JEE Main · 2025 · 02 Apr 2025 · Shift 2 · Rotational Mechanics

With the diameter equal to $r$, the ring's radius is $R=\frac{r}{2}$. About a diameter the moment of inertia is $\frac{1}{2}M{R}^2$, since the two in-plane diameters share equally the perpendicular-axis value $M{R}^2$. A tangent lying in the plane is parallel to a diameter and offset by $R$, so by the parallel-axis theorem $I=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$. Substituting $R=\frac{r}{2}$, $I=\frac{3}{2}M(\frac{r}{2}{)}^2=\frac{3}{8}M{r}^2$. The moment of inertia about the in-plane tangential axis is $\frac{3}{8}M{r}^2$.