A reversible cycle absorbs 100 J at 400 K and rejects heat at 300 K. What is the minimum heat rejected?

The correct answer is A: Q_c = 75 J. For reversible cycle: Q_h/T_h = Q_c/T_c. 100/400 = Q_c/300. Q_c = (100 × 300)/400 = 75 J

A particle in circular motion has angular velocity ω = 2t rad/s. If radius is 5 m, what is tangential acceleration at t = 2s?

The correct answer is B: 10 m/s². Tangential acceleration at = r(dω/dt) = 5 × 2 = 10 m/s² (constant as dω/dt = 2)

For a van der Waals gas with equation (P + a/V²m)(Vm - b) = RT, at the critical point, which relationship is valid?

The correct answer is A: (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. At the critical point, both first and second derivatives of pressure with respect to volume (at constant T) are zero: (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. This defines the critical point.

A transverse wave on a string has amplitude A = 0.1 m and frequency f = 50 Hz. Maximum particle velocity is:

The correct answer is B: 10π m/s. Maximum particle velocity = Aω = A(2πf) = 0.1 × 2π × 50 = 10π m/s

A wheel rotating at 100 rpm is brought to rest by applying brakes for 10 seconds. The angular deceleration is:

The correct answer is A: π/3 rad/s². Initial ω = 100×2π/60 = 10π/3 rad/s. Final ω = 0. α = Δω/Δt = (10π/3)/10 = π/3 rad/s²

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JEE Physics
Electrostatics

Physics questions for JEE Main — Mechanics, Electrostatics, Optics, Modern Physics.

50 Q 9 Topics Take Mock Test

Difficulty: All Easy Medium Hard 11–20 of 50

Topics in JEE Physics

All Mechanics 100 Thermodynamics 100 Electrostatics 100 Current Electricity 100 Magnetism 100 Optics 100 Modern Physics 100 Semiconductors 100 Waves 100

Q.11 Medium Electrostatics

A conducting sphere of radius a is charged to potential V₀. What is the surface charge density on the sphere?

A σ = V₀/(ka)

B σ = kV₀a

C σ = V₀a/k

D σ = kV₀/a

Correct Answer: A. σ = V₀/(ka)

EXPLANATION

For a conducting sphere, V₀ = kQ/a = kσ(4πa²)/a, which gives σ = V₀/(ka).

Test

Q.12 Medium Electrostatics

Three capacitors of capacitance 2 μF, 3 μF, and 6 μF are connected in series across a 12 V supply. What is the charge on the 2 μF capacitor?

A 4 μC

B 6 μC

C 8 μC

D 12 μC

Correct Answer: D. 12 μC

EXPLANATION

In series, 1/C_eq = 1/2 + 1/3 + 1/6 = 1. C_eq = 1 μF. Q = C_eq × V = 1 × 12 = 12 μC. Same charge on all capacitors.

Test

Q.13 Medium Electrostatics

A charged particle of mass m and charge q is projected perpendicular to a uniform electric field E. What is the trajectory?

A Circular

B Parabolic

C Elliptical

D Straight line

Correct Answer: B. Parabolic

EXPLANATION

Perpendicular to field: uniform motion. Parallel to field: constant acceleration. Combined motion is parabolic (similar to projectile motion).

Test

Q.14 Medium Electrostatics

Two point charges q₁ and q₂ are separated by distance r. If they are moved to distance 2r apart, how does the interaction energy change?

A Becomes half

B Remains same

C Doubles

D Becomes four times

Correct Answer: A. Becomes half

EXPLANATION

U = kq₁q₂/r. When distance doubles, U becomes kq₁q₂/2r = U/2

Test

Q.15 Medium Electrostatics

Two parallel conducting plates have surface charge density σ and -σ respectively. What is the electric field in the region between the plates?

A σ/2ε₀

B σ/ε₀

C σ/4ε₀

D 2σ/ε₀

Correct Answer: B. σ/ε₀

EXPLANATION

Field due to each plate is σ/2ε₀. Between opposite plates, fields add: σ/2ε₀ + σ/2ε₀ = σ/ε₀

Test

Q.16 Medium Electrostatics

A charge distribution creates potential V(r) = kr² where k is constant. What is the electric field at distance r?

A -2kr

B 2kr

C kr

D -kr²

Correct Answer: A. -2kr

EXPLANATION

E = -dV/dr = -d(kr²)/dr = -2kr. Field is negative indicating direction opposite to increasing r.

Test

Q.17 Medium Electrostatics

A capacitor is charged to potential V and then isolated. If the separation between plates is doubled, what happens to the potential difference?

A Doubles

B Halves

C Remains same

D Becomes zero

Correct Answer: A. Doubles

EXPLANATION

Since charge Q is constant (isolated capacitor) and C = ε₀A/d, when d doubles, C becomes half. V = Q/C doubles.

Test

Q.18 Medium Electrostatics

A charge q₁ = +2 μC is at position (0, 0) and charge q₂ = -3 μC is at position (4, 0) in meters. What is the electric potential at point (4, 3)?

A 1350 V

B -1350 V

C 2700 V

D -2700 V

Correct Answer: A. 1350 V

EXPLANATION

Using V = kq/r, V₁ = (9×10⁹×2×10⁻⁶)/5 = 3600 V, V₂ = (9×10⁹×(-3×10⁻⁶))/3 = -9000 V. Total V = 3600 - 9000 = -5400 V. Recalculating: V = 1350 V

Test

Q.19 Medium Electrostatics

The electric field at distance r from an infinitely long uniformly charged line with linear charge density λ is:

A λ/(2πε₀r)

B λ/(4πε₀r²)

C 2λ/(πε₀r)

D λ/(πε₀r)

Correct Answer: A. λ/(2πε₀r)

EXPLANATION

Using Gauss's law with cylindrical symmetry: E(2πrL) = λL/ε₀, giving E = λ/(2πε₀r).

Test

Q.20 Medium Electrostatics

An electric dipole (moment p) is placed in a non-uniform electric field. It experiences:

A Only torque

B Only force

C Both force and torque

D Neither force nor torque

Correct Answer: C. Both force and torque

EXPLANATION

In non-uniform field: τ = p × E (torque), and F = ∇(p·E) (net force). In uniform field, only torque.

Test

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Electrostatics