CBSE Class 12 Sample Paper 2025 Physics Set 1 with Solutions
Mastering the CBSE Class 12 Physics curriculum for the 2025 and 2026 board examinations requires a balance of deep conceptual understanding and structured numerical problem-solving. Practicing with the CBSE Class 12 Sample Paper 2025 Physics Set 1 with Solutions allows students to familiarize themselves with the question blueprint, the distribution of marks across chapters, and the precise step-by-step answering style demanded by CBSE evaluators. This guide breaks down the core physics principles tested in Set 1, provides complete step-by-step solutions to high-frequency board questions, and highlights key exam strategies.
CBSE Class 12 Physics Exam Pattern & Blueprint for 2025
The theory paper for CBSE Class 12 Physics carries a total of 70 marks to be completed in 3 hours. The remaining 30 marks are allocated to laboratory practicals and internal assessments. The paper blueprint follows a five-section structure designed to assess competency-based learning:
- Section A: 16 Multiple Choice Questions (MCQs) and Assertion-Reasoning items (1 mark each = 16 marks)
- Section B: 5 Very Short Answer Questions (2 marks each = 10 marks)
- Section C: 7 Short Answer Questions (3 marks each = 21 marks)
- Section D: 2 Case-Based / Source-Based Integrated Questions (4 marks each = 8 marks)
- Section E: 3 Long Answer Questions with comprehensive internal choices (5 marks each = 15 marks)
Key Concepts Tested in Physics Set 1
To secure a score of 65+ out of 70 in CBSE Physics, students must master foundational theory, vector operations, and microscopic conduction models across the NCERT curriculum.
1. Electrostatics & Electric Field Distributions
Electrostatics explores interactions between stationary electric charges. Key principles in this domain include:
- Coulomb's Law: Describes the electrostatic force between point charges. In vector form, it explicitly reflects Newton's third law of motion: &vec;F12 = -&vec;F21.
- Electric Flux (Φ): A scalar product measuring field penetration through an oriented area element Δ&vec;A.
- Gauss's Theorem: Relates the total enclosed electric flux through a Gaussian surface to the enclosed net charge qenclosed / ε0.
- Equipotential Geometries: Loci of equal potential where no work is done in displacing test charges.
2. Current Electricity & Microscopic Conduction
Electric conduction in solids is driven by free charge carriers responding to an electric potential gradient:
- Drift Velocity (vd): The tiny net average velocity acquired by conduction electrons due to an accelerating electric field against random lattice scattering.
- Ohm's Law Derivation: Linking macroscopic resistance R = V / I to microscopic parameters: number density n, relaxation time τ, cross-sectional area A, and conductor length l.
- Kirchhoff's Laws (KCL & KVL): Universal circuit laws governing conservation of charge and conservation of energy.
3. Wave & Ray Optics: Critical Angle Phenomena
Optics tests geometric ray tracing, refractive index transitions, and wave propagation:
- Total Internal Reflection (TIR): 100% reflection of incident light back into an optically denser medium when the angle of incidence exceeds the critical angle i > ic. This phenomenon forms the operational foundation for modern telecommunication optical fibers and medical endoscopes.
Detailed Concept Derivation: Microscopic Conduction and Ohm's Law
A classic 3-mark or 5-mark derivation in the board exam 12 physics paper connects drift velocity to electrical resistance.
Consider a cylindrical metallic conductor of length l, cross-sectional area A, and free electron density n (number of free electrons per unit volume). When a potential difference V is maintained across its ends, an internal uniform electric field E = V / l is established.
- Acceleration of Electron: Each electron of mass m and charge -e experiences an electrostatic force F = -eE, giving an acceleration:
a = eE / m = eV / (m · l) - Drift Velocity Expression: If τ is the average relaxation time between successive collisions:
vd = a · τ = (e · E · τ) / m = (e · V · τ) / (m · l) - Current-Drift Velocity Relation: In time Δt, charges travel a distance Δx = vd Δt. The volume swept is A · vd Δt, containing ΔN = n · A · vd Δt electrons. The total charge passing through the cross-section is:
Δq = ΔN · e = n · A · e · vd Δt
Thus, electric current is:I = Δq / Δt = n · A · e · vd - Deducing Resistance: Substituting vd = (e V τ) / (m l) into the current equation:
I = n · A · e · [ (e · V · τ) / (m · l) ] = [ (n · A · e² · τ) / (m · l) ] · V
Rearranging in terms of V / I:V / I = (m / (n · e² · τ)) · (l / A) = R
Here,ρ = m / (n · e² · τ)represents the resistivity of the material.
Important CBSE Questions with Answers (Official Set 1 Solutions)
The following questions represent core problems from the official CBSE Question Bank and Sample Paper Set 1. Each answer follows the step-by-step presentation format required by the CBSE evaluation scheme.
Question 1: Define electric flux. Write its SI unit.
Answer:
Electric Flux (Φ): Electric flux through an area is defined as the total number of electric field lines passing normally through that surface. Mathematically, it is the surface integral of the electric field intensity &vec;E over a closed or open surface area &vec;A:
Φ = ∫ &vec;E · d&vec;A = E · A · cos(θ)
where θ is the angle between the electric field vector &vec;E and the outward area normal vector &vec;A.
SI Unit: N·m²/C (Newton square meter per Coulomb) or V·m (Volt meter).
Question 2: State Coulomb's Law. Write its vector form.
Answer:
Statement: Coulomb's Law states that the magnitude of the electrostatic force 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 them, acting along the straight line joining them.
