CBSE Class 12 Physics Question Paper 2025 with Solutions

Practicing the official CBSE Class 12 Physics question paper is the single most effective strategy for students aiming to score 95+ in their board examinations. The 2025 Physics examination tested conceptual clarity, mathematical rigor, and real-world analytical skills across the revised NCERT curriculum. Whether you are analyzing the latest CBSE Class 12 Physics question paper 2025 with solutions, verifying step-marking criteria, or looking for high-yield previous year question paper CBSE resources, this comprehensive guide delivers complete question-by-question solutions, paper pattern breakdowns, difficulty analyses, and examiner marking schemes.

CBSE Class 12 Physics 2025 Question Paper Pattern & Structure

The CBSE Class 12 Physics (Theory) paper for 2025 was conducted for a total of 70 marks over a duration of 3 hours (180 minutes), with 30 marks allocated to practical examinations and internal assessments. To clear the Physics paper, students must secure a minimum of 33% marks independently in theory (23 out of 70) as well as in practicals (10 out of 30).

The question paper contains 33 compulsory questions divided into 5 distinct sections, designed in accordance with the National Education Policy (NEP) guidelines emphasizing Competency-Based Education (CBE):

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

Complete Questions with Detailed Step-by-Step Solutions

Below are essential questions from the 2025 CBSE Class 12 Physics board paper and official question bank, accompanied by detailed, step-by-step solutions and marking scheme insights.

Question 1 (Electrostatics & Electric Flux)

Question: Define electric flux. Write its SI unit.

Detailed Solution:

1. Definition: Electric flux (Φ) through a given surface placed inside an electric field is defined as the total number of electric field lines passing normally (perpendicularly) through that surface. Mathematically, it is the surface integral of the electric field vector E⃗ over the closed area vector A⃗:
   Φ = ∮ E⃗ · dA⃗ = E A cos θ where θ is the angle between the electric field vector E⃗ and the outward normal to the area vector A⃗.
2. SI Unit: The SI unit of electric field is N/C (or V/m) and area is m². Therefore, the SI unit of electric flux is N·m²/C (Newton meter squared per Coulomb) or V·m (Volt meter).

Marking Allocation: 1 Mark for accurate definition + 1 Mark for SI unit (Total: 2 Marks).

Question 2 (Coulomb's Law in Vector Form)

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

Detailed Solution:

1. Statement: Coulomb's Law states that the electrostatic force of attraction or repulsion between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. The force acts along the line joining the centers of the two charges:
   F = (1 / 4πε0) × (|q1 q2| / r2) where ε₀ is the permittivity of free space (ε₀ = 8.854 × 10^-12 C²·N^-1·m^-2).
2. Vector Form: Let r̂₂₁ be the unit vector pointing from charge q₁ to charge q₂, and r⃗₂₁ = r⃗₂ - r⃗₁. The electrostatic force F⃗₂₁ exerted on charge q₂ by charge q₁ is:
   F⃗21 = (1 / 4πε0) × (q1 q2 / |r⃗21|2) r̂21 = (1 / 4πε0) × (q1 q2 / |r⃗21|3) r⃗21 Similarly, F⃗₁₂ = -F⃗₂₁, which validates Newton's Third Law of Motion.

Marking Allocation: 1 Mark for correct statement and scalar equation + 1 Mark for vector representation with unit vector notation.

Question 3 (Equipotential Surfaces & Properties)

Question: What is an equipotential surface? List its properties.

Detailed Solution:

1. Definition: An equipotential surface is any surface over which the electric potential (V) remains constant at every single point (i.e., V_A = V_B for any two points A and B on the surface).
2. Fundamental Properties:
   • Zero Work Done: The work done (W) in moving an electric charge q between any two points on an equipotential surface is zero, since W = q(V_B - V_A) = q(0) = 0.
   • Perpendicular Electric Field: The electric field lines are always directed normal (perpendicular) to the equipotential surface at every point (since dV = -E⃗ · dr⃗ = 0 ⇒ E cos 90° = 0).
   • No Intersection: Two equipotential surfaces can never intersect each other. If they did, there would be two different values of electric potential at the point of intersection, which is physically impossible.
   • Field Strength Indication: The spacing between equipotential surfaces indicates the strength of the electric field; surfaces are closer together in regions of strong fields (since E = -dV/dr) and farther apart in weak fields.

Marking Allocation: 1 Mark for definition + 1 Mark for any two valid properties.

Question 4 (Drift Velocity & Electric Current Relation)

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

Detailed Solution:

1. Definition of Drift Velocity: Drift velocity (v_d) is defined as the average velocity with which free electrons in a conductor get drifted towards the positive terminal of the applied external electric field. It is given by:
   vd = - (e E τ / m) where e is electron charge, E is electric field strength, τ is average relaxation time, and m is the mass of an electron.
2. Derivation of Relation with Electric Current:
   • Consider a cylindrical conductor of length l and uniform cross-sectional area A.
   • Let n be the number density of free electrons (number of electrons per unit volume).
   • Total number of free electrons in the conductor = n × Volume = n × A × l.
   • Total charge contained in the conductor: Q = n × A × l × e.
   • Time taken by electrons to traverse the length l with drift velocity v_d: t = l / v_d.
   • Electric current I = Q / t = (n A l e) / (l / v_d).
   
