Class 12 Term 2 Physics Sample Paper 

Class 12 Term 2 Physics Sample Paper: CBSE has scheduled the Class 12 Term 2 Physics exam for 20th May 2022 and its high time, the students must practice and revise with Sample Paper Physics Class 12 Term 2. CBSE has released the official Physics Sample Paper Class 12 Term 2 Exam on its official website @ along with the solutions. The Class 12 Physics Term 2 Sample Paper with Solution will help the students to better understand the pattern of the exam and help them to perform their best in the examination. The direct links to download the Class 12 Physics Term 2 Sample Paper PDF has been provided in the article below. Download Class 12 Term 2 Physics Sample Paper PDF now and practice with the questions to analyse your level of preparation for the exam. 

Physics Sample Paper Class 12 Term 2 Exam

Central Board of Secondary Education has divided the Class 12 Physics Board Exam into two Terms. In Term 2 the exam is divided into three sections with 12 questions for which the exam pattern that has been followed is as follows- 

1. Section A consists of 1 to 3 questions carrying 2 marks each [Very Short Answer Questions]. 

2. Section B consists of 4 to 11 questions carrying 3 marks each [Short Answer Questions]. 

3. Section C consists of 1 question carrying 5 marks [Case Based Question].

Physics Class 12 Term 2 Answer Key 2022- Click Here

Class 12 Physics Term 2 Sample Paper and Solution

The students preparing for CBSE Class 12 Physics Term 2 Exam must download the official sample paper PDF from the below link and practice the questions as per the pattern released by CBSE. The solutions to the CBSE Class 12 Physics Term 2 Sample Paper as set by CBSE Class 12 Physics Term-2 Syllabus have been provided below along with the solution as released by CBSE- 

Class 12 Term 2 Physics Sample Paper with Solutions
Sample PaperSolution
Class 12 Physics Term 2 Sample Paper- 1 [Official]Solution Link

CBSE Class 12 Physics Term 2 Syllabus- Click to Check

Physics Class 12 Term 2 Sample Paper 

In the below section, we have provided you with some important questions as per the new exam pattern. Do go through these Class 12 Physics Questions for Term 2 and excel on your preparation done so far.

Question 1- How is a p-type semiconductor formed? Draw the energy band of p-type semiconductors and name the majority of charge carriers in it. 

Solution- If trivalent impurity atoms of B, Al, or In are doped in a pure semiconductor of Silicon or Germanium, we get a p-type semiconductor. Holes are the majority charge carriers in a p-type semiconductor. The diagram of the energy band of p-type semiconductors is given below.

Class 12 Physics Sample Paper Term 2

Related Theory- There are two types of dopants used for doping the tetravaLent Silicon or Germanium.

1. Pentavalent dopants which have five valence electrons like Arsenic, Antimony, and Phosphorus.

2. Trivalent dopants have three valence electrons like Indium, Boron, and Aluminium.

Question 2. State Bohr’s postulate to define stable orbits in hydrogen atoms. How does Louis de BrogUe’s hypothesis explain the stability of these orbits?

Solution- Bohr’s postulate for stable orbits in hydrogen atom: An electron can revolve only in those circular orbits in which its angular momentum is an integral multiple of ℎ2𝜋. where h is Planck’s constant.

If n is the principal quantum number of orbit, then an electron can revolve only in a certain orbit of definite radii. These are called stable orbits. Louis de Broglie’s explanation of the stability of orbits:

According to de Broglie. orbiting electron around the nucleus is associated with a stationary wave. An electron wave is a circular standing wave. Since destructive interference will occur if the standing wave does not close upon itself only those Louis de Brog lie waves exist for which the circumference of circular orbit Contains a whole number of wavelengths i.e., for orbit circumference of nth orbit as 2nπrn

2πrn = nλ

2πrn = n (ℎ𝑚𝑣)

mvrn = n (ℎ2𝜋)

which is the quantum condition proposed by Bohr.


A hydrogen atom initially in the ground state absorbs a photon which excites it to the n = 4 LeveL Estimate the frequency of the photon.

