Q 11-06-254JEE MainJEE Main 2025 (2 Apr, Shift 2)Easy
The moment of inertia of a circular ring of mass $M$ and diameter $r$ about a tangential axis lying in the plane of the ring is:
Answer: (B) $\dfrac38 Mr^2$
Note that $r$ is the **diameter**, so the radius is $r/2$.
About a diameter, $I_d = \tfrac12 M(r/2)^2$. A tangent in the plane is parallel to a diameter at distance $r/2$, so by the parallel axis theorem
$$I = \tfrac12 M\left(\frac r2\right)^2 + M\left(\frac r2\right)^2 = \tfrac32 M\cdot\frac{r^2}{4} = \frac38 Mr^2$$
Solution by Sreeraj P, M.Sc Physics