   I = n A e vd This establishes the fundamental microscopic relationship between macroscopically measured electric current (I) and microscopic electron drift velocity (v_d). Current density is expressed as J = I / A = n e v_d.

Marking Allocation: 1 Mark for definition + 2 Marks for step-by-step derivation (Total: 3 Marks).

Question 5 (Kirchhoff's Circuit Laws)

Question: State Kirchhoff's laws for electrical circuits.

Detailed Solution:

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

   In any electrical network, the algebraic sum of electric currents meeting at any junction is always zero:

   Σ I = 0 ⇒ Σ Ientering = Σ Ileaving

   Physical Principle: KCL is a direct consequence of the Law of Conservation of Electric Charge (charges cannot accumulate at a junction).

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

   Around any closed mesh or loop in an electrical network, the algebraic sum of changes in potential (potential differences and EMFs) is equal to zero:

   Σ ΔV = 0 ⇒ Σ E = Σ (I × R)

   Physical Principle: KVL is a direct consequence of the Law of Conservation of Energy (the net work done in moving a charge around a closed conservative path is zero).

Marking Allocation: 1 Mark for Junction Rule with physical conservation law + 1 Mark for Loop Rule with physical conservation law.

Question 6 (Total Internal Reflection & Critical Angle)

Question: What is total internal reflection? State the conditions.

Detailed Solution:

1. Definition: Total Internal Reflection (TIR) is the phenomenon in which a ray of light traveling from an optically denser medium to an optically rarer medium is completely reflected back into the denser medium at the interface without any refraction, when the angle of incidence exceeds the critical angle for the given pair of media.
2. Necessary Conditions for Total Internal Reflection:
   • Direction of Ray: Light must propagate from an optically denser medium towards an optically rarer medium (e.g., glass to air, water to air).
   • Critical Angle Threshold: The angle of incidence in the denser medium (i) must be strictly greater than the critical angle (i_c) for the pair of media: i > i_c.
3. Mathematical Relation: By Snell's Law, when i = i_c, angle of refraction r = 90°.
   μ1 sin ic = μ2 sin 90° ⇒ sin ic = μ2 / μ1 = 1 / μ where μ is the refractive index of the denser medium with respect to the rarer medium.
4. Real-World Application: TIR forms the operating foundation of high-speed fiber-optic telecommunication cables, endoscopes in biomedical engineering, and the brilliant sparkle of cut diamonds.

Marking Allocation: 1 Mark for definition + 1 Mark for stating both essential conditions.

Question 7 (Electric Field of a Charged Spherical Conductor)

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

Detailed Solution:

1. For a conducting sphere carrying a total positive charge Q distributed uniformly over its surface of radius R:
   • Inside the conducting sphere (r < R): E_in = 0 (since net enclosed charge is zero).
   • On the surface of the sphere (r = R): Applying Gauss's Law with a Gaussian surface just enclosing the sphere:
     ∮ E⃗ · dA⃗ = E × (4 π R2) = Q / ε0
Esurface = (1 / 4πε0) × (Q / R2) = σ / ε0 where σ = Q / (4πR²) is the surface charge density.
   • Outside the sphere (r > R): E_out = (1 / 4πε₀) × (Q / r²).
2. Conclusion: The electric field at the surface of a charged spherical conductor behaves as if the entire charge Q were concentrated as a point charge at the geometric center of the sphere.

Marking Allocation: 1 Mark for formula derivation + 1 Mark for surface field expression in terms of charge or surface charge density.

Question 8 (Numerical: Electric Dipole Moment)

Question: Two point charges +2 μC and -2 μC are placed 5 cm apart. What is the electric dipole moment?

Detailed Solution:

1. Given Parameters:
   • Magnitude of each charge, |q| = 2 μC = 2 × 10^-6 C
   • Separation distance between charges (2a) = 5 cm = 0.05 m = 5 × 10^-2 m
2. Formula: Electric dipole moment vector p⃗ has magnitude:
   p = |q| × (2a)
3. Calculation:
   p = (2 × 10-6 C) × (5 × 10-2 m) = 10 × 10-8 C·m = 1.0 × 10-7 C·m
4. Direction: By scientific convention, the dipole moment vector p⃗ is directed along the dipole axis from the negative charge (-2 μC) to the positive charge (+2 μC).

Final Answer: The electric dipole moment is 1.0 × 10^-7 C·m directed from -2 μC to +2 μC.

Marking Allocation: 0.5 Mark for formula + 1 Mark for correct calculation + 0.5 Mark for unit and vector direction (Total: 2 Marks).

Detailed Concept Breakdown: Microscopic Conduction & Drift Velocity Mechanism

To master high-scoring long derivations in the Class 12 Physics paper, understanding the physical mechanism of electrical conduction inside metal lattices is critical.

