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(A) The moment of inertia $I$ of the system about an axis perpendicular to the square and passing through its centre is given by $I = \sum m r^2$.The

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$\frac{l}{\sqrt{2}}$

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(A) The moment of inertia $I$ of the system about an axis perpendicular to the square and passing through its centre is given by $I = \sum m r^2$.The distance $r$ of each particle from the centre of the square is half the diagonal length,$r = \frac{\sqrt{2}l}{2} = \frac{l}{\sqrt{2}}$.Since there are four particles of mass $m$,$I = 4 	imes m 	imes (\frac{l}{\sqrt{2}})^2 = 4 	imes m 	imes \frac{l^2}{2} = 2ml^2$.The total mass of the system is $M = 4m$.The radius of gyration $k$ is defined as $k = \sqrt{\frac{I}{M}}$.Substituting the values,$k = \sqrt{\frac{2ml^2}{4m}} = \sqrt{\frac{l^2}{2}} = \frac{l}{\sqrt{2}}$.

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

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