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100+ Free Karnataka II PUC Physics (KSEAB) Practice Questions

Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Physics (Code 33) practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Karnataka II PUC Physics (KSEAB) Exam

~70 + 30

Common theory + practical mark split (total 100)

KSEAB II PUC science practical subject pattern

~3 h 15 m

Typical theory duration (often includes reading time)

KSEAB II PUC Physics model paper logistics

Code 33

Official subject code for II PUC Physics

KSEAB / DPUE Physics blueprint

~35%

Commonly reported overall pass threshold (confirm circular)

KSEAB II PUC pass criteria reporting

Mixed + practical

Official format is mixed written theory plus practical, not pure MCQ

KSEAB II PUC Physics assessment pattern

English MCQ adaptation

This free local bank is not the official mixed paper format

OpenExamPrep practice policy

KSEAB II PUC Physics (code 33) is a Class 12 board subject with ~70 theory + ~30 practical and theory lasting about 3 h 15 m. Pass is commonly ~35% overall (~30–35% careful wording) as notified. Fees are paid through colleges. This free 2026 bank is an English MCQ study adaptation — not an official paper simulation.

Sample Karnataka II PUC Physics (KSEAB) Practice Questions

Try these sample questions to test your Karnataka II PUC Physics (KSEAB) exam readiness. Each question includes a detailed explanation. Start the interactive quiz above for the full 100+ question experience with AI tutoring.

1The electrostatic force between two point charges separated by distance r in vacuum is F. If the distance is doubled and each charge is doubled, the new force is:
A.2F
B.4F
C.F/4
D.F
Explanation: F' = k(2q)(2q)/(2r)² = k(4q²)/(4r²) = kq²/r² = F. The factors cancel, so the force remains F.
2Two charges +4 μC and +9 μC are placed 30 cm apart. The electric field is zero at a point:
A.18 cm from +4 μC toward +9 μC
B.outside the segment, beyond +4 μC
C.10 cm from +4 μC toward +9 μC
D.12 cm from +4 μC toward +9 μC
Explanation: Between like charges, E = 0 divides the segment in the ratio of √|q|. Let x be from +4 μC: k(4×10⁻⁶)/x² = k(9×10⁻⁶)/(0.3−x)² ⇒ 2/x = 3/(0.3−x) ⇒ 2(0.3−x)=3x ⇒ 0.6=5x ⇒ x=0.12 m = 12 cm from +4 μC.
3An electric dipole of moment p is placed in a uniform electric field E with p parallel to E. The torque on the dipole is:
A.pE/2
B.2pE
C.pE
D.zero
Explanation: Torque τ = pE sinθ. When p is parallel to E, θ = 0°, so sinθ = 0 and τ = 0. (Net force is also zero in a uniform field.)
4According to Gauss’s law, the electric flux through a closed surface is:
A.always zero if the surface is spherical
B.independent of the enclosed charge
C.proportional to the average electric field on the surface
D.equal to q_enclosed/ε₀ in SI units
Explanation: Gauss’s law: ∮ E·dA = q_enclosed/ε₀ (SI). Flux depends only on the total charge enclosed, not on surface shape.
5A charge Q is uniformly distributed on a thin spherical shell of radius R. The electric field at distance r from the centre for r < R is:
A.kQ/(R²r)
B.kQ/r²
C.kQ/R²
D.zero
Explanation: For a point inside a uniformly charged thin shell, a Gaussian surface of radius r < R encloses zero charge, so E = 0 (Gauss’s law).
6The electric field due to an infinite plane sheet of charge density σ (SI) has magnitude:
A.σ/(4πε₀)
B.2σ/ε₀
C.σ/ε₀
D.σ/(2ε₀)
Explanation: Using a Gaussian pillbox through an infinite non-conducting sheet (or standard textbook result for a single infinite sheet), E = σ/(2ε₀), independent of distance.
7Two charges +q and +4q are separated by distance d. A third charge Q is to be placed on the line joining them so that the net force on it is zero. Q must be placed:
A.between the charges, closer to +4q
B.outside, beyond +q
C.outside, beyond +4q
D.between the charges, closer to +q
Explanation: For net force zero on Q, the forces from +q and +4q must be equal and opposite, so Q must lie between them (repulsion from both sides). Distance from +q : distance from +4q = √1 : √4 = 1 : 2, so closer to the smaller charge +q.
8An electric dipole of moment 2×10⁻⁶ C·m is placed in a uniform field of 10⁵ N/C with its axis at 60° to the field. The magnitude of torque is:
A.0.2 N·m
B.0.05 N·m
C.0.1 N·m
D.0.173 N·m
Explanation: τ = pE sinθ = (2×10⁻⁶)(1×10⁵)sin60° = 0.2 × (√3/2) = 0.1√3 ≈ 0.173 N·m.
9A charge of 2 μC is placed at the centre of a cube of side 10 cm. The electric flux through one face of the cube is:
A.2×10⁻⁶/(8ε₀)
B.zero
C.2×10⁻⁶/ε₀
D.2×10⁻⁶/(6ε₀)
Explanation: Total flux through the closed cube is q/ε₀ = 2×10⁻⁶/ε₀. By symmetry the flux is equally shared by 6 faces, so flux through one face = (2×10⁻⁶)/(6ε₀).
10The SI unit of electric field intensity is equivalent to:
A.N/C or V/m
B.J/C
C.C/N
D.N·m/C
Explanation: E = F/q so unit is N/C. Also E = −dV/dr so unit is V/m. Both are equivalent.

About the Karnataka II PUC Physics (KSEAB) Practice Questions

Verified exam format metadata for Karnataka School Examination and Assessment Board (KSEAB) II PUC Class 12 Physics (Code 33) is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.