JEE Main 2020 — Oscillations Question with Solution

The time period of oscillation of physical pendulum is given by

$T=2\pi \sqrt{\frac{I}{Mgl}} ...\left(1\right)$

Where, $I$ is the moment of inertia of ring about given axis and $l$ is the distance of centre of mass of ring from the axis of rotation.

Moment of inertia of ring in its plane is given by

$I_1=I_c+ M{d}^2=M{R}^2+M{R}^2=2M{R}^2$

Moment of inertia of ring perpendicular to its plane is given by

$I_2=I_c+ M{d}^2=\frac{M{R}^2}{2}+M{R}^2=\frac{3}{2}M{R}^2$

Now the ratio of time period is given by 

$\frac{T_1}{T_2}=\sqrt{\frac{I_1}{I_2}}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{2M{R}^2}{{\displaystyle \raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right.M{R}^2}}}$

$\Rightarrow \frac{T_1}{T_2}=\sqrt{\frac{4}{3}}=\frac{2}{\sqrt{3}}$