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

Correct Answer: 18

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:

\[ h = 18 \, \text{m} \]