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100+ Free Maharashtra HSC Physics Practice Questions

Maharashtra Higher Secondary Certificate (HSC / Class 12) Physics under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — subject code 54 practice questions are available now; exam metadata is being verified.

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2026 Statistics

Key Facts: Maharashtra HSC Physics Exam

3 hours

Typical Maharashtra HSC Physics theory paper duration

MSBSHSE HSC exam pattern reporting for Physics (code 54)

35%

Common minimum passing floor per HSC subject under MSBSHSE

Maharashtra HSC pass-mark practice (~35% each subject)

70 + 30

Common theory + practical mark split for HSC Physics (code 54)

Maharashtra HSC subject-wise mark distribution reporting

Subject code 54

Official MSBSHSE HSC subject code for Physics

MSBSHSE HSC general subject codes (mahahsscboard.org/hscsub.htm)

16 chapters

Balbharati Standard XII Physics unit structure commonly examined

Balbharati Std XII Physics textbook and MSBSHSE weightage summaries

English MCQ adaptation

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

OpenExamPrep practice policy

Maharashtra MSBSHSE HSC Physics (code 54) is a Class 12 public-exam science subject of about 3 hours theory (commonly 70+30 with practical) with ~35% pass floor, covering Balbharati themes from rotational dynamics through semiconductor devices — not a pure MCQ board paper. This free 2026 bank is an English MCQ study adaptation for concepts and calculation fluency.

Sample Maharashtra HSC Physics Practice Questions

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

1For a particle in uniform circular motion of radius r with constant speed v, the magnitude of centripetal acceleration is:
A.v/r
B.v²/r
C.vr
D.v²r
Explanation: Centripetal acceleration has magnitude a = v²/r (or ω²r) and always points toward the centre, continuously changing only the direction of velocity.
2Moment of inertia of a rigid body about an axis is defined as the sum of mᵢrᵢ² of its mass elements. Its SI unit is:
A.kg m/s
B.kg m²
C.N m
D.kg/m²
Explanation: I = Σ m r² has dimensions mass × length², so the SI unit is kilogram metre squared (kg m²).
3According to the parallel-axis theorem, if I_cm is the moment of inertia about an axis through the centre of mass and M is total mass, then MOI about a parallel axis at distance h is:
A.I_cm − M h²
B.I_cm + M h²
C.I_cm + M h
D.I_cm / M h²
Explanation: The parallel-axis theorem states I = I_cm + M h² for an axis parallel to the one through the CM, provided the CM axis is used as the reference.
4A particle of mass m moves in a vertical circle of radius R on a light string. The minimum speed at the lowest point so that it just completes the circle is:
A.√(gR)
B.√(2gR)
C.√(5gR)
D.√(3gR)
Explanation: At the top, minimum speed is √(gR) so tension can be zero with mg providing centripetal force. Energy conservation then gives u_bottom² = v_top² + 4gR = 5gR, so u = √(5gR).
5Torque τ and angular acceleration α of a rigid body about a fixed axis are related by:
A.τ = I / α
B.τ = I α
C.τ = α / I
D.τ = I² α
Explanation: Newton’s second law for rotation about a fixed axis (or principal axis through CM) is τ = I α, analogous to F = m a.
6A solid sphere of mass M and radius R rolls without slipping with centre-of-mass speed v. Its total kinetic energy is:
A.(1/2) M v²
B.(1/5) M v²
C.(7/10) M v²
D.(2/5) M v²
Explanation: KE_total = (1/2)Mv² + (1/2)Iω² with I = (2/5)MR² and v = ωR, so rotational KE = (1/5)Mv² and total = (7/10)Mv².
7A disc of moment of inertia 0.2 kg m² is spinning at 10 rad/s. If a torque of 0.4 N m acts for 5 s about the spin axis, the final angular speed is:
A.10 rad/s
B.15 rad/s
C.20 rad/s
D.30 rad/s
Explanation: α = τ/I = 0.4/0.2 = 2 rad/s². Then ω = ω₀ + αt = 10 + 2×5 = 20 rad/s (assuming torque increases the spin).
8Pressure in a fluid at rest increases with depth h according to (ρ density, g gravity):
A.P = ρ g / h
B.P = ρ g h (gauge, from free surface)
C.P = ρ / (g h)
D.P = g / (ρ h)
Explanation: Hydrostatic gauge pressure at depth h below a free surface is ΔP = ρgh for an incompressible fluid of density ρ.
9Bernoulli’s equation for steady, incompressible, non-viscous flow along a streamline is based primarily on:
A.Conservation of charge
B.Conservation of energy (work–energy for fluid)
C.Coulomb’s law
D.Ampere’s law
Explanation: Bernoulli’s relation P + ρgh + (1/2)ρv² = constant follows from work done by pressure and gravity changing kinetic and potential energy of a fluid element.
10Stokes’ law for the viscous drag on a small sphere of radius r moving with speed v in a fluid of viscosity η is:
A.F = 6 π η r v
B.F = 4 π η r² v
C.F = 2 π η r / v
D.F = 6 π η r / v
Explanation: For low Reynolds number, viscous drag on a sphere is F_d = 6πηrv, used for terminal velocity of raindrops and Millikan-type setups.

About the Maharashtra HSC Physics Practice Questions

Verified exam format metadata for Maharashtra Higher Secondary Certificate (HSC / Class 12) Physics under the Maharashtra State Board of Secondary and Higher Secondary Education (MSBSHSE), Pune — subject code 54 is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.