Question

Find the elastic potential energy per unit volume of water (in $\times 10^{3}J{m}^{-3}$ ) at a depth of $1km$. Given, compressibility of water $=5\times 10^{10}$ SI units and density of water = $1000kg{m}^{-3}$.

Answer 1

Answer: 24

Energy per unit volume $=\frac{1}{2}$ (stress)(strain)
$u=\frac{U}{V}=\frac{1}{2}($ stress $)\left(\frac{\text{ stress }}{\text{ Bulk modulus }}\right)$
$=\frac{1}{2}(\rho gh{)}^2$ (Compressibility)
$=\frac{1}{2}(1000\times 9.8\times 1000{)}^2\left(5\times 10^{-10}\right)$
$=2.4\times 10^{4}J{m}^{-3}=24\times 10^{3}J{m}^{-3}$