F = (1 / 4πε0) · (|q1 · q2| / r²)
Vector Form: Let &vec;r1 and &vec;r2 be the position vectors of charges q1 and q2. The force &vec;F12 exerted on charge q1 by charge q2 is given by:
&vec;F12 = (1 / 4πε0) · [ (q1 · q2) / |&vec;r12|² ] · r̂21 = (1 / 4πε0) · [ (q1 · q2) / |&vec;r1 - &vec;r2|³ ] · (&vec;r1 - &vec;r2)
where r̂21 is the unit vector pointing from charge q2 to q1.
Question 3: What is an equipotential surface? List its properties.
Answer:
Definition: An equipotential surface is an imaginary or real surface over which the electric potential remains constant at every point (i.e., VA - VB = 0 for any two points A and B on the surface).
Key Properties:
- No Work Done: No work is done in moving a test charge from one point to another on an equipotential surface (
W = q ΔV = q(0) = 0). - Perpendicular Electric Field: The electric field is always perpendicular to the equipotential surface at every point. (If it were not normal, a parallel component of &vec;E would exist, requiring non-zero work).
- Non-Intersecting: 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: Equipotential surfaces are spaced closer together in regions of strong electric fields and farther apart in regions of weak fields (since
E = -dV / dr).
Question 4: Define drift velocity. How is it related to current?
Answer:
Definition: Drift velocity (vd) is the average velocity with which free conduction electrons inside a metallic conductor drift towards the positive terminal of the power source under the influence of an applied external electric field.
Relation with Current:
I = n · A · e · vd
Where:
- I = Electric current flowing through the conductor (in Amperes)
- n = Number density of free electrons (electrons per m³)
- A = Area of cross-section of the conductor (in m²)
- e = Magnitude of electronic charge (
1.6 × 10⁻¹⁹ C) - vd = Drift velocity (in m/s)
Question 5: State Kirchhoff's laws for electrical circuits.
Answer:
- Kirchhoff's First Law (Junction Rule / KCL): The algebraic sum of all electric currents entering and leaving any junction in an electrical circuit is equal to zero.
∑ Iin = ∑ Iout &implies; ∑ I = 0
Physical Basis: KCL is based on the Law of Conservation of Electric Charge. - Kirchhoff's Second Law (Loop Rule / KVL): The algebraic sum of changes in potential (potential differences across resistors and EMFs of sources) around any closed loop in an electrical network is zero.
∑ ΔV = 0 &implies; ∑ Ε = ∑ (I · R)
Physical Basis: KVL is based on the Law of Conservation of Energy.
Question 6: What is total internal reflection? State the conditions.
Answer:
Definition: Total Internal Reflection (TIR) is the optical phenomenon in which a ray of light travelling from an optically denser medium into an optically rarer medium is completely reflected back into the denser medium at the interface without any transmission or refraction.
Necessary Conditions for TIR:
- The light ray must travel from an optically denser medium to an optically rarer medium (e.g., from glass to air or water to air).
- The angle of incidence in the denser medium must be strictly greater than the critical angle (ic) for the given pair of media (
i > ic, wheresin(ic) = μrarer / μdenser).
Question 7: What is the electric field at the surface of a charged spherical conductor of radius R and charge Q?
Answer:
For an isolated spherical conductor of radius R carrying a uniform surface charge Q:
Electric Field at Surface:
E = (1 / 4πε0) · (Q / R²) = k · Q / R²
Alternatively, expressing this in terms of surface charge density σ = Q / (4πR²):
E = σ / ε0
Key Explanation: By Gauss's Law, the electric field at any point on or outside a uniformly charged spherical conductor behaves as though the entire net charge Q were concentrated at the geometric centre of the sphere. Inside the conductor (r < R), the electrostatic field is zero (E = 0).
Question 8: Two point charges +2μC and -2μC are placed 5 cm apart. What is the electric dipole moment?
Answer:
Given:
- Magnitude of charge:
q = 2 μC = 2 × 10⁻⁶ C - Separation distance:
2a = 5 cm = 0.05 m = 5 × 10⁻² m
Formula:
p = q · (2a)
Calculation:
p = (2 × 10⁻⁶ C) × (0.05 m) = 1.0 × 10⁻⁷ C·m
Result: The magnitude of the electric dipole moment is 1.0 × 10⁻⁷ Coulomb-meter (C·m).
Direction: By convention, the dipole moment vector points along the dipole axis from the negative charge (-2 μC) to the positive charge (+2 μC).
How to Prepare for CBSE Class 12 Physics 2025–2026
Achieving top marks in the CBSE Class 12 board examinations requires focused revision techniques tailored to physics theory and numericals:
- Maintain a Derivation Ledger: NCERT derivations account for 15 to 20 marks in the board exam. Create an indexed notebook containing clean derivations for Gauss's law applications, Biot-Savart law, cyclotron/solenoid field, AC resonance, lens maker's formula, and astronomical telescopes.
- Adhere to SI Units and Significant Figures: Always state final numerical answers with appropriate standard units (e.g.,
C·m,V/m,T·m/A). A missing unit results in a mandatory 0.5-mark deduction per numerical step. - Practice Circuit and Ray Diagrams: In optics and current electricity, accurate diagrams carry dedicated marks. Always indicate directional arrows for incident and refracted rays in ray optics, as well as current branch directions in Kirchhoff's loop diagrams.
- Solve Official Sample Papers Under Timed Conditions: Attempt full 70-mark sample papers within a strict 3-hour window to improve time management across 33 questions.
Where to Practice More
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