JEE Main 2025 · 29 January · morning · Single correct · +4 / −1
An electric dipole of mass m, charge q, and length l is placed in a uniform electric field E⃗ = E₀î. When the dipole is rotated slightly from its equilibrium position and released, the time period of its oscillations will be :
- (a) 1/2π √(ml/2qE₀)
- (b) 2π √(ml/2qE₀)
- (c) 2π √(ml/qE₀)
- (d) 1/(2 π) √((2 m l)/(q E₀))
Answer and solution
Answer (b) 2π √(ml/2qE₀)
Solution
Physics sir · 3 steps
Treat the dipole like a pendulum: restoring torque over moment of inertia gives you ω².
- Restoring torque. The dipole moment is p = ql. When it is turned by a small angle θ, the field gives a torque τ = pE₀ sin θ ≈ qlE₀ θ that pulls it back toward equilibrium.
- Moment of inertia. Take mass m at each charge. The dipole rotates about its centre, so each mass sits at l/2 from the axis. That gives I = 2m(l/2)² = ml²/2.
ω² = qlE₀/I = qlE₀/(ml²/2) = 2qE₀/ml
- So. T = 2π/ω = 2π√(ml/2qE₀), which is option (b).
The idea to keep. For any small angular oscillation, write τ = -kθ and use T = 2π√(I/k). Measure distances from the rotation axis, which for a dipole is its centre.
Answer from the NTA final answer key. Solution by PrepWiser.