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A body of volume $\mathrm{V}$ floats on water with $\frac{1}{3}$ of its volume above the surface. The volume of the object above the surface when floating on a liquid of specific gravity 1.5 is
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The correct answer is:
$\frac{5 \mathrm{~V}}{9}$
We have
$\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\rho}{\rho \hat{l}}$,
$\mathrm{V}^{\prime}=$ Volume of substance inside the liquid
$\begin{aligned} & \text { So, } \frac{2}{3}=\frac{\rho}{1000} \\ & \rho=\frac{2000}{3} \mathrm{~kg} / \mathrm{m}^3\end{aligned}$
When, $\rho_{\hat{l}}=1500 \mathrm{~kg} / \mathrm{m}^3, \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{2000}{3 \times 1500} \Rightarrow \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{4}{9}$
$\Rightarrow \mathrm{V}^{\prime}=-\mathrm{V}$
So, volume above the surface
$\mathrm{V}-\frac{4}{9} \mathrm{~V}=\frac{5}{9} \mathrm{~V}$
$\frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{\rho}{\rho \hat{l}}$,
$\mathrm{V}^{\prime}=$ Volume of substance inside the liquid
$\begin{aligned} & \text { So, } \frac{2}{3}=\frac{\rho}{1000} \\ & \rho=\frac{2000}{3} \mathrm{~kg} / \mathrm{m}^3\end{aligned}$
When, $\rho_{\hat{l}}=1500 \mathrm{~kg} / \mathrm{m}^3, \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{2000}{3 \times 1500} \Rightarrow \frac{\mathrm{V}^{\prime}}{\mathrm{V}}=\frac{4}{9}$
$\Rightarrow \mathrm{V}^{\prime}=-\mathrm{V}$
So, volume above the surface
$\mathrm{V}-\frac{4}{9} \mathrm{~V}=\frac{5}{9} \mathrm{~V}$
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