JEE Main 2025 — Rotational Motion Question with Solution

The moment of inertia of a circular ring with mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is calculated as follows:

Given the diameter $r$ of the ring, the radius $R$ is $\frac{r}{2}$.

The formula for the moment of inertia about a tangential axis in the plane of the ring is:

$I_{\text{tangential}}=\frac{3}{2}M{\left(\frac{R}{2}\right)}^2$

Substitute the value of $R=\frac{r}{2}$ into the equation:

$I_{\text{tangential}}=\frac{3}{2}M{\left(\frac{r}{2}\right)}^2=\frac{3}{8}M{r}^2$

Therefore, the moment of inertia of the ring about the specified axis is $\frac{3}{8}M{r}^2$.