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(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} mR^2$,which implies $mR^2 = \frac{5}{2} I$.For the system,

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$\frac{9 I}{2}$

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(C) The moment of inertia of a solid sphere about its diameter is given by $I = \frac{2}{5} mR^2$,which implies $mR^2 = \frac{5}{2} I$.For the system,the total moment of inertia about axis $AB$ is $I_{AB} = I_1 + I_2$.For the first sphere,the axis $AB$ passes through its center,so its moment of inertia is $I_1 = I$.For the second sphere,the axis $AB$ is tangent to it. Using the parallel axis theorem,$I_2 = I_{com} + mR^2 = I + mR^2$.Substituting $mR^2 = \frac{5}{2} I$,we get $I_2 = I + \frac{5}{2} I = \frac{7}{2} I$.Therefore,the total moment of inertia is $I_{AB} = I + \frac{7}{2} I = \frac{9}{2} I$.

The moment of inertia of a solid sphere about its diameter is $I$. Two such spheres are arranged as shown in the figure. The moment of inertia of the system about axis $AB$ will be

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