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(D) To find the moment of inertia of the square frame about an axis passing through its center $O$ and perpendicular to its plane,we use the parallel

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$\frac{4 ML^2}{3}$

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(D) To find the moment of inertia of the square frame about an axis passing through its center $O$ and perpendicular to its plane,we use the parallel axis theorem for each rod.For a single rod of mass $M$ and length $L$,the moment of inertia about an axis passing through its center of mass and perpendicular to its length is $I_{cm} = \frac{ML^2}{12}$.The distance from the center of the square to the center of each rod is $d = \frac{L}{2}$.Using the parallel axis theorem,the moment of inertia of one rod about the axis passing through the center of the square is $I_{rod} = I_{cm} + Md^2 = \frac{ML^2}{12} + M(\frac{L}{2})^2 = \frac{ML^2}{12} + \frac{ML^2}{4} = \frac{ML^2 + 3ML^2}{12} = \frac{4ML^2}{12} = \frac{ML^2}{3}$.Since there are four such rods,the total moment of inertia of the system is $I_{total} = 4 	imes I_{rod} = 4 	imes \frac{ML^2}{3} = \frac{4ML^2}{3}$.

Four thin metal rods,each of mass $M$ and length $L$,are welded end to end to form a square. The moment of inertia of the system about an axis passing through the centre of the square and perpendicular to its plane is

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