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A block of wood of volume $V$ floats in water with half of its volume submerged. The same block floats in an oil with $0.8 \mathrm{~V}$ volume submerged. If the density of water is $1000 \mathrm{~kg} \mathrm{~m}^{-3}$, then the density of the oil is
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Verified Answer
The correct answer is:
$625 \mathrm{~kg} \mathrm{~m}^{-3}$
Given, density of water, $\rho_w=1000 \mathrm{kgm}^{-3}$
For the block of wood in water, volume of immersed part
Total volume of block of wood
$$
\begin{aligned}
& =\frac{\text { Density of wood }\left(\rho_s\right)}{\text { density of water }\left(\rho_w\right)} \\
\Rightarrow \frac{V}{V} & =\frac{\rho_s}{1000} \Rightarrow \frac{1}{2}=\frac{\rho_s}{1000} \\
\Rightarrow \quad \rho_s & =500 \mathrm{kgm}^{-3}
\end{aligned}
$$
Similarly, for the block of wood in oil
$$
\begin{aligned}
\frac{0.8 \mathrm{~V}}{V} & =\frac{\rho_s}{\rho_{\text {oil }}} \\
\Rightarrow \quad \rho_{\text {oil }} & =\frac{\rho_s}{0.8}=\frac{500}{0.8} \\
& =625 \mathrm{kgm}^{-3}
\end{aligned}
$$
For the block of wood in water, volume of immersed part
Total volume of block of wood
$$
\begin{aligned}
& =\frac{\text { Density of wood }\left(\rho_s\right)}{\text { density of water }\left(\rho_w\right)} \\
\Rightarrow \frac{V}{V} & =\frac{\rho_s}{1000} \Rightarrow \frac{1}{2}=\frac{\rho_s}{1000} \\
\Rightarrow \quad \rho_s & =500 \mathrm{kgm}^{-3}
\end{aligned}
$$
Similarly, for the block of wood in oil
$$
\begin{aligned}
\frac{0.8 \mathrm{~V}}{V} & =\frac{\rho_s}{\rho_{\text {oil }}} \\
\Rightarrow \quad \rho_{\text {oil }} & =\frac{\rho_s}{0.8}=\frac{500}{0.8} \\
& =625 \mathrm{kgm}^{-3}
\end{aligned}
$$
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