The moment of inertia (M.I.) of four bodies,h | vedclass

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(C) For a thin circular ring of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{1} = \frac{1}{2} MR^{2}$.For a circular disc o

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$I_{1} = I_{2} = I_{3} > I_{4}$

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(C) For a thin circular ring of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{1} = \frac{1}{2} MR^{2}$.For a circular disc of mass $M$ and radius $R$,the moment of inertia about an axis perpendicular to the disc and passing through the centre is $I_{2} = \frac{1}{2} MR^{2}$.For a solid cylinder of mass $M$ and radius $R$,the moment of inertia about its central axis is $I_{3} = \frac{1}{2} MR^{2}$.For a solid sphere of mass $M$ and radius $R$,the moment of inertia about its diameter is $I_{4} = \frac{2}{5} MR^{2}$.Comparing these values,we see that $I_{1} = I_{2} = I_{3} = 0.5 MR^{2}$ and $I_{4} = 0.4 MR^{2}$.Therefore,$I_{1} = I_{2} = I_{3} > I_{4}$.

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