Electric Dipole question from JEE Main 2025 (29 Jan, Shift 1)

Answer: (b)

1. Take each charge of mass $m$ at $\frac{l}{2}$ from the centre: $I = 2m\left(\frac{l}{2}\right)^2 = \frac{ml^2}{2}$.
2. Restoring torque $\tau = pE_0\sin\theta \approx qlE_0\,\theta$.
3. $\omega^2 = \frac{qlE_0}{ml^2/2} = \frac{2qE_0}{ml}$
4. $T = 2\pi\sqrt{\frac{ml}{2qE_0}}$, so the answer is (b).