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 axis of rotation passes perpendicularly through the center of the square with side length $L$. Each particle of mass $M$ lies at distance $r = L/\sqrt{2}$ from the center, equal to half the diagonal length.

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

Using the relationship $I_{\text{total}} = (4M)K^2$ for the system’s total mass $4M$ and radius of gyration $K$, we substitute $2ML^2 = 4MK^2$.

Solving yields $K^2 = L^2/2$, so $K = L/\sqrt{2}$.

The radius of gyration is $\boxed{\text{B}}$.