A drop of water of radius 0.0015 mm is falling in air. If the coefficient of viscosity of air is 1.8 $\times {10}^{-5} \mathrm{kg}/(\mathrm{m}/\mathrm{s}),$what will be the terminal velocity of the drop? (density of water = $1.0\times {10}^3 \mathrm{kg}/{\mathrm{m}}^2$ and g = $9.8 \mathrm{m}/{\mathrm{s}}^2)$ Density of air can be neglected.

By Stokes' law, the terminal velocity of a water drop of radius r is given by

v = $\frac{2}{9}\frac{{\mathrm{r}}^2(\mathrm{\rho }-\mathrm{\sigma })\mathrm{g}}{\mathrm{\eta }}$

where $\mathrm{\rho }$ is the density of water, $\mathrm{\sigma } \mathrm{is} \mathrm{the} \mathrm{density} \mathrm{of} \mathrm{air} \mathrm{and} \mathrm{\eta } \mathrm{is} \mathrm{the} \mathrm{coefficient} \mathrm{of} \mathrm{viscosity} \mathrm{of} \mathrm{air}.$$\mathrm{Here} \mathrm{\sigma } \mathrm{is} \mathrm{negligible} \mathrm{and} \mathrm{r} = 0.0015 \mathrm{mm} = 1.5\times {10}^3\mathrm{mm} = 1.5\times {10}^{-6} \mathrm{m}. \mathrm{Substituting} \mathrm{the} \mathrm{values}:$

$\mathrm{v} = \frac{2}{9}\times \frac{{(1.5\times {10}^{-6})}^2\times (1.0\times {10}^3)\times 9.8}{1.8\times {10}^{-5}}$

   = $2.72 \times {10}^{-4} \mathrm{m}/\mathrm{s}$