Question

Calculate the approximate change in density of water(in $kg/m^3$) in a lake at a depth of $400m$ below the surface. The density of water at the surface is $1030kg/m^3$ and bulk modulus of water is $2\times {10}^{9}N/m^2$.

Answer 1

${\rho }^{{}^{\prime }}\frac{\rho }{1-\frac{dp}{B}}=\rho (1+\frac{dp}{B})$
$\therefore \Delta \rho ={\rho }^{{}^{\prime }}-\rho =\frac{\rho \left(dp\right)}{B}$
$=\frac{\rho \left(\rho gh\right)}{B}$
$=\frac{(1030{)}^2\times 9.8\times 400}{2\times {10}^9}$
$=2.0kg/m^3$