A disc of radius R and thickness frac{R}{6} h | vedclass

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(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} MR^2$ ... $(i

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$\frac{I}{5}$

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(A) The moment of inertia of a disc about an axis passing through its centre and perpendicular to its plane is given by $I = \frac{1}{2} MR^2$ ... $(i)$ The volume of the disc is $V = \pi R^2 	imes 	ext{thickness} = \pi R^2 	imes \frac{R}{6} = \frac{\pi R^3}{6}$. When the disc is recast into a solid sphere of radius $R_s$,the volume remains constant: $\frac{\pi R^3}{6} = \frac{4}{3} \pi R_s^3$ $R_s^3 = \frac{R^3}{8} \implies R_s = \frac{R}{2}$ ... (ii) The moment of inertia of a solid sphere about its diameter is $I_{	ext{sphere}} = \frac{2}{5} MR_s^2$. Substituting $R_s = \frac{R}{2}$ into the expression for $I_{	ext{sphere}}$: $I_{	ext{sphere}} = \frac{2}{5} M \left(\frac{R}{2}ight)^2 = \frac{2}{5} 	imes \frac{1}{4} MR^2 = \frac{1}{5} \left(\frac{1}{2} MR^2ight)$. Using equation $(i)$,we get $I_{	ext{sphere}} = \frac{I}{5}$.

$A$ disc of radius $R$ and thickness $\frac{R}{6}$ has moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. The disc is melted and recast into a solid sphere. The moment of inertia of the sphere about its diameter is

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