CBSE Class 12 Physics Sample Paper 2025 Standard Set 1 with Solutions

Practicing with a verified CBSE Class 12 Physics Sample Paper is the single most effective strategy to master the latest board examination blueprint, refine problem-solving speed, and eliminate conceptual blind spots. The Central Board of Secondary Education (CBSE) designs Class 12 Physics to evaluate conceptual depth, analytical reasoning, and mathematical rigor across electrostatics, current electricity, magnetism, optics, and modern physics. This comprehensive CBSE sample paper standard set provides an authentic board-style model paper with step-by-step solutions, detailed marking schemes, and expert difficulty analysis to help you achieve a 95+ score in your CBSE Class 12 Physics board examination.

CBSE Class 12 Physics Exam Pattern & Paper Structure

The theory paper for CBSE Class 12 Physics carries 70 marks with a total duration of 3 hours (180 minutes). The remaining 30 marks are allocated to internal assessments and practical examinations. According to the latest board guidelines, the question paper comprises 33 compulsory questions categorized into five distinct sections, featuring a balanced distribution of objective, conceptual, numerical, and case-based evaluation.

Section	Question Type	Number of Questions	Marks Per Question	Total Marks
Section A	Multiple Choice Questions (MCQs) & Assertion-Reason	16 (Q1 to Q16)	1 Mark	16 Marks
Section B	Very Short Answer (VSA) Questions	5 (Q17 to Q21)	2 Marks	10 Marks
Section C	Short Answer (SA) Questions	7 (Q22 to Q28)	3 Marks	21 Marks
Section D	Case-Based / Source-Integrated Questions	2 (Q29 to Q30)	4 Marks	8 Marks
Section E	Long Answer (LA) Questions	3 (Q31 to Q33)	5 Marks	15 Marks
Total		33 Questions	—	70 Marks

There is no overall choice in the paper. However, internal choices are provided in two questions of Section B, two questions of Section C, one question in each Case-Based problem of Section D, and all three questions of Section E.

Complete CBSE Physics Sample Paper with Solutions

Below is the complete CBSE Physics sample paper with solutions, curated in exact alignment with NCERT curriculum guidelines and CBSE official question banks. Use this standard set under simulated exam conditions before reviewing the detailed solutions.

Section A: Objective Type Questions (1 Mark Each)

Question 1

Q: Define electric flux. Write its SI unit.

Solution:

Electric Flux (Φ): Electric flux through a given surface placed inside an electric field is defined as the total number of electric field lines crossing normally through that surface. Mathematically, it is the surface integral of the electric field vector over a closed or open surface area:

Φ = ∫ E̅ ⋅ dA̅ = E A cos θ

where E is the magnitude of the electric field, A is the surface area, and θ is the angle between the electric field vector E̅ and the outward area normal vector A̅.

SI Unit: The SI unit of electric flux is N⋅m²/C (Newton meter squared per Coulomb) or V⋅m (Volt meter).

Question 2

Q: What is the electric field at the surface of a charged spherical conductor of radius R and total charge Q?

Solution:

For a charged conducting sphere of radius R carrying a total charge Q, all electrostatic charge resides entirely on the outer surface. Applying Gauss's Law just outside the surface gives:

E = (1 / 4πε₀) × (Q / R²) = kQ / R²

Alternatively, in terms of surface charge density σ = Q / (4πR²), the electric field is expressed as:

E = σ / ε₀

Key Insight: The electric field outside and at the surface of a spherical conductor behaves as if the entire charge Q were concentrated as a point charge at its center. Inside the conducting sphere (r < R), the electrostatic field is strictly zero (E = 0).

Question 3

Q: State Coulomb's Law. Write its vector form.

Solution:

Coulomb's Law: The magnitude of the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the scalar product of the magnitudes of the charges and inversely proportional to the square of the distance separating them. The force acts along the straight line joining the two charges.

F = (1 / 4πε₀) × (|q₁ q₂| / r²)

Vector Form: Let charge q₁ be located at position vector r̅₁ and charge q₂ at r̅₂. The force exerted on charge q₁ by charge q₂ is:

F̅₁₂ = (1 / 4πε₀) × [ (q₁ q₂) / |r̅₁ − r̅₂|³ ] (r̅₁ − r̅₂)

In unit vector notation where r̂₂₁ is the unit vector pointing from q₂ toward q₁:

F̅₁₂ = (1 / 4πε₀) × [ (q₁ q₂) / r² ] r̂₂₁

Note that F̅₁₂ = −F̅₂₁, which proves that electrostatic forces obey Newton's Third Law of Motion.

Question 4

Q: Define drift velocity. How is it related to electric current?

Solution:

Drift Velocity (v_d): Drift velocity is defined as the average velocity with which conduction electrons move through a conductor opposite to the direction of the applied external electric field.

v_d = − (e E τ) / m

where e is the elementary electron charge, E is the applied electric field intensity, τ is the average relaxation time between successive electron collisions, and m is the mass of an electron.

