=(h/p)=λ, by de -Broglie hypothesis

Therefore, (2πr/n)=λ

Now for second excited state

λ2= (2πr3/3)  ...(i)

and for third excited state

λ3 =(2πr4/4) ...(ii)

On dividing equation (i) by (ii),

diagram20

22. Here V = 100 volts.

The de-Broglie wavelength λ is

diagram21

23. Full wave rectifier :

diagram 22
Explanation : In positive half cycle of AC, end A becomes positive and D1 becomes forward biased and D2 is reverse biased, so D1 conducts and D2 doesn’t. So conventional current flows through D1, RL and upper half of secondary winding. Similarly, during negative half cycle of AC, diode D2 becomes forward biased and D1 is reverse biased, current flows through D2, RL and lower half of secondary winding. Thus, current flows in same direction in both half cycles of input AC voltage.

OR

(a) Reverse biased.

(b) Full wave Rectifier :

Diode conducts only when the junction is forward biased. Hence, during first half cycle of input A.C., D1 will conduct while D2 will not and current in RL will flow from A to B. During second half cycle of input A.C., diode D2 will conduct while D1 will not conduct and current will again flow from A to B in RL. Hence, complete cycle will become unidirectional.

24. Wave theory cannot explain the following laws of photoelectric effect.

(i) The instantaneous emission of photoelectrons.

(ii) Existence of threshold frequency for metal surface.

(iii) K.E. of emitted electrons is independent of intensity of light and depends on frequency.

Photo theory can explain the photoelectric effect :

(i) Increasing the intensity means increasing the number of photons that does not change the maximum K.E. but changes the number of ejected electron.

(ii) E = hn means energy is dependent on the frequency, i.e., threshold frequency is needed to be matched for photon emission.

(iii) As photons strikes metallic surface it is absorbed and hence electron is ejected instantaneously.

Therefore, the charge on the sphere is 1.447 × 10–3 C.

25. (a) Radius of the sphere, r = 1.2 m.

Surface charge density, σ = 80.0 μC/M2
= 80 × 10–6 C/m2

Total charge on the surface of the sphere.

Q = Charge density × Surface area

= σ × 4πr2
= 80 × 10–6 × 4 × 3.14 × (1.2)2
= 1.447 × 10–3 C
Therefore, the charge on the sphere is 1.447 × 10–3 C.
diagram 25

Section-C

26. Electric flux through the cylindrical surface,

diagram 26
diagram 27

= – pE cos 60°

= – 8 × (1/2)

= – 4J.

28. Let us consider a Wheatstone bridge arrangement as shown below :

Wheatstone bridge is a special bridge type circuit which consists of four resistances, a galvanometer and a battery. It is used to determine unknown resistance.

diagram 28
In figure four resistance P, Q, R and S are connected in the form of four arms of a quadrilateral. Let the current given by battery in the balanced position be I. This current on reaching point A is divided into two parts I1 and I2. As there is no current in galvanometer in balanced state, therefore, current in resistances P and Q is I1 and in resistances R and S it is I2.
diagram 29

Here, σ is the conductivity of the material.

The above relation is equivalent to Ohm’s law.

Now, for a given material, s is a constant.

Hence, the plot of j versus E will be a straight line starting from the origin.

The slope of the graph gives the conductivity σ of the material.

The graph with a greater slope is a better conductor and the graph with a lesser slope is a poor conductor than the other.

diagram 30

(a) To make a standard resistance : A resistor should allow only a limited current to flow through it. Hence, from the graph, we come to know that the material whose plot is similar to plot (2) in graph should be used to make a resistor.

(b) To make a connecting wire : A wire should allow all the current to flow through it without resisting it. So, it should have a higher conductivity. Hence, from the graph, we come to know that the material whose plot is similar to plot (1) should be used to make a wire.

Electron drift speed is estimated to be of the order of mm s– 1. However the current set-up in the wires is of the order of few amperes. This is because the electron density is very large in a material. It is of the order 1028/m3 of the wire. Hence, all these electrons contribute to the total current. Therefore, despite having small drift speeds, the current set-up in wires is large.

29. The phenomenon of total internal reflection occurs when,

(i) Angle of incidence is equal or greater than critical angle.

i ≥ C

(ii) When light travels from more denser medium to less denser medium.

In case of right angle isosceles triangle if light rays fall normally on AB then light incident on face AC with angle of incidence > critical angle.

Hence, total internal reflection will occur with normal to the surface of BC.

OR

Magnifying Power: The magnifying power of a compound microscope is defined as the ratio of the angle subtended at the eye by the final image to the angle subtended at the eye by the object, when both the final image and the object are situated at the least distance of distinct vision from the eye.

Magnifying power of the compound microscope when the final image is at infinity,

diagram 32

From the above equation, we can see that to achieve a large magnification, the objective and eyepiece should have small focal lengths.

30. The slope of the cut-off voltage (V) versus frequency (n) of an incident light is given by

(V/v)=4.12 × 10– 15 Vs

V is related to frequency by

hv = eV

Here, Charge on an electron (e) = 1.6 × 10– 19 C

Planck’s constant (h) = ?

∴ h = e ×(V/v)

= 1.6 × 10– 19 × 4.12 × 10– 15
= 6.592 × 10– 34 Js
Hence, the value of Planck’s constant is 6.592 × 10– 34 Js.

