The moment of inertia of a sphere (mass M and | vedclass

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(D) Given,the moment of inertia $(MOI)$ of a sphere about its diameter is $I = \frac{2}{5} M R^{2}$.In the given arrangement,two spheres are placed su

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$9$

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(D) Given,the moment of inertia $(MOI)$ of a sphere about its diameter is $I = \frac{2}{5} M R^{2}$.In the given arrangement,two spheres are placed such that their centers lie on the axis $XX'$. For these two spheres,the axis $XX'$ passes through their diameters. Thus,the $MOI$ for each of these two spheres is $I = \frac{2}{5} M R^{2}$.The other two spheres are placed such that their centers are at a distance $R$ from the axis $XX'$. Using the parallel axis theorem,the $MOI$ of each of these two spheres about the axis $XX'$ is $I' = I_{cm} + M R^{2} = \frac{2}{5} M R^{2} + M R^{2} = \frac{7}{5} M R^{2}$.Since $I = \frac{2}{5} M R^{2}$,we have $M R^{2} = \frac{5}{2} I$. Therefore,$I' = \frac{7}{5} 	imes (\frac{5}{2} I) = \frac{7}{2} I$.The total moment of inertia of the system is $I_{total} = 2 	imes I + 2 	imes I' = 2I + 2 	imes (\frac{7}{2} I) = 2I + 7I = 9I$.

The moment of inertia of a sphere (mass $M$ and radius $R$) about its diameter is $I$. Four such spheres are arranged as shown in the figure. The moment of inertia of the system about the axis $XX'$ will be (in $,I$)

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