Relation with Electric Current: For a conductor of cross-sectional area A with free electron number density n:

I = n A e v_d

Section B: Very Short Answer Questions (2 Marks Each)

Question 5

Q: Two point charges +2 μC and −2 μC are placed 5 cm apart in air. Calculate the magnitude of the electric dipole moment and specify its direction.

Solution:

Given Parameters:

• Magnitude of each charge: q = 2 μC = 2 × 10⁻⁶ C
• Separation distance (dipole length): 2a = 5 cm = 0.05 m = 5 × 10⁻² m

Formula & Calculation:

p = q × (2a)

p = (2 × 10⁻⁶ C) × (5 × 10⁻² m) = 1.0 × 10⁻⁷ C⋅m

Direction: By scientific convention, the electric dipole moment vector p̅ is directed along the dipole axis from the negative charge (−2 μC) to the positive charge (+2 μC).

Final Answer: Magnitude = 1.0 × 10⁻⁷ C⋅m; Direction = From −2 μC to +2 μC.

Question 6

Q: What is Total Internal Reflection (TIR)? State the two essential conditions required for it to occur.

Solution:

Total Internal Reflection: Total Internal Reflection is the optical phenomenon wherein a ray of light propagating in an optically denser medium and striking the interface of an optically rarer medium at an angle of incidence greater than the critical angle is reflected entirely back into the denser medium without any refraction.

Essential Conditions for TIR:

1. The light ray must travel from an optically denser medium to an optically rarer medium (e.g., from glass/water into air).
2. The angle of incidence (i) inside the denser medium must be strictly greater than the critical angle (i_c) for the given pair of media, where sin i_c = μ_rarer / μ_denser = 1 / μ.

Question 7

Q: What is an equipotential surface? List its two characteristic properties.

Solution:

Equipotential Surface: An equipotential surface is any surface over which the electric potential (V) has the same constant value at every point (V_A = V_B = \text{constant}).

Characteristic Properties:

1. Zero Work Done: No work is done by the electrostatic field in moving a test charge q₀ between any two points lying on an equipotential surface, since W = q₀(V_B − V_A) = 0.
2. Perpendicular Electric Field Lines: Electric field lines are always directed perpendicular (normal) to the equipotential surface at every point. If there were a tangential component, work would be required to move a charge along the surface, violating the definition of an equipotential surface.

Section C: Short Answer Questions (3 Marks Each)

Question 8

Q: State Kirchhoff's Laws for electrical networks. Explicitly name the fundamental conservation law on which each rule is based.

Solution:

1. Kirchhoff's First Law — Current Law (KCL) / Junction Rule:

The algebraic sum of all electric currents entering and leaving any node or junction in an electrical circuit is equal to zero:

∑ I = 0   ⇒   ∑ I_entering = ∑ I_leaving

Conservation Principle: KCL is directly based on the Law of Conservation of Electric Charge. In a steady-state circuit, electric charge cannot accumulate or deplete at a junction.

2. Kirchhoff's Second Law — Voltage Law (KVL) / Loop Rule:

In any closed mesh or loop of an electrical network, the algebraic sum of the changes in electric potential (including electromotive forces and potential drops across resistances) is equal to zero:

∑ ΔV = 0   ⇒   ∑ Ε = ∑ I R

Conservation Principle: KVL is based on the Law of Conservation of Energy. Because the electrostatic force is conservative, the net work done in carrying a unit test charge completely around any closed closed loop must be zero.

Detailed Concept Breakdown & High-Yield Derivations

To score full marks on 3-mark and 5-mark theoretical questions in your CBSE sample paper 2025 Physics exam, structured derivations with explicit statements, vector definitions, and boundary conditions are required. Below is the full mathematical derivation linking macroscopic current to microscopic electron transport.

Deduction of Ohm's Law from Electron Drift Velocity

Consider a cylindrical conductor of length l, uniform cross-sectional area A, and free electron density n subjected to a potential difference V across its ends.

1. Applied Electric Field: The uniform electric field set up inside the conductor is:

   E = V / l

2. Acceleration of Free Electrons: Under the influence of E, each electron experiences an electrostatic force F = −eE. The resulting acceleration is:

   a = − (e E) / m = − (e V) / (m l)

3. Average Drift Velocity: In terms of average relaxation time τ:

   v_d = |a| τ = (e V τ) / (m l)

4. Total Charge Traversing Conductor: In time interval Δt, all electrons within a distance v_d Δt pass through area A. The volume element is A v_d Δt, and the total charge transferred is:

   Δq = n × (A v_d Δt) × e

5. Current Expression:

   I = Δq / Δt = n A e v_d

6. Substituting v_d into the Current Equation:

   I = n A e [ (e V τ) / (m l) ] = [ (n A e² τ) / (m l) ] V

   V / I = (m / (n e² τ)) × (l / A)

7. Verification of Ohm's Law: At constant physical conditions (constant temperature and pressure), parameters m, n, e, and τ remain constant for a given metallic material. Defining resistivity ρ = m / (n e² τ) and resistance R = ρ l / A:

   V = I R

This rigorous mathematical proof demonstrates that electrical resistance depends directly on material parameters (electron density n and relaxation time τ) and conductor geometry (length l and area A).

