CBSE 12 Physics Board Exam 2025 : Sample Practice Questions for Last Minute Revision (with Solution)

CBSE 12th exams are underway and your CBSE 12th Physics exam is scheduled on 21th Feb, 2025. You have just a few hours left for CBSE 12th Physics exam.

We know how important it is to focus on the right topics and questions during revision, so this article provides important CBSE 12th Physics Sample Practice Questions for last minute revision.

Check out this practice paper or sample paper prepared by experts to prepare for CBSE Class 12 Physics Theory Exam 2025.

Answers is also provided along with the questions for easy and quick preparation. The question paper given here is solved, and so all you have to do is read the question, solve it, and match it with the given answer.

CBSE Class 12 Physics Practice Sample Question Paper 2025

SECTION A: Multiple Choice Questions

Q1. Two positive ions, each carrying a charge q, are separated by a distance d. If F is the force of repulsion between the ions, the number of electrons missing from each ion is:

(A) q/e
(B) F/q
(C) d/q
(D) q/d

Answer: (A) q/e

Q2. The total electric flux emanating from an alpha particle is:

(A) 2e/𝞮₀
(B) e/𝞮₀
(C) 4e/𝞮₀
(D) e²/𝞮₀

Answer: (C) 4e/𝞮₀

Q3. The electric field due to a point charge at a given distance varies:

(A) Inversely with distance
(B) Inversely with square of the distance
(C) Directly with distance
(D) Does not vary

Answer: (B) Inversely with square of the distance

Q4. Which of the following does not obey Ohm’s Law?

(A) Copper wire
(B) Silicon diode
(C) Resistor
(D) Nichrome wire

Answer: (B) Silicon diode

Q5. The capacitance of a parallel plate capacitor increases when:

(A) The plate area increases
(B) The separation between plates increases
(C) A conductor is placed between the plates
(D) Air is removed from between the plates

Answer: (A) The plate area increases

Q6. The force experienced by a charged particle moving perpendicular to a uniform magnetic field is given by:

(A) qE
(B) qvB
(C) qvB sinθ
(D) qvB cosθ

Answer: (C) qvB sinθ

Q7. A transformer works on the principle of:

(A) Electrostatic Induction
(B) Electromagnetic Induction
(C) Magnetic Effect of Current
(D) Conservation of Charge

Answer: (B) Electromagnetic Induction

Q8. The energy of a photon of light is directly proportional to its:

(A) Wavelength
(B) Frequency
(C) Amplitude
(D) Speed

Answer: (B) Frequency

Q9. The power loss in a transformer is mainly due to:

(A) Mechanical Friction
(B) Heat Dissipation in Windings
(C) Magnetic Flux Leakage
(D) Both B and C

Answer: (D) Both B and C

Q10. In an intrinsic semiconductor, the number of electrons and holes:

(A) Are equal
(B) Electrons are more
(C) Holes are more
(D) None of the above

Answer: (A) Are equal

Q11. The minimum energy required to remove an electron from the surface of a metal is called:

(A) Kinetic Energy
(B) Potential Energy
(C) Work Function
(D) Photoelectric Energy

Answer: (C) Work Function

Q12. Which of the following logic gates outputs 1 when any one of its inputs is 1?

(A) AND Gate
(B) OR Gate
(C) NAND Gate
(D) NOR Gate

Answer: (B) OR Gate

Assertion-Reasoning Based Questions

For the following questions, choose the correct answer:

(A) Both A and R are true, and R is the correct explanation of A
(B) Both A and R are true, but R is NOT the correct explanation of A
(C) A is true, but R is false
(D) A is false, but R is true

Q13. Assertion (A): The charge on a body can only be changed in integral multiples of the elementary charge e.
Reason (R): Charge is quantized.

Answer: (A) Both A and R are true, and R is the correct explanation of A

Q14. Assertion (A): Coulomb’s force is the dominating force in the universe.
Reason (R): Coulomb’s force is weaker than gravitational force.

Answer: (D) A is false, but R is true Q

15. Assertion (A): A metallic shield in the form of a hollow conductor can block an external electric field.
Reason (R): In a hollow spherical conductor, the electric field inside is always zero.

Answer: (A) Both A and R are true, and R is the correct explanation of A

Q16. Assertion (A): The range of gravitational force and Coulomb force is infinite.
Reason (R): The Coulomb force is stronger than the gravitational force.

Answer: (B) Both A and R are true, but R is NOT the correct explanation of A

Section B: 2 Marks Questions

Q17. Define magnetic susceptibility of a material. Name two elements, one having positive susceptibility and the other having negative susceptibility. What does negative susceptibility signify?

