Physics - Rotation Question with Solution | TestHub

Let the mass of the rod be $M$ and its length be $L$.

The moment of inertia of the rod about a perpendicular axis passing through its centre is:
$I=\frac{M{L}^2}{12}$

When the rod is bent into a ring, its length becomes the circumference of the ring. Let the radius of the ring be $R$.
$L=2\pi R\Rightarrow R=\frac{L}{2\pi }$

The moment of inertia of the ring about its diameter is:
$I_1=\frac{M{R}^2}{2}$

Substituting the value of $R$:
$I_1=\frac{M}{2}{\left(\frac{L}{2\pi }\right)}^2=\frac{M{L}^2}{8{\pi }^2}$

The ratio of the moments of inertia is:
$\frac{I}{I_1}=\frac{\frac{M{L}^2}{12}}{\frac{M{L}^2}{8{\pi }^2}}=\frac{8{\pi }^2}{12}=\frac{2{\pi }^2}{3}$

Answer: $\frac{2{\pi }^2}{3}$