JEE Main 2025 Apr 2 Shift 2, Physics Q3: Moment of Inertia

The given quantity $r$ is the diameter, so the ring's actual radius is $R=\frac{r}{2}$.

For a ring of radius $R$ about a diameter, $I_{\text{dia}}=\frac{1}{2}M{R}^2$. Using the parallel axis theorem to shift to a tangent in the plane of the ring (distance $R$ from the centre):

$I_{\text{tan}}=I_{\text{dia}}+M{R}^2=\frac{1}{2}M{R}^2+M{R}^2=\frac{3}{2}M{R}^2$.

Substituting $R=r/2$: $I_{\text{tan}}=\frac{3}{2}M{\left(\frac{r}{2}\right)}^2=\frac{3}{8}M{r}^2$.