Q 11-13-029JEE MainHard
A uniform rod of length $L$ is pivoted at one end and oscillates in a vertical plane as a compound pendulum. The length of a simple pendulum with the same period is
Answer: (C) $\dfrac{2L}{3}$
$$T = 2\pi\sqrt{\frac{I}{mgd}} = 2\pi\sqrt{\frac{mL^2/3}{mg(L/2)}} = 2\pi\sqrt{\frac{2L}{3g}}$$
This equals a simple pendulum of length $\dfrac{2L}{3}$.
Solution by Sreeraj P, M.Sc Physics