Question

Find the depth of a lake( in $\text{km}$ ) at which the density of water is $1\%$ greater than at the surface, if compressibility of water is $50\times {10}^{-6}/\text{atm}$

Answer 1

${\rho }^{\prime }=\frac{\rho }{1-\frac{\Delta P}{B}}$
$\Rightarrow \frac{{\rho }^{\prime }}{\rho }\times 100=\frac{100}{1-\Delta P\times 50\times {10}^{-6}}$

$\Rightarrow \frac{101}{100}\times 100=\frac{100}{1-1000\times g\times h\times {10}^{-5}\times 50\times {10}^{-6}}$
$\Rightarrow \frac{101}{100}=\frac{1}{1-5\times h\times {10}^{-6}}$
$1-5\times h\times {10}^{-6}=\frac{100}{101}$
$\Rightarrow h=\frac{{10}^6}{5\times 101}\text{m}=1.98\text{km}$