Question

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

Answer 1

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