Four identical solid spheres each of mass ‘m’ and radius ‘a’ are placed with their centres on the four corners of a square of side ‘b’. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is :

The moment of inertia I could be derived from the figure as,

Since the axis of rotation is parallel to the plane of the square, we must determine the system's moment of inertia, I, about one side of the square.

We are aware of the sphere's moment of inertia (radius a and mass m)

Thus, the moment of inertia I is equal to the sum of the moments of inertia of the two spheres and the moments of inertia along the axis of rotation b.

$\mathrm{I}=\frac{2}{5}{\mathrm{ma}}^2+\frac{2}{5}{\mathrm{ma}}^2+\left[\frac{2}{5}{\mathrm{ma}}^2+{\mathrm{mb}}^2\right]+\left[\frac{2}{5}{\mathrm{ma}}^2+{\mathrm{mb}}^2\right]$

$\mathrm{I}=4\times \frac{2}{5}{\mathrm{ma}}^2+2{\mathrm{mb}}^2$

$\mathrm{I}=\frac{8}{5}{\mathrm{ma}}^2+2{\mathrm{mb}}^2$

Hence the correct answer is $\frac{8}{5}{\mathrm{ma}}^2+2{\mathrm{mb}}^2.$