Three bodies have equal masses m. Body A is a | vedclass

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(B) The moment of inertia for each body about an axis passing through its center of mass is as follows:For body $A$ (solid cylinder of radius $R$): $I

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$B$

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(B) The moment of inertia for each body about an axis passing through its center of mass is as follows:For body $A$ (solid cylinder of radius $R$): $I_{A} = \frac{1}{2} mR^{2} = 0.5 mR^{2}$.For body $B$ (square lamina of side $R$): The moment of inertia about an axis passing through the center and perpendicular to the plane is $I_{B} = \frac{m}{12}(R^{2} + R^{2}) = \frac{2mR^{2}}{12} = \frac{1}{6} mR^{2} \approx 0.167 mR^{2}$.For body $C$ (solid sphere of radius $R$): $I_{C} = \frac{2}{5} mR^{2} = 0.4 mR^{2}$.Comparing the values: $0.5 mR^{2} > 0.4 mR^{2} > 0.167 mR^{2}$.Thus,$I_{A} > I_{C} > I_{B}$.Therefore,body $B$ has the smallest moment of inertia.

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