Q 11-06-162JEE MainJEE Main 2021 (26 Feb, Shift 1)Medium
Four identical solid spheres each of mass $m$ and radius $a$ are placed with their centres on the four corners of a square of side $b$. The moment of inertia of the system about one side of the square where the axis of rotation is parallel to the plane of the square is:
Answer: (C) $\frac{8}{5}ma^2 + 2mb^2$
Two spheres have their centres on the axis: each contributes $\frac{2}{5}ma^2$.
The other two are at distance $b$ from the axis: each contributes $\frac{2}{5}ma^2 + mb^2$ (parallel axis theorem).
$$I = 4\cdot\frac{2}{5}ma^2 + 2mb^2 = \frac{8}{5}ma^2 + 2mb^2$$
Solution by Sreeraj P, M.Sc Physics