A circular disc of radius R and thickness R/8 | vedclass

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(C) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R_d^2$. Since the mass $M$ remains constant,we equate the volumes: $V

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$\frac{I}{5}$

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(C) The moment of inertia of a disc about its central axis is $I = \frac{1}{2} M R_d^2$. Since the mass $M$ remains constant,we equate the volumes: $V_{	ext{disc}} = V_{	ext{sphere}}$ $\pi R_d^2 	imes (R_d/8) = \frac{4}{3} \pi R_s^3$ $\frac{R_d^3}{8} = \frac{4}{3} R_s^3 \implies R_s^3 = \frac{3}{32} R_d^3 \implies R_s^2 = \left(\frac{3}{32}ight)^{2/3} R_d^2$. The moment of inertia of a solid sphere about its diameter is $I_{	ext{sphere}} = \frac{2}{5} M R_s^2$. Substituting $R_s^2$: $I_{	ext{sphere}} = \frac{2}{5} M \left(\frac{3}{32}ight)^{2/3} R_d^2$. Since $M R_d^2 = 2I$,we have: $I_{	ext{sphere}} = \frac{2}{5} (2I) \left(\frac{3}{32}ight)^{2/3} = \frac{4I}{5} \left(\frac{9}{1024}ight)^{1/3} \approx \frac{I}{5}$.

$A$ circular disc of radius $R$ and thickness $R/8$ has a moment of inertia $I$ about an axis passing through its centre and perpendicular to its plane. It is melted and recast into a solid sphere. The moment of inertia of the sphere about an axis passing through its diameter is:

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