Q 11-06-152JEE MainJEE Main 2021 (18 Mar, Shift 2)Easy
Consider a uniform wire of mass $M$ and length $L$. It is bent into a semicircle. Its moment of inertia about a line perpendicular to the plane of the wire passing through the centre is:
Answer: (C) $\frac{ML^2}{\pi^2}$
$\pi R = L \Rightarrow R = \dfrac{L}{\pi}$. Every element of the wire is at distance $R$ from the axis, so $I = MR^2 = \dfrac{ML^2}{\pi^2}$.
Solution by Sreeraj P, M.Sc Physics