Answer: ○ Magnetic susceptibility (χ) is the measure of how much a material will become magnetized in an applied magnetic field.
○ Element with positive susceptibility: Iron (paramagnetic/ferromagnetic)
○ Element with negative susceptibility: Copper (diamagnetic)
○ Negative susceptibility signifies that the material develops a weak magnetic moment in the direction opposite to the applied field (diamagnetic behavior).

Q18. Why do electrostatic field lines not form closed loops?

Answer: ○ Electrostatic field lines do not form closed loops because they originate from positive charges and terminate at negative charges.
○ This indicates that electrostatic forces are conservative, meaning work done in moving a charge in a closed loop is zero.

Differentiate between n-type and p-type semiconductors based on their charge carriers.

Answer: ● n-type semiconductor: Majority charge carriers are electrons, and doping is done with pentavalent atoms (e.g., phosphorus).
● p-type semiconductor: Majority charge carriers are holes, and doping is done with trivalent atoms (e.g., boron).

Q19. A convex lens is made of glass with refractive index 1.5. It is immersed in a liquid with refractive index 1.6. How will the behavior of the lens change?

Answer: ○ Since the refractive index of the liquid is higher than that of the lens, the convex lens will behave as a diverging lens instead of a converging one.

Q20. Two point charges placed at a distance r in air exert a force F on each other. At what distance will these charges experience the same force F in a medium of dielectric constant k?

Answer: The force between two point charges q1 and q2 separated by a distance r in air (or vacuum) is given by Coulomb’s law:

When the charges are placed in a medium of dielectric constant kkk, the permittivity of the medium becomes ϵ=kϵ0. The force in the medium is then:

Substituting ϵ=kϵ0:
For the force to remain the same (F′=F), we equate the two expressions:

Q21. P and Q are two identical charged particles each of mass 4 × 10^–26 kg and charge 4.8 × 10^–19 C, each moving with the same speed of 2.4 × 105 m/s as shown in the figure. The two particles are equidistant (0.5 m) from the vertical Y -axis. At some instant, a magnetic field B is switched on so that the two particles undergo head-on collision.

Find – (I) the direction of the magnetic field and
(II) the magnitude of the magnetic field applied in the region.

👉 CBSE Class 12 Study Materials

CBSE Class 12 Syllabus 2024-25	CBSE Class 12 Previous Year Papers
NCERT Books For Class 12 Books	NCERT Class 12 Solutions
CBSE Class 12 Full Study Material	CBSE Class 12 Sample Paper 2024-25

Section C: 3 Marks Questions

Q1. Using Gauss’s law, derive an expression for the electric field intensity at any point outside a uniformly charged thin spherical shell of radius RRR and charge density σ\sigmaσ C/m². Draw the graph of electric field EEE versus distance rrr.

Answer: ○ Consider a point P outside the shell at a distance rrr. The Gaussian surface is a sphere of radius rrr centered at the shell.
○ Applying Gauss’s law: ∮E⋅dA=q_enc /ϵ0
○ Since q_enc=σ4πR² , the electric field outside is: E=σR² /ϵ0 r²
○ The graph of E vs. r shows that E decreases as r increases for the electric field at a point on the equatorial plane of an electric dipole.

Q2. Explain how of a compound microscope changes when:
○ (i) The frequency of the incident light on the objective lens is increased.
○ (ii) The focal length of the objective lens is increased.
○ (iii) The aperture of the objective lens is increased.

Answer: ○ (i) Increases, since resolving power R∝1λR \propto \frac{1}{\lambda}R∝λ1, and higher frequency means lower wavelength.
○ (ii) Decreases, as a longer focal length reduces the numerical aperture (NA), decreasing resolution.
○ (iii) Increases, as a larger aperture allows more light, increasing NA and resolving power.

Q3. State Huyg it to explain the refraction of light.

Answer: ○ Huygens' Principle: Every point on a wavefront acts as a source of secondary wavelets, which spread in all directions.
○ When a wave passes from one medium to another, the wavelets in the second medium travel at a different speed, bending the wavefront and causing refraction.
○ Using Snell’s law: sin i/sin r=v1/v2=n21
○ This explains how light bends towards or away from the normal .

Q4. Derive an expression for the double-slit experiment and explain how it changes if:
○ (i) Light of a smaller frequency is used.
○ (ii) The distance between the slits is decreased.

Answer: ○ Fringe width: β=λD/d
○ (i) Increases, since lower frequency means longer wavelength λ.
○ (ii) Increases, since reducing d increases β.

