Find the depth of lake at which density of water is $1\%$ greater than at the surface. Given compressibility ........... $km$ . $K = 50\times10^{-6}/atm$ ?
Medium
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$\frac{1}{\mathrm{K}}=\frac{-\Delta \mathrm{P}}{\frac{\Delta \mathrm{V}}{\mathrm{V}}}$ $\left(\therefore \frac{\Delta \mathrm{V}}{\mathrm{V}}=-\frac{\Delta \rho}{\rho}\right)$
$\frac{10^{5}}{50 \times 10^{-6}}=\frac{\rho \mathrm{gh} \times 100}{1}$
$h=2 \mathrm{km}$
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