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(A) The moment of inertia of a single sphere about its diameter is given as $I = \frac{2}{5} M (2R)^2$,where $2R$ is the diameter of the sphere. From

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$9$

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(A) The moment of inertia of a single sphere about its diameter is given as $I = \frac{2}{5} M (2R)^2$,where $2R$ is the diameter of the sphere. From this,we have $M (2R)^2 = \frac{5}{2} I$.The system consists of four spheres. Two spheres are centered on the axis $XX'$,so their moment of inertia about $XX'$ is simply their moment of inertia about their diameter,which is $I$ each.The other two spheres are located at a distance of $2R$ from the axis $XX'$. According to the parallel axis theorem,the moment of inertia of each of these spheres about the axis $XX'$ is $I_{sphere} = I_{cm} + M d^2$,where $I_{cm} = I$ and $d = 2R$.Thus,$I_{sphere} = I + M (2R)^2 = I + \frac{5}{2} I = \frac{7}{2} I$.The total moment of inertia of the system is $I_{system} = I + I + (I + M(2R)^2) + (I + M(2R)^2) = 2I + 2(I + \frac{5}{2} I) = 2I + 2(\frac{7}{2} I) = 2I + 7I = 9I$.

The moment of inertia of a sphere about its diameter is $I$. Four such spheres are arranged as shown in the figure. Calculate the moment of inertia of the system about the axis $XX'$. (in $I$)

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