Question

Find the depth of a lake at which density of water is 1% greater than at the surface. Given compressibility K = 50 x ${10}^{-6}$ /atm

Answer 1

$B=\frac{\Delta p}{-\frac{\Delta V}{V}}$ $\frac{\Delta V}{V}=-\frac{\Delta P}{B}$
m = $\rho$V = const.
$d\rho V+dV.\rho =0$ $\Rightarrow$ $\frac{d\rho }{\rho }=-\frac{dV}{V}$
i.e. $\frac{\Delta \rho }{\rho }=\frac{\Delta P}{B}$ $\Rightarrow$ $\frac{\Delta \rho }{\rho }=\frac{1}{100}$
$\frac{1}{100}=\frac{h\rho g}{B}$ [assuming $\rho$ = const.]
h$\rho$g=$\frac{B}{100}=\frac{1}{100K}$ $\Rightarrow$ h$\rho$ g = $\frac{1\times {10}^5}{100\times 50\times {10}^{-6}}$

h = $\frac{{10}^5}{5000\times {10}^{-6}\times 1000\times 10}$ = $\frac{100\times {10}^3}{50}=2$ km