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:

Four identical solid spheres, each of mass $$m$$ and radius $$a$$, are placed with their centres at the four corners of a square of side $$b$$. We need to find the moment of inertia about one side of the square.

Consider the axis along one side of the square. Two spheres have their centres on this axis (at distance 0 from it), and two spheres have their centres at a perpendicular distance $$b$$ from this axis.

The moment of inertia of a solid sphere about its own diameter is $$I_0 = \frac{2}{5}ma^2$$.

For the two spheres whose centres lie on the axis, by the parallel axis theorem their contribution is simply $$2 \times \frac{2}{5}ma^2 = \frac{4}{5}ma^2$$ (no additional parallel axis term since the distance is zero).

For the two spheres whose centres are at distance $$b$$ from the axis, using the parallel axis theorem: each contributes $$\frac{2}{5}ma^2 + mb^2$$. Together they give $$2\left(\frac{2}{5}ma^2 + mb^2\right) = \frac{4}{5}ma^2 + 2mb^2$$.

The total moment of inertia is $$I = \frac{4}{5}ma^2 + \frac{4}{5}ma^2 + 2mb^2 = \frac{8}{5}ma^2 + 2mb^2$$.

Hence the correct answer is Option 3: $$\frac{8}{5}ma^2 + 2mb^2$$.