CBSE Class 12 Physics Sample Paper

Sample Paper Class 12

We have provided CBSE Class 12 Physics Sample Paper with solutions below. These Sample guess papers have been prepared as per the latest examination guidelines and paper pattern issued by CBSE. Students of Class 12 should practice all practice papers for Class 12 Physics given below as it will help them top improve their understanding of the subject. Please click on the links below to access free sample paper for Physics Class 12.

Sample Papers for Class 12 Physics

CBSE Class 12 Physics Sample Paper Set A
CBSE Class 12 Physics Term 1 Sample Paper Set A
CBSE Class 12 Physics Term 1 Sample Paper Set B
CBSE Class 12 Physics Term 1 Sample Paper Set C
CBSE Class 12 Physics Term 1 Sample Paper Set D
CBSE Class 12 Physics Term 1 Sample Paper Set F
CBSE Class 12 Physics Term 2 Sample Paper Set A

CBSE Class 12 Physics Term 1 Sample Paper Set A


1. Explain how the width of depletion layer in a P-n junction diode changes when the junction is (i) forward biased
(ii) reverse biased.
Answer : (i) In forward bias, positive terminal of the battery is connected to p side and negative terminal to n-side of the p-n junction. So the holes from p side which are majority carrier are pushed towards the junction and electron from n-side are pushed towards the junction. This results in reduction of the width of depletion layer.
(ii) in reverse bias, n-side is connected with positive terminal of the battery and p-side to negative terminal of the battery. The minority carriers i.e., holes from n-side are pushed towards the junction whereas electrons which are minority carriers in p-side are pushed towards the junction. The majority carrier from p as well as n-side move away from the junction. This increases the width of depletion layer.
Forward Biased: 

AA′ → Original depletion layer
BB′ → Depletion layer after forward biasing
Reverse Biased     

AA′ → Original depletion layer
BB′ → Depletion layer after reverse biasing

2. You are given two nuclei 3X7 and 3Y4. Explain giving reasons, as to which one of the two nuclei is likely to be more stable?

Calculate the binding energy per nucleon of 26Fe56 nucleus. Given that mass of 26Fe56 = 55.934939 u, mass of proton = 1.007825 u, mass of neutron = 1.008665 u and 1u = 931 MeV.
Answer : In case of 3X7

No. of neutron /No. of proton = 7-3 / 3 =4/3 = 1.33

In case of 3Y4 ,     No. of neutron/No. of proton  = 4-3/3 = 1/3 = 0.33

For stability, this ratio has to be close to one. Obviously, nucleus 3X7 is more stable than the nucleus 3Y4.

For 26Fe56, no. of protons = 26 = Z
No. of neutrons = A – Z = 56 –26 = 30
Dm = [Zmp + (A – Z) mn – MN]

∴ Mass defect = 26 mp + 30 mn – MFe
= 26 × 1.007825 + 30 × 1.008665 – 55.934939
= 26.20345 + 30.25995 – 55.934939
= 0.528461 u.
B.E./nucleon = 0.528461 × 931 /56 = 8.79 MeV.

3. What do the acronyms LASER and LED stand for? Name the factor which determines (i) frequency and (ii) intensity of light emitted by LED.
Answer : LASER stands for Light Amplification by Stimulated Emission of Radiation.
LED stands for Light Emitting Diode
(i) The frequency of light emitted by LED is determined by the band gap of the semiconductor used in LED.
(ii) Intensity of light emitted by LED depends on the doping level of the semiconductor and forward bias voltage.


4. (a) In photoelectric effect, do all the electrons that absorb a photon come out as photoelectrons irrespective of their location? Explain.
(b) A source of light, of frequency greater than the threshold frequency, is placed at a distance ‘d’ from the cathode of a photocell. The stopping potential is found to be V. If the distance of the light source is reduced to d/n (where n > 1), explain the changes that are likely to be observed in the (i) photoelectric current and (ii) stopping potential.
Answer : (a) No, it is not necessary that if the energy supplied to an electron is more than the work function, it will come out.
The electron after receiving energy, may lose energy to the metal due to collisions with the atoms of the metal.
Therefore, most electrons get scattered into the metal. Only a few electrons near the surface may come out of the surface of the metal for whom the incident energy is greater than the work function of the metal.
(b) on reducing the distance, intensity increases.
Photoelectric current increases with the increase in intensity. Stopping potential is independent of intensity, and therefore remains unchanged.

5. (a) Draw the necessary energy band diagrams to distinguish between conductors, semiconductors and insulators.
(b) How does the change in temperature affect the behaviour of these materials? Explain briefly.
Answer : (a) 

(b) With increase in temperature, resistivity of a conductor increases because of increase in no. of collisions. So relaxation time decreases which increases resistivity.
With increase in temperature, resistivity of a semiconductor decreases. Because of increase in temperature more number of electrons gain energy to jump to conduction band and are available for conducting current.
In case of insulator for moderate change of temperature, there is no change in behaviour.

6. Binding energy per nucleon versus mass number curve is as shown.
AZS , Z1A1W ,  Z2 A2X , Z3 A3Y and  are four nuclei indicated on the curve.
Based on the graph:
(a) Arrange X, W and S in the increasing order of stability.
(b) Write the relation between the relevant A and Z values for the following nuclear reaction.   (Img 20)
S —→ X + W

Answer : (a) S, W, X
(b) Z = Z1 + Z2
A = A1 + A2
(c) Reason for low binding energy:
In heavier nuclei, the Coulombian repulsive effects can increase considerably and can match/ offset the attractive effects of the nuclear forces. This can result in such nuclei being unstable. 

