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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$. For a square of side $L$,the distance of each c

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

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(B) The moment of inertia $I$ of a system of particles about an axis is given by $I = \sum m_i r_i^2$. For a square of side $L$,the distance of each corner from the centre is $r = \frac{L}{\sqrt{2}}$. Since there are four particles of mass $M$,the total moment of inertia about an axis perpendicular to the plane and passing through the centre 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 $I = Mk_{total} k^2$,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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