Q 11-06-205JEE MainJEE Main 2019 (10 Jan, Shift 2)Medium
Two identical spherical balls of mass $M$ and radius $R$ each are stuck on two ends of a rod of length $2R$ and mass $M$ (see figure). The moment of inertia of the system about the axis passing perpendicularly through the centre of the rod is
Answer: (C) $\dfrac{137}{15}MR^2$
Rod: $\dfrac{M(2R)^2}{12} = \dfrac{MR^2}{3}$.
Each ball's centre is $R + R = 2R$ from the axis. By the parallel axis theorem each ball contributes
$$\frac25MR^2 + M(2R)^2 = \frac{22}{5}MR^2$$
$$I = \frac{MR^2}{3} + 2\cdot\frac{22}{5}MR^2 = \frac{5 + 132}{15}MR^2 = \frac{137}{15}MR^2$$
Solution by Sreeraj P, M.Sc Physics