Section-Wise Difficulty Analysis & Chapter Weightage

A strategic breakdown of this CBSE mock paper reveals how marks are distributed across core units, highlighting high-yield focus areas for your board revision.

Unit / Chapter	Theoretical Weightage	Numerical & Case Study Focus	Difficulty Level
Unit 1: Electrostatics (Electric Charges, Fields, Potential & Capacitance)	16 Marks (Combined with Current Electricity)	Electric dipole moments, flux calculations, spherical conductors, capacitance combinations	Moderate to High
Unit 2: Current Electricity	Covered above	Drift velocity derivations, Kirchhoff's loop rules, Wheatstone bridge, internal resistance	Moderate
Unit 3: Magnetic Effects of Current & Magnetism	17 Marks (Combined with EMI & AC)	Biot-Savart Law, Ampere's circuital law, moving coil galvanometer sensitivity	Moderate
Unit 4: Electromagnetic Induction & Alternating Currents	Covered above	LCR series resonance, quality factor, phasor diagrams, transformer power losses	High
Unit 5 & 6: Optics (Ray & Wave Optics)	18 Marks	Total internal reflection, lens maker formula, prism dispersion, Young's double slit derivation	High
Unit 7 & 8: Modern Physics (Dual Nature, Atoms, Nuclei)	12 Marks	Photoelectric effect graphs, de Broglie wavelength, Bohr atomic energy levels, mass defect	Easy to Moderate
Unit 9: Electronic Devices (Semiconductor Physics)	7 Marks	Energy band diagrams, p-n junction diode forward/reverse V-I characteristics, full wave rectifier	Easy to Moderate

Cognitive Skill Distribution in Class 12 Physics

• Remembering & Understanding (approx. 38% — 27 Marks): Direct definitions (electric flux, equipotential surfaces, TIR), principle statements, and fundamental vector notations.
• Application & Problem Solving (approx. 32% — 22 Marks): Multi-step numerical computations (dipole moment, circuit networks, optical prism angles).
• Analyzing, Evaluating & Creating (approx. 30% — 21 Marks): Case-based analysis, experimental data evaluation, graphical deduction, and derivations from first principles.

How to Practice with This CBSE Practice Paper

To translate practice into top board exam percentiles, implement this structured 3-phase revision framework:

1. Strict 3-Hour Time Management Strategy

Divide your 180-minute examination window systematically to avoid rushing during lengthy 5-mark derivations:

• Reading Time (First 15 minutes): Scan through all sections. Select your internal choice questions immediately in Sections B, C, and E. Mark questions with numerical computations to calculate rough values methodically.
• Section A (16 Marks — 25 Minutes): Spend approximately 1.5 minutes per MCQ. Write the option letter along with the complete answer statement for clarity.
• Section B & C (31 Marks — 65 Minutes): Allocate 6 minutes for each 2-mark question and 9 minutes for each 3-mark question. Ensure all standard circuit diagrams, ray diagrams, and vector symbols are drawn neatly.
• Section D & E (23 Marks — 65 Minutes): Dedicate 15 minutes to each 4-mark case study and 18 minutes to each 5-mark long answer question. Break answers into labeled sub-parts (Principle, Diagram, Working, Mathematical Proof).
• Revision & Unit Checking (Final 10 Minutes): Verify SI units for every final numerical value (e.g., C⋅m, N⋅m²/C, Ω, V) and ensure vector arrows (E̅, p̅) are clearly indicated.

2. Self-Assessment & Marking Scheme Evaluation

Grade your practice paper strictly against CBSE step-marking criteria:

• Award ½ mark for writing the correct formula before substituting numerical data.
• Award 1 mark for correct intermediate algebraic simplification with SI units maintained throughout.
• Deduct ½ mark if the final numerical answer is missing proper SI units or lacks the specified vector direction.
• Ensure all optical ray diagrams include arrowheads showing the exact direction of light propagation; diagrams without arrowheads receive zero marks in board evaluations.

3. Pinpoint and Rectify Concept Deficits

Maintain an error log categorizing mistakes into three buckets:

1. Calculation & Unit Conversion Errors: Forgetting to convert centimeters to meters or microcoulombs to coulombs (e.g., 5 cm = 0.05 m; 2 μC = 2 × 10⁻⁶ C).
2. Formula Misapplication: Confusing electric potential (scalar) with electric field (vector), or missing the factor of 1/μ in critical angle equations.
3. Incomplete Step Justification: Omitting the underlying conservation principle when applying Kirchhoff's Junction Rule (Charge conservation) or Loop Rule (Energy conservation).

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