Q 11-06-264JEE MainJEE Main 2025 (8 Apr, Shift 2)Easy
A rod of linear mass density $\lambda$ and length $L$ is bent to form a ring of radius $R$. The moment of inertia of the ring about any of its diameters is:
Answer: (D) $\dfrac{\lambda L^3}{8\pi^2}$
$M = \lambda L$ and $L = 2\pi R$, so $R = \dfrac{L}{2\pi}$.
$$I_{\text{diameter}} = \frac{MR^2}{2} = \frac{\lambda L}{2}\cdot\frac{L^2}{4\pi^2} = \frac{\lambda L^3}{8\pi^2}$$
Solution by Sreeraj P, M.Sc Physics