All Practice Exams

100+ Free RBSE SS Physics Practice Questions

Rajasthan RBSE Senior Secondary (Class 12) Physics — Code 040 practice questions are available now; exam metadata is being verified.

✓ No registration✓ No credit card✓ No hidden fees✓ Start practicing immediately
See RBSE subject-wise statistics for the current cycle on statistics2026.htm. Pass Rate
100+ Questions
100% Free

Loading practice questions...

2026 Statistics

Key Facts: RBSE SS Physics Exam

040

RBSE subject code for Physics

RBSE 2026 subject-wise statistics

Theory 56 + Sessional 14

Typical full marks / assessment pattern

RBSE syllabus examination scheme

3 h 15 m

Typical theory paper duration (where applicable)

RBSE Class 12 / Praveshika schemes

33%

Common minimum pass threshold under RBSE regulations

RBSE examination regulations (confirm circular)

2026

Main examination cycle evidenced in official subject-wise statistics

rajeduboard.rajasthan.gov.in/statistics2026.htm

MCQ study aid

Local bank adapts knowledge to four-option MCQs; official paper is mixed/performance format

OpenExamPrep assessment-format policy

RBSE Senior Secondary (Class 12) Physics (code 040): Theory 56 + Sessional 14 = 70 and Practical 30 = 100; 3 hours 15 minutes (theory). Pass about 33% per RBSE rules (confirm circular). Fee as per board notification. Free English MCQ study aid — not a full official-format simulation.

Sample RBSE SS Physics Practice Questions

Try these sample questions to test your RBSE 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 like point charges repel each other with force F when separated by distance r in vacuum. If the separation is doubled, the repulsive force becomes:
A.F/2
B.2F
C.4F
D.F/4
Explanation: Coulomb's law states F proportional to 1/r squared. Doubling the separation multiplies the force by 1/4, so the new force is F/4. The charges remain like, so the force is still repulsive but weaker.
2Two point charges +3 μC and −2 μC are placed 2 m apart in vacuum. The magnitude of the electrostatic force between them is (k = 9×10^9 N m^2 C^−2):
A.1.35×10^−2 N
B.2.7×10^−2 N
C.6.75×10^−3 N
D.1.35×10^−3 N
Explanation: F = k|q1 q2|/r^2 = (9×10^9)(3×10^−6)(2×10^−6)/(2)^2 = (9×10^9)(6×10^−12)/4 = 54×10^−3/4 = 1.35×10^−2 N. Opposite signs mean attraction, but magnitude is positive.
3Electric field intensity at a point is defined as the electrostatic force experienced by a:
A.unit magnetic pole placed at that point
B.unit positive charge placed at that point
C.unit negative charge placed at that point
D.unit mass placed at that point
Explanation: By definition E = F/q0 for a small positive test charge q0. The direction of E is the direction of force on a positive charge. Unit mass defines gravitational field strength, not electric field.
4The electric field due to a point charge of 2 μC at a distance of 3 m in vacuum is (k = 9×10^9 SI units):
A.2×10^4 N/C
B.9×10^3 N/C
C.2×10^3 N/C
D.6×10^3 N/C
Explanation: E = kq/r^2 = (9×10^9)(2×10^−6)/(3)^2 = 18×10^3/9 = 2×10^3 N/C. Direction is radially outward because the charge is positive.
5Charges +q and +4q are fixed 30 cm apart. A third charge is placed on the line joining them so that the net force on it is zero. Its distance from +q is:
A.20 cm (between the charges)
B.6 cm (between the charges)
C.15 cm (midpoint)
D.10 cm (between the charges)
Explanation: Let the null point be at distance x from +q. Then kqQ/x^2 = k(4q)Q/(0.30−x)^2 gives 1/x = 2/(0.30−x), so 0.30−x = 2x and x = 0.10 m = 10 cm. For like charges the null point lies between them, closer to the smaller charge.
6The SI unit of electric potential is the:
A.volt
B.newton
C.coulomb
D.farad
Explanation: Electric potential V equals work done per unit charge; 1 V = 1 J/C. Newton is force, coulomb is charge, and farad is capacitance.
7Electric potential due to a charge of 5 μC at a distance of 9 m from it in vacuum is (k = 9×10^9):
A.9×10^3 V
B.5×10^3 V
C.4.5×10^4 V
D.5×10^2 V
Explanation: V = kq/r = (9×10^9)(5×10^−6)/9 = 45×10^3/9 = 5×10^3 V. Potential is a scalar and falls as 1/r, not 1/r^2.
8Three identical capacitors each of capacitance C are connected in series. Their equivalent capacitance is:
A.C
B.C/2
C.C/3
D.3C
Explanation: For n equal capacitors in series, 1/Ceq = n/C so Ceq = C/n. With n = 3, Ceq = C/3. Series connection reduces equivalent capacitance below any single capacitor.
9A 2 μF parallel-plate capacitor is charged to 100 V and then disconnected from the battery. If the plate separation is doubled (dielectric still air), the stored energy becomes:
A.5.0×10^−3 J
B.1.0×10^−2 J
C.4.0×10^−2 J
D.2.0×10^−2 J
Explanation: Initially U = (1/2)CV^2 = (1/2)(2×10^−6)(100)^2 = 1.0×10^−2 J. After disconnection Q is constant. Doubling d halves C. With Q fixed, U = Q^2/(2C) doubles to 2.0×10^−2 J. Energy is supplied by the work done in separating the plates against attraction.
10Gauss's law states that the net electric flux through a closed surface is equal to:
A.q_enclosed/ε0
B.q_enclosed × ε0
C.E × A only
D.zero always
Explanation: Gauss's law: closed surface integral of E·dA equals q_encl/ε0 in SI. Flux depends only on the enclosed charge, not on charges outside the Gaussian surface.

About the RBSE SS Physics Practice Questions

Verified exam format metadata for Rajasthan RBSE Senior Secondary (Class 12) Physics — Code 040 is pending. The practice questions above remain available while official exam length, timing, passing score, fee, and administrator details are reviewed.