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(D) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the

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$\frac{L}{\sqrt{2}}$

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(D) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$,where $r_i$ is the perpendicular distance of the $i$-th particle from the axis.For a square of side $L$,the distance of each corner from the centre is $r = \frac{	ext{diagonal}}{2} = \frac{\sqrt{2}L}{2} = \frac{L}{\sqrt{2}}$.Since there are four particles each of mass $M$,the total moment of inertia $I$ about the axis passing through the centre and perpendicular to the plane is:$I = 4 	imes M 	imes r^2 = 4 	imes M 	imes (\frac{L}{\sqrt{2}})^2 = 4 	imes M 	imes \frac{L^2}{2} = 2ML^2$.The radius of gyration $k$ is defined by the relation $I = Mk_{total} k^2$,where $M_{total} = 4M$.So,$2ML^2 = (4M)k^2$.$k^2 = \frac{2ML^2}{4M} = \frac{L^2}{2}$.$k = \frac{L}{\sqrt{2}}$.

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:

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