Question

The depth below the surface of the sea to which a rubber ball is taken so as to decrease its volume by 0.02% is _____ m. (Take density of sea water $= 10^3 \, \text{kg m}^{-3}$, Bulk modulus of rubber $= 9 \times 10^8 \, \text{N m}^{-2}$, and $g = 10 \, \text{m s}^{-2}$)

Answer 1

Given:

$\beta = -\frac{\Delta P}{\Delta V / V}$ $\Delta P = -\beta \frac{\Delta V}{V}$

Pressure difference due to sea water:

$\rho g h = -\beta \frac{\Delta V}{V}$

Substituting the given values:

$10^3 \times 10 \times h = -9 \times 10^8 \times \left(-\frac{0.02}{100}\right)$

Simplifying:

$h = 18 \, \text{m}$