A solid sphere of mass M and radius R is divi | vedclass

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(B) Density $\rho = \frac{M}{\frac{4}{3}\pi R^3}$.Mass of smaller part $m_1 = \frac{M}{8}$. Since $\rho = \frac{m_1}{V_1} = \frac{m_1}{\frac{4}{3}\pi

Vedclass · Accepted Answer

$70$

Vedclass · Answer

(B) Density $ho = \frac{M}{\frac{4}{3}\pi R^3}$.Mass of smaller part $m_1 = \frac{M}{8}$. Since $ho = \frac{m_1}{V_1} = \frac{m_1}{\frac{4}{3}\pi r^3}$, we have $\frac{M}{\frac{4}{3}\pi R^3} = \frac{M/8}{\frac{4}{3}\pi r^3} \Rightarrow r^3 = \frac{R^3}{8} \Rightarrow r = \frac{R}{2}$.$I_1$ (moment of inertia of sphere about its centre) $= \frac{2}{5}m_1r^2 = \frac{2}{5}(\frac{M}{8})(\frac{R}{2})^2 = \frac{2}{5} 	imes \frac{M}{8} 	imes \frac{R^2}{4} = \frac{M R^2}{80}$.Mass of larger part $m_2 = M - \frac{M}{8} = \frac{7M}{8}$.$I_2$ (moment of inertia of disc about its diameter) $= \frac{m_2 R_D^2}{4}$, where $R_D = 2R$.$I_2 = \frac{(\frac{7M}{8})(2R)^2}{4} = \frac{(\frac{7M}{8})(4R^2)}{4} = \frac{7M R^2}{8}$.Ratio $\frac{I_2}{I_1} = \frac{7M R^2 / 8}{M R^2 / 80} = \frac{7}{8} 	imes 80 = 70$.

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