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 $\dfrac{T_1}{T_2}$ will be:
Answer: (A) $\dfrac{2}{\sqrt3}$
For a physical pendulum $T = 2\pi\sqrt{\dfrac{I}{mgR}}$, with the centre of mass at distance $R$ from the nail.
(i) Oscillation in its plane: axis perpendicular to the plane through a point on the rim, $I_1 = mR^2 + mR^2 = 2mR^2$.
(ii) Oscillation perpendicular to its plane: axis is a tangent in the plane of the ring, $I_2 = \tfrac12 mR^2 + mR^2 = \tfrac32 mR^2$.
$$\frac{T_1}{T_2} = \sqrt{\frac{I_1}{I_2}} = \sqrt{\frac{2}{3/2}} = \frac{2}{\sqrt3}$$
Solution by Sreeraj P, M.Sc Physics