7. In the figure given below, three light rays red (R), green (G) and blue (B) are incident on an isosceles right-angled prism abc at face ab. Explain with reason, which ray of light will be transmitted through the face ac. The refractive index of the prism for red, green, blue light are 1.39, 1.44, 1.47 respectively.
Trace the path of rays after passing through face ab. 

Answer : 

Then total internal reflection will take place.

8. (a) A giant refracting telescope at an observatory has an objective lens of focal length 15 m. If an eye piece of focal length 1.0 cm is used, what is the angular magnification of the telescope

(b) If this telescope is used to view the moon, what is the diameter of the image of the moon formed by the objective lens. The diameter of the moon is 3.48 × 106 m and radius of lunar orbit is 3.8 × 108 m.
(a) Draw the labelled diagram for the formation of image by a compound microscope.
(b) Define the magnifying power of compound microscope.
(c) Explain why both the objective and the eyepiece of a compound microscope must have short focal length.

Answer : (a) fo = 15 m, fe = 1 cm = 10–2 m

Angular magnification
m = f0 /fe = -15 / 10-2 = -1500   

–ve sign shows that image is inverted.
(b) Let D be the diameter of the moon in meters. r is radius of the lunar orbit. Then Objective d is the diameter of the image of the moon.

∝ = D/r = d/f0 ⇒ 3.48 x 106 / 3.8 x 108 = d/15
d = 3.48 x 106 x15 / 3.8 x 108 = .1373 m = 13.73 cm


(b) The magnifying power of a compound microscope is defined as the ratio of angle subtended at the eye by the final virtual image to the angle subtended at the eye by the object when both are at the teast distance of distinct vision from the eye.

(c) (i) If the focal lengths are less than their magnifying power will be more.
(ii) To avoid any aberration in refraction due to larger bend on passing through the eye piece.

9. (a) If one of two identical slits producing interference in Young’s experiment is covered with glass, so that the light intensity passing through it is reduced to 50%, find the ratio of the maximum and minimum intensity of the fringe in the interference pattern.
(b) What kind of fringes do you expect to observe if white light is used instead of monochromatic light?
Answer : (a) Let the maximum intensity be Imax and minimum intensity Imin 

(b) Time varying fringes patterns in which relative positions of fringes changes with time because of the presence of various wavelengths in white light.
The interferences patterns due to different constituent colours of white light overlap (incoherently) coloured fringes of different width are obtained.

10. Given the ground state energy E0 = – 13.6 eV and Bohr radius a0 = 0.53 Å. Find out how de-Broglie wavelength associated with the electron orbiting in the ground state would change when it jumps into the first excited state.
Answer : As de-Broglie wavelength   λ = h/p = h/√2mE

For ground state n = 1
For First excited state n = 2   

11. In a plane electromagnetic wave, the electric field oscillates sinusoidally at a frequency of 2.0 × 1010 Hz and amplitude 48 Vm–1.
(a) What is the wavelength of a wave?
(b) What is the amplitude of the oscillating magnetic field?
(c) Show that the average energy density of the E field equals the average energy density of the B-field [c = 3 × 108 ms–1]
Name the parts of electromagnetic spectrum which is:
(a) Suitable for radar system used in air craft navigation
(b) Used to treat muscular strain.
(c) Used as a diagnostic tool in medicine.
Write in brief, how these waves can be produced.

Answer : Given : v = 2 × 1010 Hz, E0 = 48 Vm–1
(a) λ = c/v = 3 x 108/2 x 1010 = 1.5 x 10-2 m
(b) B0 = E0/c = 48/3×108 = 1.6 x 10-2 T

(c) To show that the energy density of electric field = average energy density of the magnetic field.
Energy density of E. field = ue = 1/ε0 E2
Energy density of M-field = ue = 1/2μ0 B2

(a) Microwaves
Production: Klystron/magnetron
(b) Infrared radiation
Production: Hot bodies/vibrations of atoms and molecules
(c) X-rays
Production: Bombarding high energy electrons on metal target.


12. A prism is a portion of a transparent medium bounded by two plane faces inclined to each other at a suitable angle. 
A ray of light suffers two refractions on passing through a prism and hence deviates through a certain angle from its original path. The angle of deviation of a prism is, d = (m – 1) A, through which a ray deviates on passing through a thin prism of small refracting angle A.

If m is refractive index of the material of the prism, then prism formula is, μ = sin(A + δ) /2 /(sin A/2)

Based on the above facts, answer the following questions.
(i) The angle of deviation is minimum for
(a) red colour
(b) yellow colour
(c) violet colour
(d) blue colour

Answer : A

(ii) When white light moves through vacuum
(a) all colours have same speed
(b) different colours have different speeds
(c) violet has more speed than red
(d) red has more speed than violet

Answer : A

(iii) The deviation through a prism is maximum when angle of incidence is
(a) 45°
(b) 70°
(c) 90°
(d) 60°

Answer : C

(iv) What is the deviation produced by a prism of angle 6°? (Refractive index of the material of the prism is 1.644)
(a) 3.864°
(b) 4.595°
(c) 7.259°
(d) 1.252°

Answer : A

(v) A ray of light falling at an angle of 50° is refracted through a prism and suffers minimum deviation. If the angle of prism is 60°, then the angle of minimum deviation is
(a) 45°
(b) 75°
(c) 50°
(d) 40°

Answer : D

CBSE Class 12 Physics Term 1 Sample Paper Set A