Question

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is $I_{1}$. The same rod is bent into a ring and its moment of inertia about a diameter is $I _{2}$. If $\frac{ I _{1}}{ I _{2}}$ is $\frac{ x \pi^{2}}{3}$, then the value of $x$ will be_______

Answer 1

$\ell=2 \pi r \Rightarrow \frac{\ell}{ r }=2 \pi$

$\frac{ I _{1}}{ I _{2}}=\frac{2}{3}\left(\frac{\ell}{ r }\right)^{2}$
$=\frac{2}{3} \times 4 \pi^{2}=\frac{8 \pi^{2}}{3}$
$x =8$