Section-D

31. (a) In equilibrium, a magnetic dipole always aligns itself along the direction of the magnetic field. When the magnetic dipole is displaced from the equilibrium position, a restoring torque acts on the dipole to bring it back. Therefore, to place the dipole at some inclination with the field, work has to be done against the restoring torque. This work done is stored in the dipole as its potential energy.

(b) (i) The behaviour of magnetic field lines in the presence of a paramagnetic substance is shown below :

(ii) The behaviour of magnetic field lines in the presence of a diamagnetic substance is shown below :

This distinguishing feature is because of the difference in their relative permeability. The relative permeability of diamagnetic substance is less than 1, so the magnetic lines of force do not prefer passing through the substance whereas the relative permeability of a paramagnetic substance is greater than 1, so the magnetic lines of force prefer passing t hrough the substance.

OR

(a) The material having positive and small susceptibility is paramagnetic material.

Properties :

(i) They have tendency to move from a region of weak magnetic field to strong magnetic field, i.e., they get weakly attracted to a magnet.

(ii) When a paramagnetic material is placed in an external field the field lines get concentrated inside the material, and the field inside is enhanced.

(b) (i) Consider a ring of radius r having n turns per metre. If n is the number of turns per metre, then total number of turns in the ring = 2πrn = N

diagram 35
32. (a) (i) No, however strong the magnet may be, current can be induced only by changing the magnetic flux through the loop.

(ii) No current is induced in either case. Current can not be induced by changing the electric flux.

(iii) The induced emf is expected to be constant only in the case of the rectangular loop. In the case of circular loop, the rate of change of area of the loop during its passage out of the field region is not constant, hence induced emf will vary accordingly.

(b) The ratio of magnetic flux through the solenoid to the current passing through it is called self-inductance of a solenoid. It is given by L = Φ/I

Energy stored in an inductor : When a current grows through an inductor, a back e.m.f. is set up which opposes the growth of current. So work needs to be done against back e.m.f. (e) in building up the current. This work done is stored as magnetic potential energy.

diagram 36

When the north pole of a bar magnet is pushed towards the coil, the amount of magnetic flux linked with the coil increase. Current is reduced in the coil from a direction such that it opposes the increase in magnetic flux. This is possible only when the current induced in the coil is in anti-clockwise direction, with respect to an observer. The magnetic moment M associated with this induced emf has north polarity, towards the north pole of the approaching bar magnet. Similarly, when the north pole of the bar magnet is moved away from the coil, the magnetic flux linked with the coil decreases. To counter this decrease in magnetic flux, current is induced in the coil in clockwise direction so that its south pole faces the receding north pole of the bar magnet. This would result in an attractive force which opposes the motion of the magnet and the corresponding decrease in magnetic flux.

33. (a) If instead of monochromatic light, white light is used, then the central fringe will be white and the fringes on either side will be coloured. Blue colour will be nearer to central fringe and red will be farther away. The path difference at the centre on perpendicular bisector of slits will be zero for all colours and each colour produces a bright fringe thus resulting in white fringe. Farther, the shortest visible wave, blue, produces a bright fringe first.

(b) (i) Two characteristic features distinguishing the diffraction pattern from the interference fringes obtained in Young’s double slit experiment are:

(1) The interference fringes may or may not be of the same width whereas the fringes of diffraction pattern are always of varying width.

(2) In interference the bright fringes are of same intensity whereas in diffraction pattern the intensity falls as we go to successive maxima away from the centre, on either side.

(ii) Wavelength of the light beam, λ1 = 590 nm = 5.9 × 10−7 m
Wavelength of another light beam, λ2 = 596 nm = 5.96 × 10−7 m

Distance of the slits from the screen, D = 1.8 m

Distance between the two slits = 1 × 10−4 m

For the first secondary maxima,

Now, each of these infinitesimally small slit sends zero intensity in direction q. Hence, the combination of these slits will give zero intensity.

(b) (i) Huygen’s Principle:

(a) Each point on the primary wave front acts as a source of secondary wavelets, transferring out disturbance in all directions in the same way as the original source of light does.

(b) The new position of the wave front at any instant is the envelope of the secondary wavelets at that instant.

Refraction on the basis of wave theory

1. Consider any point Q on the incident wave front.

diagram 39

2. Suppose when disturbance from point P on incident wave front reaches point P’ on the refracted wave front, the disturbance from point Q reaches Q’ on the refracting surface XY.

3. Since P’A’ represents the refracted wave front, the time taken by light to travel from a point on incident wave front to the corresponding point on refracted wave front should always be the same.
Now, time taken by light to go from Q to Q′ will be

diagram 40

(ii) (1) The frequency of reflected and refracted light remains constant as the frequency of incident light because frequency only depends on the source of light.

(2) Since the frequency remains same, hence there is no reduction in energy.

Section-E

34. (i) Here fo = 2.0, fo = 6.25 cm, u0 = ?

When the final image is obtained at the least distance of distinct vision :

ve = – 25 cm
diagram 41
diagram 41
diagram 42

(iii) The intermediate image formed by the objective of a compound microscope is real, inverted and magnified.

CBSE 36 Sample Question Papers Science Stream (PCM)

All Subjects Combined for Class 12 Exam 2023

CBSE 36 Sample Question Papers Science Stream (PCB)

All Subjects Combined for Class 12 Exam 2023

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