In ground state, n = 1

In excited state, n = 4

1𝜆=𝑅[1(1)2−1(4)2] where R is Rydberg constant

1𝜆 = 1.097 x 107 x 1516

= 1.028 x 107 m-1

Frequency, v = 𝑐𝜆 = 3 x 108 x 1.028 x 107

= 3.09 x 1015 Hz.

Question 3- What is meant by the size of the nucleus? A nucleus of mass number 225 splits into two fresh nucLei having radius R = 1.1 x 10^-15 A^1/3 m, find the radii of the new nuclei formed.

Solution- The extremely small central core of the atom in which the whole of the positive charge and practically the entire mass is confined is known as the nucleus. Different nuclei differ in size. The radius of nuclei is directly proportional to A^1/3, where A is its mass number.

Here A = 225, R = 1.1 x 10^-15 A^1/3 m

Let A1 and A2 be the respective mass number of two new nuclei formed.

A1= 3(3+2) A

A1= 3/(3+2) A

= 3/5 x 225

= 135


A2 = 3/(3+2) A

=2/5 x 225 = 90

R1= 1.1 x 10-15 A^1/3

= 1.1 x 10^-15 x 135^1/3

= 5.643 x 10^-15 m

R2 = 1.1 x 10^-15 A^1/3

= 1.1 x 10^-15 x 90^1/3

= 4.93 x 10^-15 m.

Question 4- When p-n junction Is said to be reverse-biased, what will be the direction of flow of majority charge carriers?

Solution- A p-n junctión ¡s set to be reverse biased when its p-region is maintained at lower potential with respect to its n-region. i.e, p-end is connected with negative terminal, and n-end is connected with positive terminal In a reverse-biased p-n junction. Majority charge carriers move away from the junction.

Question 5- Which semiconductors are preferred to make LED and why? Give two advantages of using LED.

Solution- Materials like gallium phosphide (Gap) Gallium arsenide (GaAs), etc., are used. They emit the maximum amount of energy in form of tight radiation.

The following are the advantages of LEDs:

(i) They are operational at two voltages.

(ii) They are quick in action and their power consumption is Low.

Question 6- Monochromatic light of frequency 6.0 × 10^14 Hz is produced by a laser. The power emitted is 2.0 × 10^-3 W. Estimate the number of photons emitted per second on an average by the source.

Solution- The energy of a photon of frequency v is
E = hv

= (6.63 × 10^-34 Js) × (6 × 10^14 s^-1)

= 3.98 × 10^-19 J = 4 × 10^-19 J

If n be the number of photons emitted by the source per second, then the power P transmitted in the beam is given by, P = nE

∴ n = 𝑃𝐸

⇒ n = 2×10^−34×10^−19

5 × 10^15 photons/sec

Related Theory
Photoelectric current is directly proportional to the intensity of incident radiation.

Question 7- Write the relationship between angle of incidence ‘F, angle of prism ‘A’ and angle of minimum deviations δm for a triangular prism. A ray of light suffers minimum deviation while passing through a prism of refractive index 1.5 and refracting angle 60°. Calculate the angle of deviation and angle of incidence. 

Solution- The relation between the angle of incidence i, angle of prism A, and the angle of minimum deviation δm, for a triangular prism is given by i = (𝐴+𝛿𝑚)/2

μ = 1.5, A = 60°, δm = ?

In the position of minimum deviation

I = 𝐴+𝛿𝑚/2,

r = A/2

As, μ = sin𝑖/sin𝑟

∴ 1.5 = sin𝑖/sin30∘

⇒ sin i = 1.5 × 0.5

= 0.75

⇒ i = sin^-1 (0.75)

= 48.60°

δm = 2i – A

= 2 × 48.6 – 60 = 37.2°

Question 8- The ground state energy of hydrogen atom is -13.6 eV. If an electron makes a transition from an energy level -0.85 eV to -1.51 eV, calculate the wavelength of the spectral line emitted. To which series of hydrogen spectrum does this wavelength belong?

Solution- As we know, En = −13.6/𝑛²eV …………. (i)

For n = 1, E1 = – 13.6 eV

When electron undergoes transition from EA = – 0.85 eV Then, from equation (i),

– 0.85 = – 13.6/𝑛²𝐴 ⇒ nA = 4

Similarly, -1.51 = −13.6,𝑛²𝐵 ⇒ nB

Hence, electron transits from n = 4 to n = 3 which corresponds to Paschen series of hydrogen atom.

