A metallic sphere of radius $1.0 \times {10}^{-3} \mathrm{m} \mathrm{and} \mathrm{density} 1.0\times {10}^4 \mathrm{kg}/{\mathrm{m}}^3$ enter a tank of water, after a free fall through a distance of h in the earth's gravitational field, if its velocity remains unchanged after entering water, determine the value of h. Given: coefficient of viscosity of water = $1.0 \times {10}^{-3} \mathrm{N}-\mathrm{s}/{\mathrm{m}}^2, \mathrm{g} = 10 \mathrm{m}/{\mathrm{s}}^2 \mathrm{and} \mathrm{density} \mathrm{of} \mathrm{water} = 1.0\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3.$

The velocity attained by the sphere in falling freely from a height h is 

                  v = $\sqrt{2\mathrm{gh}}-----\left(\mathrm{i}\right)$

This is the terminal velocity of the sphere in water. Hence by Stoke's law, we have

               v = $\frac{2}{9}\frac{{\mathrm{r}}^2(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}}{\mathrm{\eta }}$

where r is the radius of the sphere, $\rho$ is the density of the material of the sphere

$\mathrm{\sigma }(=1.0\times {10}^3\mathrm{kg}/{\mathrm{m}}^3)$ is the density of water and $\eta$ is coefficient of viscosity of water.

v = $\frac{2\times {(1.0\times {10}^{-3})}^2(1.0\times {10}^4-1.0\times {10}^3)\times 10}{9\times 1.0\times {10}^{-3}}$

  = 20 m/s

from equation (i), we have

h = $\frac{{\mathrm{v}}^2}{2\mathrm{g}} = \frac{20\times 20}{2\times 10} = 20 \mathrm{m}$