A ring is hung on a nail. It can oscillate, without slipping or...

The time period of oscillation of physical pendulum is given by

$T = 2\pi\sqrt{\dfrac{I}{Mgl}} . . . ( 1 )$

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

${I}_{2} = {I}_{c} + M{d}^{2} = \dfrac{M{R}^{2}}{2} + M{R}^{2} = \dfrac{3}{2}M{R}^{2}$

Now the ratio of time period is given by

$\dfrac{{T}_{1}}{{T}_{2}} = \sqrt{\dfrac{{I}_{1}}{{I}_{2}}}$

$⇒ \dfrac{{T}_{1}}{{T}_{2}} = \sqrt{\dfrac{2M{R}^{2}}{\dfrac{3}{2}M{R}^{2}}}$

$⇒ \dfrac{{T}_{1}}{{T}_{2}} = \sqrt{\dfrac{4}{3}} = \dfrac{2}{\sqrt{3}}$