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(D) The moment of inertia of a hollow sphere of mass $M$ and radius $r$ about its diameter is $I = \frac{2}{3} Mr^2$.Here,the radius of each sphere is

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$\frac{13}{3} MR^2$

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(D) The moment of inertia of a hollow sphere of mass $M$ and radius $r$ about its diameter is $I = \frac{2}{3} Mr^2$.Here,the radius of each sphere is $r = \frac{R}{2}$.For the sphere through which the axis $yy'$ passes,the moment of inertia about $yy'$ is $I_1 = \frac{2}{3} M(\frac{R}{2})^2 = \frac{2}{3} M(\frac{R^2}{4}) = \frac{1}{6} MR^2$.For the other sphere,the axis $yy'$ is at a distance $d = 2R$ from its center. Using the parallel axis theorem,$I_2 = I_{cm} + Md^2 = \frac{2}{3} M(\frac{R}{2})^2 + M(2R)^2$.$I_2 = \frac{1}{6} MR^2 + 4MR^2 = \frac{1 + 24}{6} MR^2 = \frac{25}{6} MR^2$.The total moment of inertia of the system is $I_{total} = I_1 + I_2 = \frac{1}{6} MR^2 + \frac{25}{6} MR^2 = \frac{26}{6} MR^2 = \frac{13}{3} MR^2$.

Two hollow spheres each of mass $M$ and radius $\frac{R}{2}$ are connected with a massless rod of length $2R$ as shown in the figure. What will be the moment of inertia of the system about an axis $yy'$ passing through the center of one of the spheres and perpendicular to the rod?

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