JEE Main 2019 — Rotational Motion Question with Solution

Using parallel axis theorem, moment of inertia about the given axis in the figure below will be

$\begin{array}{rl}& I_1=\frac{2}{5}M{R}^2+M(2R{)}^2\\ & \\ & I_1=\frac{22}{5}M{R}^2\end{array}$

Considering both spheres at equal distance from the axis, moment of inertia due to both spheres about this axis will be

$$2I_1=2\times 22/5M{R}^2$$ = $$\frac{44}{5}$$MR²

Moment of inertia of rod

Here, given that L = 2R

$\therefore I_2=\frac{1}{12}M(2R{)}^2=\frac{1}{3}M{R}^2$

So, total moment of inertia of the system is

$$\begin{array}{rl}I& =2I_1+I_2=2\times \frac{22}{5}M{R}^2+\frac{1}{3}M{R}^2\\ & \\ \Rightarrow I& =\left(\frac{44}{5}+\frac{1}{3}\right)M{R}^2=\frac{137}{15}M{R}^2\end{array}$$