QJEEM19Phadb5: JEE Main 2019 Physics - Class 11 JEEM19Phadb5 Solution: Two identical spherical balls of mass M and radius R each are…

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

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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

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

So, total moment of inertia of the system is

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