Question

If $I_1$ is the moment of inertia of a thin rod about an axis perpendicular to its length and passing through its centre of mass and $I_2$ the moment of inertia of the ring formed by the same rod about an axis passing through the centre of mass of the ring and perpendicular to the plane of the ring. If the ratio $\frac{I_1}{I_2}is\frac{{\pi }^2}{x}$. Find x.

Answer 1

$l=2\pi R\Rightarrow R=\frac{l}{2\pi }$
$\frac{I_1}{I_2}=\frac{m{l}^2/12}{m{R}^2}$
$=\frac{l^2}{12R^2}$
$=\frac{l^2}{12(l/2\pi {)}^2}=\frac{{\pi }^2}{3}$