Find the viscous force on a steel ball of 2 mm radius ( density 8 g/cc ) acquires a terminal velocity of 4 cm/s in falling freely in the tank of glycerine ( density of glycerine= 1.3 g / cc).

The formula for terminal velocity v is given by
$V=\frac{2g{r}^2\left({\rho }_{sphere}-{\rho }_{medium}\right)}{9\eta }$
where r is the radius of the still ball = 2 mm = $2\times {10}^{-3}$ m, ${\rho }_{sphere}$ is the density of the still ball = 8 g/cc = 8000 kg/m³, and ${\rho }_{medium }$ is density of the medium = 1.3 g/cc = 1300 kg/m³, terminal velocity v is 4 cm/s = 0.04 m/s and $\eta$ is the viscosity.
So viscosity is
$\eta =\frac{2g{r}^2\left({\rho }_{sphere}-{\rho }_{medium}\right)}{9v}=\frac{2\times 9.8\times {\left(2\times {10}^{-3}\right)}^2\times \left(8000-1300\right)}{9\times 0.04}=1.46$ pa s
Now the viscus force is given by $f=6\mathrm{\pi \eta rv}=6\mathrm{\pi }\times 1.46\times 2\times {10}^{-3}\times 0.04=0.002\mathrm{N}$