A steel ball of radius 0.05cm and density $7.8gc{m}^{-3}$ is dropped into a tank of water. The terminal velocity of the steel ball is (Density of water = $1gc{m}^{-3}$ and viscosity of water=0.001 Pa-s)

Given r = 0.05cm $=5\times {10}^{-4}m$
Density of ball $\rho =7.8gc{m}^{-3}=7.8\times {10}^{3}kg/m^3$
Density of water $\sigma =1gc{m}^{-3}={10}^{3}kg{m}^{-3}$
Coefficient of viscosity, $\eta =0.001Pa-s={10}^{-3}Pa-s$
$\therefore$  terminal velocity if ball
$v=\frac{2g{r}^2(\rho -\sigma )}{9\eta }=\frac{2\times 10\times {(5\times {10}^{-4})}^2(7.8\times {10}^3-{10}^3)}{9\times {10}^{-3}}=3.77m{s}^{-1}$