Q5. A convex lens of focal length 10 cm is placed in airo its focal length if it is immersed in water (Refractive Index of Water = 1.33, Glass = 1.5)?

Answer: ○ Lens Maker’s Formula: 1/f=(n−1)(1/R₁−1/R₂ )
○ In water, effective refractive index n′=1.5/1.33≈1.13
○ Since n′<1.5 f will increase, meaning the lens will become less converging.

Q6. A parallel plate capacitor is connected across a ba happens to:

Section D: Case Study-Based Questions

Case Study 1: Rutherford’s Atomic Model

Ernest Rutherford conducted the famous gold foil experiment in 1909, which led to the discovery of the atomic nucleus. He directed a narrow beam of alpha particles at a thin gold foil and observed their scattering pattern. Most particles passed straight through, some deflected at small angles, and a few bounced back at large angles. This led to the conclusion that most of the atom's mass is concentrated in a tiny, dense, positively charged nucleus, contradicting the Plum Pudding Model of J.J. Thomson.

Answer the following questions:
1. Why did most of the alpha particles pass through the gold foil without any deflection?
2. What conclusion did Rutherford draw from the fact that some alpha particles were deflected at large angles?
3. How did the experiment disprove J.J. Thomson’s atomic model?
4. If an alpha particle moves directly toward the nucleus, what will happen to its trajectory? Justify your answer.

5. How did Rutherford’s experiment lead to the discovery of the nucleus? Explain.

Case Study 2: Transformer Efficiency

A transformer is a device used to step up or step down AC voltages based on the principle of mutual induction. It consists of a primary coil and a secondary coil wound on a soft iron core. The alternating current in the primary coil produces a changing magnetic flux, which induces an alternating EMF in the secondary coil. The efficiency of a transformer is given by:

η=Power output/Power input×100%

A transformer has 2000 turns in the primary coil and 50 turns in the secondary coil. The primary coil is connected to a 120V AC source, and the secondary coil is connected to a bulb of resistance 0.6Ω. The efficiency of the transformer is 80%.

Answer the following questions:
1. Is this a step-up or step-down transformer? Justify your answer.
2. Calculate the voltage across the secondary coil.
3. Determine the current in the secondary coil.
4. Find the power output of the transformer.

5. How does the efficiency of a transformer affect power transmission over long distances?

Section E: Long Answer Questions

Q1. Derivation-Based Question on Electromagnetic Induction

(a) State Faraday’s laws of electromagnetic induction and derive an expression for the induced emf in a coil rotating in a uniform magnetic field.
(b) A coil with 200 turns and area A = 0.02 m² is rotating at 50 Hz in a uniform magnetic field B = 0.5 T. Find the maximum induced emf in the coil.

(b) Explain Lenz’s Law and illustrate it with a suitable example.

Answer: ● Faraday’s First Law: A changing magnetic flux induces an emf in a conductor.

● Lenz’s Law Example: A falling magnet induces a current in a loop that opposes its motion.

Q2. Ray Optics: Lens and Optical Instruments

(a) Derive the lens maker’s formula for a thin convex lens in terms of refractive indices and radii of curvature.
(b) A convex lens has focal length f = 20 cm in air. It is placed in water (n=1.33)(n = 1.33)(n=1.33). Find the new focal length of the lens.

(b) Explain the working of a compound microscope and derive its magnification formula.

Answer:

Q3. Semiconductor Devices and Diode Characteristics

(a) Explain the V-I characteristics of a p-n junction diode in forward and reverse bias with a labeled diagram.
(b) A zener diode is used for voltage regulation. Explain its working and derive an expression for breakdown voltage.

(b) Compare the working of a solar cell and LED based on energy band diagrams.

Answer: ● V-I Characteristics: In forward bias, current increases exponentially after a threshold voltage (~0.7V for silicon). In reverse bias, a small leakage current flows until breakdown voltage is reached.
● Zener Breakdown: At breakdown voltage Vz, the reverse current increases sharply, stabilizing the output voltage. Vout=Vzener
● Solar Cell vs. LED:
○ Solar Cell: Converts light energy to electrical energy (Photovoltaic effect).
○ LED: Converts electrical energy to light energy (Electroluminescence).

👉 CBSE Class 12 Study Materials

CBSE Class 12 Syllabus 2024-25	CBSE Class 12 Previous Year Papers
NCERT Books For Class 12 Books	NCERT Class 12 Solutions
CBSE Class 12 Full Study Material	CBSE Class 12 Sample Paper 2024-25