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(A) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{Mgd}}$,where $I$ is the moment of inertia about the pivot point,$M$ i

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$\frac{2}{\sqrt{3}}$

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(A) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{Mgd}}$,where $I$ is the moment of inertia about the pivot point,$M$ is the mass,$g$ is the acceleration due to gravity,and $d$ is the distance from the pivot to the center of mass. For a ring of radius $R$,$d = R$.Case $(i)$: The ring oscillates in its own plane. The axis of rotation is at the rim,perpendicular to the plane of the ring. By the parallel axis theorem,$I_{1} = I_{cm} + MR^{2} = MR^{2} + MR^{2} = 2MR^{2}$.Thus,$T_{1} = 2\pi \sqrt{\frac{2MR^{2}}{MgR}} = 2\pi \sqrt{\frac{2R}{g}}$.Case $(ii)$: The ring oscillates back and forth in a direction perpendicular to its plane. The axis of rotation is at the rim,lying in the plane of the ring. By the parallel axis theorem,$I_{2} = I_{cm} + MR^{2} = \frac{1}{2}MR^{2} + MR^{2} = \frac{3}{2}MR^{2}$.Thus,$T_{2} = 2\pi \sqrt{\frac{\frac{3}{2}MR^{2}}{MgR}} = 2\pi \sqrt{\frac{3R}{2g}}$.Taking the ratio: $\frac{T_{1}}{T_{2}} = \sqrt{\frac{2MR^{2}}{\frac{3}{2}MR^{2}}} = \sqrt{\frac{2}{3/2}} = \sqrt{\frac{4}{3}} = \frac{2}{\sqrt{3}}$.

$A$ ring is hung on a nail. It can oscillate,without slipping or sliding $(i)$ in its plane with a time period $T_{1}$ and,$(ii)$ back and forth in a direction perpendicular to its plane,with a period $T_{2}$. The ratio $\frac{T_{1}}{T_{2}}$ will be

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