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

For an extended object of mass M, having moment of inertia I about an axis then its radius of gyration about that axis is given by, $\mathrm{k}=\sqrt{\frac{\mathrm{I}}{\mathrm{M}}}$

Moment of inertia of system of particles is : $\mathrm{I}=\sum _{\mathrm{i}} {\mathrm{m}}_{\mathrm{i}}{\mathrm{r}}_{\mathrm{i}}^2$

For the given system,

$\mathrm{I}=4\mathrm{m}{\left(\frac{\mathrm{l}}{\sqrt{2}}\right)}^2$

$\Rightarrow \mathrm{I}=\frac{4{\mathrm{ml}}^2}{2}$

$\Rightarrow I=2m{l}^2$

Now $I=\left(4m\right)k^2$

$\Rightarrow 4{\mathrm{mk}}^2=2{\mathrm{ml}}^2$

$\Rightarrow \mathrm{k}=\frac{\mathrm{l}}{\sqrt{2}}$