In the absence of an applied electric field, free conduction electrons move randomly in all possible directions inside the metallic crystal lattice due to thermal energy at room temperature (thermal velocities reach ~10⁵ to 10⁶ m/s). Because this thermal motion is completely isotropic, the vector sum of thermal velocities is zero:

u⃗avg = (1 / N) ∑ u⃗i = 0

When an external potential difference V is applied across a conductor of length l, a uniform electric field E⃗ = -V / l is established. Each electron experiences an electrostatic force F⃗ = -eE⃗, imparting an acceleration:

a⃗ = F⃗ / m = - (e E⃗ / m)

As electrons accelerate, they collide frequently with heavy positive lattice ions. Between consecutive collisions, an electron gains velocity for a mean duration called the relaxation time (τ). The final velocity of the i-th electron just before collision is v⃗_i = u⃗_i + a⃗τ_i. Averaging over all N electrons:

v⃗d = (1 / N) ∑ (u⃗i + a⃗ τi) = 0 + a⃗ τ = - (e E⃗ τ / m)

Combining this with the current relation I = n A e v_d and replacing E = V / l:

I = n A e (e V τ / m l) = (n A e2 τ / m l) V ⇒ V / I = (m / n e2 τ) × (l / A) = R

This microscopic derivation rigorously validates Ohm's Law, defining electrical resistivity as ρ = m / (n e² τ).

Section-Wise Difficulty Level & Paper Analysis

Based on comprehensive feedback from senior CBSE evaluators and students who appeared for the 2025 board examination, the Physics paper was rated Moderate to Difficult with an emphasis on numerical applications and analytical thinking.

Section	Typology	Difficulty Level	Key Observations & Exam Trends
Section A	16 MCQs & Assertion-Reason	Moderate	Assertion-reason questions required deep NCERT textbook reading; 3 MCQs included direct formula-based numerical calculations.
Section B	5 VSA (2 Marks each)	Easy to Moderate	Standard theoretical questions from Electrostatics, Wave Optics, and Modern Physics. Direct scoring section.
Section C	7 SA (3 Marks each)	Moderate to Difficult	Lengthy derivations coupled with multi-step numericals (AC circuits, Ray Optics lens combinations, and Moving Charges).
Section D	2 Case Studies (4 Marks each)	Moderate	Passages were based on Semiconductor Devices (p-n junction diodes) and Electromagnetic Induction (transformers/eddy currents).
Section E	3 LA (5 Marks each)	Moderate	Internal choices were balanced. Questions tested standard derivations (Wavefront Huygens principle, Galvanometer to Ammeter conversion, Gauss's law).

Unit-Wise Marks Distribution

• Unit 1 & 2 (Electrostatics & Current Electricity): 16 Marks (~23% weightage)
• Unit 3 & 4 (Magnetic Effects & EMI / AC): 17 Marks (~24% weightage)
• Unit 5 & 6 (Electromagnetic Waves & Optics): 18 Marks (~26% weightage)
• Unit 7 & 8 (Dual Nature, Atoms & Nuclei): 12 Marks (~17% weightage)
• Unit 9 (Electronic Devices - Semiconductors): 7 Marks (~10% weightage)

CBSE Class 12 Physics Official Marking Scheme & Step-Marking Guidelines

CBSE awards marks strictly through step-marking criteria. Evaluators look for specific structural elements in each answer:

• Formula & Given Data (0.5 to 1 Mark): Explicitly write the relevant formula and identify given values with appropriate SI units before substituting numbers.
• Circuit & Ray Diagrams (0.5 to 1 Mark): Diagrams in Optics (with proper directional arrows) and Circuit Laws (with current loop labels) carry independent marks. A derivation with an incorrect or missing diagram receives zero marks for the diagram component.
• Intermediate Calculation Steps (1 to 2 Marks): Showing algebraic simplification and correct power-of-ten exponents prevents complete loss of marks if a minor arithmetic error occurs at the end.
• Final Answer with SI Units (0.5 Mark): Writing the final calculated value without SI units or with incorrect units results in an automatic deduction of 0.5 marks per numerical.
• Vector Directions & Sign Conventions: Always state the directional vector (e.g., dipole moment from -q to +q, magnetic force direction using Fleming's Left-Hand Rule).

How to Use This Question Paper for Board Exam Preparation

1. Simulate Authentic Exam Conditions: Print out the question paper, set a strict 3-hour countdown timer, and solve it in an isolated environment without reference books or internet access.
2. Perform Self-Assessment with Step-Marking: Compare your handwritten answer sheet against the official step-by-step solutions provided above. Calculate your realistic score based on step-wise deductions.
3. Target Weak Numerical & Derivation Areas: Maintain an error logbook. If you lost marks in Ray Optics lens formulas or Kirchhoff's loop rules, re-derive those specific NCERT concepts immediately.
4. Master Time Management: Allocate 30 minutes for Section A, 25 minutes for Section B, 50 minutes for Section C, 30 minutes for Section D, 35 minutes for Section E, and preserve the remaining 10 minutes for final revision and diagram labeling.

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