Four particles each of mass M are placed at the corners of a square of side L . The radius of gyration of the system about an axis perpendicular to the square and passing through its centre is

The system consists of four particles of mass $M$ arranged at the corners of a square of side $L$. The rotation axis is perpendicular to the square's plane and passes through its center.

The distance $r$ from each corner to the center is half the diagonal length. The diagonal is $L\sqrt{2}$, so $r = L/\sqrt{2}$.

The moment of inertia for one particle about this axis is $Mr^2 = M(L^2 / 2) = ML^2 / 2$. For four identical particles, the total moment of inertia is $I_{\text{total}} = 4(ML^2 / 2) = 2ML^2$.

The total mass is $M_{\text{total}} = 4M$. The radius of gyration $k$ satisfies $I_{\text{total}} = M_{\text{total}} k^2$, so $2ML^2 = 4M k^2$.

Solving gives $k^2 = L^2 / 2$, hence $k = L / \sqrt{2}$.

The radius of gyration is $L / \sqrt{2}$.