A metallic shpere of radius $1.0\times 10^{-3}m$ and density $1.0\times 10^{4}kg/m^3$ enters a tank of water after a free fall, falling 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 = $1.0\times 10^{-3}N-s/m^2,g=10m/s^2$ and density of water = $1.0\times 10^{3}kg/m^3$]

The velocity attained by the sphere in falling freely from a height h is
$v=\sqrt{2gh}$                             ............(i)
This is the terminal velocity of the sphere in water Hence by Stoke's law we have 
$v=\frac{2}{9}\frac{r^2(\rho -\sigma )}{\eta }$
Where r is the radius of the sphere $\rho$ is the density of the material of the sphere $\sigma (=1.0\times 10^{3}kg/m^3)$ is the density of water and $\eta$ is coefficient of viscosity of water
$\therefore$$v=\frac{2\times (1.0\times 10^{-3}{)}^2(1.0\times 10^4-1.0\times 10)}{9\times 1.0\times 10^{-3}}=20m/s$
from equation (i) we have
$h=\frac{v^2}{2g}=\frac{20\times 20}{2\times 10}=20m$