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(D) The mass of the first part is $M_1 = \frac{7M}{8}$ and it is converted into a disc of radius $R_1 = 2R$.The moment of inertia of a disc about its

Vedclass · Accepted Answer

$140$

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(D) The mass of the first part is $M_1 = \frac{7M}{8}$ and it is converted into a disc of radius $R_1 = 2R$.The moment of inertia of a disc about its central axis is $I_1 = \frac{1}{2} M_1 R_1^2$.Substituting the values: $I_1 = \frac{1}{2} \left( \frac{7M}{8} ight) (2R)^2 = \frac{1}{2} 	imes \frac{7M}{8} 	imes 4R^2 = \frac{7MR^2}{4}$.The mass of the second part is $M_2 = M - \frac{7M}{8} = \frac{M}{8}$.Since the density remains constant,the volume $V_2 = \frac{4}{3} \pi R_2^3 = \frac{1}{8} V_{total} = \frac{1}{8} (\frac{4}{3} \pi R^3)$,which implies $R_2^3 = \frac{R^3}{8}$,so $R_2 = \frac{R}{2}$.The moment of inertia of a solid sphere about its axis is $I_2 = \frac{2}{5} M_2 R_2^2$.Substituting the values: $I_2 = \frac{2}{5} \left( \frac{M}{8} ight) \left( \frac{R}{2} ight)^2 = \frac{2}{5} 	imes \frac{M}{8} 	imes \frac{R^2}{4} = \frac{MR^2}{80}$.Finally,the ratio is $\frac{I_1}{I_2} = \frac{7MR^2 / 4}{MR^2 / 80} = \frac{7}{4} 	imes 80 = 7 	imes 20 = 140$.

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