As, 1𝜆 = R(1/𝑛²𝐵−1/𝑛²𝐴)

Here nA = 4, n²B = 3, R = 1.0974 × 107

Then 1/𝜆 = 1.097 × 107(1/3²−1/4²)m^-1

⇒ λ = 1875 nm

Question 9- Give reasons for the following:

(A) Why semiconductor diodes are preferred over vacuum diodes?

Solution- In semiconductor diodes, no cathode heating is required in junction diode for production of charge carriers. Voltage drop across a junction diode is much less compared to that across a vacuum diode. Semiconductor diodes can be used for much higher frequencies.

(B) A photodiode, when used as a detector of optical signals is operated under reverse bias.

Solution- When operated under reverse bias, the photodiode can detect changes in current with changes in light intensity more easily because the fractional change in minority charge carriers is more than that in majority charge carriers.

(C) The band gap of the semiconductor used for fabrication of visible LEDs must be at least 1.8 eV. 

Solution- The Photon energy, visible light photons vary from about 1.8 eV to 3 eV. Hence for visible LEDs, the semiconductor must have a band gap of 1.8 eV.

Question 10- State principle of reversibility of light. Define lateral shift.

Solution- Principle of reversibility of light states that if the final path of ray of light after it has suffered several reflections and refractions is reversed, it retraces its path exactly.

Lateral shift is the perpendicular distance between the incident and emergent rays when light is incident obliquely on a refracting slab with parallel faces.

Question 11- (A) In a double slit experiment using light of wavelength 600 nm, the angular width of the fringe formed on a distant screen is 0.1°. Find the spacing between the two slits.

Solution- Angular width (θ) of fringe in double-slit experiment is given by

θ = 𝜆/𝑑

Where d = Spacing between the slits.

Given: Wavelength of light,

λ = 600 nm

Angular width of fringe,

θ = 0.1°

= 0.1 × 𝜆/180 rad

= 0.0018 rad

∴ d = 𝜆/𝜃

d = 600×10^−9/18×10^−4 

= 33.33 × 10^-5

⇒ d = 0.33 × 10^-3 m

(B) Light of wavelength 5000 A propagating in air gets partly reflected from the surface of water. How will the wavelengths and frequencies of the reflected and refracted light be affected?

Solution- The frequency and wavelength of reflected wave will not change. The refracted wave wiLl have same frequency, only the wavelength will change. The velocity of light v in water is given by

v = λf

where v = Velocity of light, 

f = Frequency of light, 

λ = Wavelength of light 

As light rays in travelling from a rarer (air) medium to a denser medium, their speed will decrease. Hence wavelength (λ) will also decrease.

Related Theory- When a light ray travels from a rarer medium to a denser medium, then its speed and wavelength both will decrease.

Steps to download CBSE Class 12 Physics Sample Paper 2022

Download the Sample Paper Class 12 Physics Term 2 Exam with solutions for the term-2 exam by following the below-mentioned steps :

Step I- Visit the official website of CBSE Academic @ or Click on the CBSE Class 12 Physics Sample Paper with Solutions for Term 2 Exam mentioned above. 

Step II- Click on the notification appearing in the academic section- “Sample Question Papers of Classes XII for Term 2 Exams 2022”.

Step III- Now click on the link mentioned under “Sample Papers Class XII”.

Step IV- The list of all subjects “Class XII Sample Question Paper & Marking Scheme for Exam 2022” appears on the screen. 

Step V- Click on “SQP” for “Physics” and download CBSE Class 12 Physics Sample Paper 2022 Term-2 pdf along with Solution. 

Step VI- Check the marking scheme after attempting each subject CBSE Class 12 Physics Sample Paper 2022 Term-2.

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CBSE Class 12 English Syllabus (Term 1+Term 2)CBSE Class 12 Physics Syllabus (Term 1+Term 2)
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Class 12 Physics Term 2 Sample Paper: FAQs

Ans. You can download Class 12 Term 2 Physics Sample Paper PDF from the article above.

Ans. Class 12 Physics Term 2 Exam has been scheduled for 20th May 2022.

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