Physics - Rotation Question with Solution | TestHub

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

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 of radius $R$, its length becomes the circumference of the ring.

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

Moment of inertia of the ring about its diameter is $I_1=\frac{M{R}^2}{2}$.

Substituting the value of $R$, we get:

$I_1=\frac{M}{2}{\left(\frac{L}{2\pi }\right)}^2=\frac{M{L}^2}{8{\pi }^2}$

Given $I_1=xI$, we have:

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

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