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A solid spherical ball and a hollow spherical ball of two different materials of densities $\rho_{1}$ and $\rho_{2}$ respectively have same outer radii and same mass. What will be the ratio, the moment of inertia (about an axis passing through the centre) of the hollow sphere to that of the solid sphere?
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The correct answer is:
$\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$
$\therefore \quad M_{1}=\rho_{1} \frac{4}{3} \pi R^{3}$ and $M_{2}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\therefore \quad \rho_{1} \frac{4}{3} \pi R^{3}=\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \rho_{1} R^{3}=\rho_{2}\left(R^{3}-r^{3}\right)$
$\Rightarrow \quad \frac{\rho_{1}}{\rho_{2}}=1-\frac{r^{3}}{R^{3}}$
$\therefore \frac{r}{R}=\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{1}{3}}$
Moment of inertia, $\begin{aligned} \frac{I_{M}}{I_{S}} &=\frac{\frac{2}{5} M_{2}\left[\frac{R^{5}-r^{3}}{R^{3}-r^{3}}\right]}{\frac{2}{5} M_{1} R^{2}} \\ &=\frac{\rho_{2} \frac{4}{3} \pi\left(R^{3}-r^{3}\right)\left[\frac{R^{5}-r^{5}}{\left(R^{3}-r^{3}\right)}\right]}{\rho_{1} \frac{4}{3} \pi R^{3} R^{2}} \\ &=\frac{\rho_{2}}{\rho_{1}}\left[\frac{R^{5}-r^{5}}{R^{5}}\right]=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(\frac{r}{R}\right)^{3}\right] \end{aligned}$
$\therefore \quad \frac{I_{H}}{I_{s}}=\frac{\rho_{2}}{\rho_{1}}\left[1-\left(1-\frac{\rho_{1}}{\rho_{2}}\right)^{\frac{5}{3}}\right]$
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