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(B) The moment of inertia $(M.I.)$ of a single sphere of mass $M$ and radius $r$ about its diameter is given by $I_{cm} = \frac{2}{5}Mr^2$.The distanc

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

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(B) The moment of inertia $(M.I.)$ of a single sphere of mass $M$ and radius $r$ about its diameter is given by $I_{cm} = \frac{2}{5}Mr^2$.The distance of the centre of each sphere from the centre of the square $(O)$ is $d = \frac{R}{\sqrt{2}}$.Using the parallel axis theorem,the moment of inertia of one sphere about an axis passing through the centre of the square and perpendicular to its plane is:$I_O = I_{cm} + Md^2 = \frac{2}{5}Mr^2 + M\left(\frac{R}{\sqrt{2}}ight)^2 = \frac{2}{5}Mr^2 + \frac{MR^2}{2}$.Since there are four such spheres,the total moment of inertia of the system is:$I_{total} = 4 	imes I_O = 4 \left( \frac{2}{5}Mr^2 + \frac{MR^2}{2} ight) = \frac{8}{5}Mr^2 + 2MR^2$.Factoring out $\frac{2}{5}M$,we get:$I_{total} = \frac{2}{5}M(4r^2 + 5R^2)$.

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

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