A metallic sphere 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 through a distance of $h$ in the earth's gravitational field. If its velocity remains unchanged after entering water, determine the value of $h$ (in $m$ ).
Given: coefficient of viscosity of water $=1.0\times {10}^{-3}Ns/m^2$, g=10~m/s^2 and density of water $=1.0\times {10}^{3}kg/m^3$.

Given, radius of the sphere,
$r=1.0\times {10}^{-3}m$
Density of the material of the sphere,
$\sigma =1.0\times {10}^{3}kg/m^3$
coefficient of viscosity of water,
$\eta =1.0\times {10}^{-3}N-s/m^2$
and density of water,
$\rho =1.0\times {10}^{3}kg/m^3$
The velocity attained by the sphere in falling freely from a height $h$ is
$v=\sqrt{2gh}\dots \left(i\right)$
This is the terminal velocity of the sphere in water. Hence by Stokes's law, we have
$v=\frac{2}{9}\frac{r^2(\rho -\sigma )g}{\eta }$
$v=\frac{2\times (1.0\times {10}^{-3}{)}^2(1.0\times {10}^4-1\cdot 0\times {10}^3)\times 10}{9\times 1.0\times {10}^{-3}}$
$v=20m/s$
from equation $\left(i\right)$, we have
$h=\frac{v^2}{2g}=\frac{20\times 20}{2\times 10}$
$h=20m$
Final answer: $\left(20\right)$