A sphere of radius R is cut from a larger sol | vedclass

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(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \fr

Vedclass · Accepted Answer

$\frac{7}{57}$

Vedclass · Answer

(C) Let the density of the sphere be $\rho$. The mass of the larger sphere of radius $2R$ is $M = \rho \cdot \frac{4}{3} \pi (2R)^3 = 8 \rho \cdot \frac{4}{3} \pi R^3$. Let $m_0 = \rho \cdot \frac{4}{3} \pi R^3$ be the mass of the smaller sphere. Then $M = 8m_0$,so $m_0 = M/8$.The moment of inertia of the larger sphere about the $Y$-axis (passing through its center) is $I_1 = \frac{2}{5} M (2R)^2 = \frac{8}{5} MR^2$.The smaller sphere has radius $R$ and its center is at $x = R$. Its moment of inertia about the $Y$-axis is calculated using the parallel axis theorem: $I_2 = I_{cm} + m_0 d^2 = \frac{2}{5} m_0 R^2 + m_0 R^2 = \frac{7}{5} m_0 R^2$.Substituting $m_0 = M/8$,we get $I_2 = \frac{7}{5} (M/8) R^2 = \frac{7}{40} MR^2$.The moment of inertia of the remaining part is $I_{rem} = I_1 - I_2 = \frac{8}{5} MR^2 - \frac{7}{40} MR^2 = \frac{64 - 7}{40} MR^2 = \frac{57}{40} MR^2$.The ratio of the moment of inertia of the smaller sphere to that of the rest part is $\frac{I_2}{I_{rem}} = \frac{7/40 MR^2}{57/40 MR^2} = \frac{7}{57}$.

$A$ sphere of radius $R$ is cut from a larger solid sphere of radius $2R$ as shown in the figure. The ratio of the moment of inertia of the smaller sphere to that of the rest part of the sphere about the $Y$-axis is

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