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 square where the axis of rotation is parallel to the plane of the square is :

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$\frac{4}{5} m a^{2} + 2 m b^{2}$

$\frac{8}{5} m a^{2} + m b^{2}$

$\frac{4}{5} m a^{2}$

$\frac{8}{5} m a^{2} + 2 m b^{2}$

answer is C.

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Detailed Solution

Moment of inertia of a sphere $= \frac{2}{5} m a^{2}$

Using the parallel axis theorem of moment of inertia, we have

$I = 2 \times \frac{2}{5} m a^{2} + 2 (\frac{2}{5} m a^{2} + m b^{2})$

$\Rightarrow I = \frac{2}{5} m a^{2} \times 4 + 2 \times m b^{2}$

$= \frac{8}{5} m a^{2} + 2 m b^{2}$

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