Four spheres, each of mass M and radius r are situated at the four corners of square of side R. The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its center will be

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$\frac{5}{2} M (4 r^{2} + 5 R^{2})$

$\frac{2}{5} M (4 r^{2} + 5 R^{2})$

$\frac{2}{5} M (4 r^{2} + 5 r^{2})$

$\frac{5}{2} M (4 r^{2} + 5 r^{2})$

answer is B.

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Detailed Solution

MI of sphere A about its diameter $I_{d} = \frac{2}{5} M r^{2}$
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Now MI of sphere A about an axis perpendicular to the plane of square and passing through its center will be
$I_{0} = I_{d} + M {(\frac{R}{\sqrt{2}})}^{2}$

$= \frac{2}{5} M r^{2} + \frac{M R^{2}}{2}$ [by the theorem of parallel axis]
Moment of inertia of system (i.e. four sphere)
$= 4 l_{0} = 4 [\frac{2}{5} M r^{2} + \frac{M R^{2}}{2}] = \frac{2}{5} M [4 r^{2} + 5 R^{2}]$