Question

A ring of diameter 2m oscillates as a compound pendulum about a horizontal axis passing through a point of its rim. It oscillates such that its centre move in a plane which is perpendicular to the plane of the ring. The equivalent length of the simple pendulum is :

• A. 2 m
• B. 4 m
• C. 1.5 m
• D. 3 m

Answer 1

The correct option is D 3 m

$T=2\pi \sqrt{\frac{3M{R}^2}{2gMR}}$

$=2\pi \sqrt{\frac{3R}{2g}}$

$=\frac{3\times 2}{2}=3m$

Hence,

option $\left(D\right)$ is correct answer.