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 of a solid sphere about an axis passing through its center is given by (2/5)mr^2. However, in this case, the axis of rotation is not passing through the center of the spheres. Therefore, we need to use the parallel axis theorem which states that the moment of inertia about any axis parallel to and a distance d away from an axis through the center of mass is given by I = I_cm + md^2.

Step 1: Calculate the moment of inertia of one sphere about its center. I_cm = (2/5)ma^2

Step 2: Calculate the distance from the center of the sphere to the axis of rotation. The spheres are placed at the corners of a square of side 'b'. Therefore, the distance d from the center of the sphere to the side of the square (which is the axis of rotation) is d = b/2.

Step 3: Use the parallel axis theorem to calculate the moment of inertia of one sphere about the axis of rotation. I = I_cm + md^2 I = (2/5)ma^2 + m(b/2)^2

Step 4: Since there are four identical spheres, the total moment of inertia of the system is four times the moment of inertia of one sphere. I_total = 4I I_total = 4[(2/5)ma^2 + m(b/2)^2]

So, the moment of inertia of the system about one side of the square where the axis of rotation is parallel to the plane of the square is 4[(2/5)ma^2 + m(b/2)^2].