The period of a physical pendulum is
$\begin{align*}T = 2 \pi\sqrt{\frac{I}{m g d}}\end{align*}$
where $I$ is the moment of inertia about the pivot point and $d$ is the distance from the pivot to the center of mass. Let $M$ be the mass of hoop $Y$ and let $R$ be the radius of hoop $Y$ . Then

$\begin{align*}I_X &= 2 \cdot 4M \cdot (4R)^2 = 128 M R^2 \\I_Y &= 2 M R^2 \\d_X &= 4 R \\d_Y &= R\end{align*}$
Therefore,
$\begin{align*}\frac{T_Y}{T_X} = \frac{\displaystyle 2 \pi\sqrt{\frac{I_Y}{m_Y g d_Y}}}{ \displaystyle 2 \pi\sqrt{\frac{I_X}{...$
In other words,
$\begin{align*}T_Y = \frac{T_X}{2} = \boxed{\frac{T}{2}}\end{align*}$
Therefore, answer (B) is correct.

Solution to 1992 Problem 74