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100+ Free PSEB SS Physics Practice Questions

Punjab School Education Board (PSEB) Senior Secondary Physics (Class 12 — Physics, subject code 052) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: PSEB SS Physics Exam

70 + 25 + 5

PSEB Senior Secondary Physics Theory + Practical + Project/INA mark split

PSEB Class 12 Scheme of Studies 2026–27

3 hours

Typical PSEB Senior Secondary Physics theory exam duration

Common Class 12 theory logistics / confirm on pseb.ac.in

33%

Separate pass floor in Theory, Practical, INA, and aggregate

PSEB Class 12 Scheme of Studies 2026–27

Code 052

Physics subject code on PSEB Senior Secondary materials

PSEB Scheme of Studies / syllabus

English MCQ adaptation

This free local bank is not the official Senior Secondary paper format

OpenExamPrep practice policy

PSEB Senior Secondary Physics is a Class 12 public-exam subject with Theory 70 + Practical 25 + Project/INA 05 (about 3 hours theory; code 052) and 33% pass separately in theory, practical, INA, and aggregate — not a pure MCQ board paper. This free 2026 bank is an English MCQ study adaptation for NCERT/PSEB-aligned Class 12 Physics concepts and numericals.

Sample PSEB SS Physics Practice Questions

Try these sample questions to test your PSEB SS Physics exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1Two point charges +4 μC and −2 μC are placed 3 m apart in vacuum. The magnitude of the force between them is (take 1/(4πε₀) = 9×10⁹ N m² C⁻²):
A.8×10⁻³ N
B.2.4×10⁻² N
C.8×10⁻² N
D.4×10⁻³ N
Explanation: F = k|q₁q₂|/r² = (9×10⁹)(4×10⁻⁶)(2×10⁻⁶)/(3)² = (9×10⁹)(8×10⁻¹²)/9 = 72×10⁻³/9 = 8×10⁻³ N.
2The electric field due to a uniformly charged infinite non-conducting plane sheet of surface charge density σ is:
A.σ/(2ε₀), independent of distance from the sheet
B.σ/ε₀, falling as 1/r
C.σ/(4πε₀r²)
D.σ/(2ε₀r)
Explanation: By Gauss’s law with a pillbox straddling an infinite non-conducting sheet, E = σ/(2ε₀) on each side and is independent of distance.
3A parallel-plate capacitor has capacitance 5 μF in air. When a dielectric of dielectric constant κ = 3 completely fills the gap, the capacitance becomes:
A.5/3 μF
B.5 μF
C.15 μF
D.45 μF
Explanation: With dielectric filling the gap, C′ = κC₀ = 3 × 5 μF = 15 μF.
4Three capacitors of 2 μF, 3 μF and 6 μF are connected in series. Their equivalent capacitance is:
A.1 μF
B.11 μF
C.0.5 μF
D.2 μF
Explanation: 1/C = 1/2 + 1/3 + 1/6 = 3/6 + 2/6 + 1/6 = 6/6 = 1 ⇒ C = 1 μF.
5Work done by an external agent in slowly moving a +2 μC charge from a point at potential 10 V to a point at potential 30 V (no KE change) is:
A.40 μJ
B.20 μJ
C.−40 μJ
D.60 μJ
Explanation: W_ext = qΔV = (2×10⁻⁶)(30−10) = 4×10⁻⁵ J = 40 μJ.
6The electric field on the axial line of a short electric dipole of moment p at distance r (r ≫ dipole length) is:
A.(1/(4πε₀))·(2p/r³) along the dipole axis
B.(1/(4πε₀))·(p/r³) opposite to p
C.(1/(4πε₀))·(p/r²)
D.(1/(4πε₀))·(2p/r²)
Explanation: For a short dipole, axial field E_axial = (1/(4πε₀))(2p/r³) in the direction of p; equatorial field is half that magnitude and opposite to p.
7A charge of 8 μC is placed at the centre of a cube of side 10 cm. The electric flux through one face of the cube is:
A.q/(6ε₀) = (8×10⁻⁶)/(6ε₀)
B.q/ε₀
C.q/(24ε₀)
D.zero
Explanation: Total flux through a closed surface = q_enc/ε₀. By symmetry the cube has 6 identical faces, so flux through one face = (q/ε₀)/6 = q/(6ε₀).
8Energy stored in a 4 μF capacitor charged to 100 V is:
A.0.02 J
B.0.04 J
C.0.2 J
D.2 J
Explanation: U = (1/2)CV² = (1/2)(4×10⁻⁶)(100)² = (2×10⁻⁶)(10⁴) = 0.02 J.
9Two identical capacitors each of capacitance C are connected in series and charged by a battery of voltage V, then disconnected from the battery and reconnected in parallel with plates of like polarity joined. The final energy stored is:
A.CV²/4
B.CV²/2
C.CV²
D.CV²/8
Explanation: Series equivalent C/2 takes charge Q_stack = (C/2)V from the battery; each capacitor holds charge (C/2)V. After isolation and parallel reconnection of like plates, total charge on the combined positive side is CV and C_eq = 2C, so V_f = V/2. Final energy = ½(2C)(V/2)² = CV²/4.
10If the distance between two point charges is doubled and each charge is doubled, the electrostatic force becomes:
A.unchanged
B.doubled
C.halved
D.four times
Explanation: F′/F = (k(2q₁)(2q₂)/(2r)²)/(kq₁q₂/r²) = 4/4 = 1. Force is unchanged.

About the PSEB SS Physics Practice Questions

Verified exam format metadata for Punjab School Education Board (PSEB) Senior Secondary Physics (Class 12 — Physics, subject code 052) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.