Find the elastic potential energy per unit volume of water (in x103 Jm-3) at a depth of 1 km. Given, compressibility of water = 5x10-10 SI units and density of water = 1000 kg m-3.

Solution:

$Energy per unit volume = \frac{1}{2} (stress)(strain)$

$\begin{array}{l} u = \frac{U}{V} = \frac{1}{2} (stress) (\frac{stress}{Bulk modulus}) \\ = \frac{1}{2} (ρ g h)^{2} (Compressibility) \\ = \frac{1}{2} (1000 \times 9.8 \times 1000)^{2} (5 \times 10^{- 10}) \\ = 2.4 \times 10^{4} {Jm}^{- 3} = 24 \times 10^{3} {Jm}^{- 3} \end{array}$

Find the elastic potential energy per unit volume of water (in x10³ Jm^-3) at a depth of 1 km. Given, compressibility of water = 5x10^-10 SI units and density of water = 1000 kg m^-3.

Solution:

Related content