A solid sphere of radius R has a moment of in | vedclass

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(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is $I = \frac{2}{5} MR^2$.When it is melted into a disc of m

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$\frac{2}{\sqrt{15}} R$

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(A) The moment of inertia of a solid sphere of mass $M$ and radius $R$ about its diameter is $I = \frac{2}{5} MR^2$.When it is melted into a disc of mass $M$ and radius $r$,the moment of inertia of the disc about its central axis (perpendicular to the plane) is $I_{cm} = \frac{1}{2} Mr^2$.According to the parallel axis theorem,the moment of inertia about a tangential axis perpendicular to the plane of the disc is $I_{tangent} = I_{cm} + Mr^2 = \frac{1}{2} Mr^2 + Mr^2 = \frac{3}{2} Mr^2$.Given that $I_{tangent} = I$,we have $\frac{3}{2} Mr^2 = \frac{2}{5} MR^2$.Solving for $r$: $r^2 = \frac{2}{5} 	imes \frac{2}{3} R^2 = \frac{4}{15} R^2$.Therefore,$r = \sqrt{\frac{4}{15}} R = \frac{2}{\sqrt{15}} R$.

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