Physics - Rotation Question with Solution | TestHub

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

The moment of inertia of the rod about an axis passing through its centre and perpendicular to its length is:

$I_1=\frac{M{L}^2}{12}$

When the rod is bent into a ring, its circumference is equal to the length of the rod.

$2\pi R=L\Rightarrow R=\frac{L}{2\pi }$

The moment of inertia of the ring about its diameter is:

$I_2=\frac{M{R}^2}{2}$

Substituting the value of $R$:

$I_2=\frac{M}{2}{\left(\frac{L}{2\pi }\right)}^2=\frac{M{L}^2}{8{\pi }^2}$

Taking the ratio of $I_1$ to $I_2$:

$\frac{I_1}{I_2}=\frac{\frac{M{L}^2}{12}}{\frac{M{L}^2}{8{\pi }^2}}=\frac{8{\pi }^2}{12}=\frac{2{\pi }^2}{3}$

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