Four small spheres each of radius ‘r’ and mass ‘m’ are placed with their centres at the four corners of a square of side ‘L’. The M.I. of the system about a diagonal of the square is

There are 4 spheres at the corners of the square.

Moment of inertia (M.I.) has two parts:The sphere’s own M.I. = (2/5)mr² (about its center)

If it's away from the axis, we add md² using the parallel axis theorem.

Now, we take the diagonal of the square as the axis.

Two spheres lie on the diagonal → distance d = 0, only self M.I. counts.

Two spheres lie off the diagonal → distance d = L/√2

Calculation:

For 2 spheres on the diagonal:M.I. = 2 × (self M.I.) = 2 × (5/16)mr² = (5/8)mr²

For 2 spheres off the diagonal:Distance from diagonal = L/√2

M.I. = 2 × m × (L²/2) = mL²

Total Moment of Inertia:

M.I. = (5/8)mr² + mL²