A steel ball of radius 0.05cm & density $7.8g/c{m}^{-3}$ is dropped into a tank of water the terminal velocity of the stell ball is [Density of water = $1gc{m}^{-3}\&$ viscosity of water = 0.001pas]

$V_T=\frac{2r^2(\rho -{\rho }_0)g}{9n}$ 
$n=0.001={10}^{-3}$ pa-s
$r=0.05\times {10}^{-2}m=5\times {10}^{-4}m$
$\begin{array}{l}{P}_{steel}=7.8\times {10}^{3}kg/m^3\\ P_w={10}^{3}kg/m^3\end{array}$
$V_T=\frac{2{(5\times {10}^{-4})}^2(7.8-1){10}^3\times 10}{9\times {10}^{-3}}$
$=\frac{2\times 25\times {10}^{-8}6.8\times {10}^4}{9\times {10}^{-3}}$
$=\frac{50\times {10}^{-4}\times 6.8}{9\times {10}^{-3}}=\frac{5\times 68\times {10}^{-4}}{9\times {10}^{-3}}=\frac{340}{9}\times {10}^{-1}$
$V_T=3.77m/s$