In Millikan’s oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0\times {10}^{-5}\mathrm{m}$ and density $1.2\times {10}^3{ \mathrm{kg} \mathrm{m}}^{-3}?$ Take viscosity of liquid $=1.8\times {10}^{-5}{\mathrm{Nsm}}^{-2}$. (Neglect buoyancy due to air).

Given, radius of oil drop, $\mathrm{r}=2.0\times {10}^{-5} \mathrm{m}$

Density of oil drop, $\mathrm{\rho }=1.2\times {10}^3{ \mathrm{kg} \mathrm{m}}^{-3}$

Viscosity of liquid, $\mathrm{\eta }=1.8\times {10}^{-5}{ \mathrm{Ns} \mathrm{m}}^{-2}$

To neglect buoyancy, consider the density of air, ${\mathrm{\rho }}_{\mathrm{a}}=0$

Viscous force acting on drop can be given as

        $\mathrm{F}=6\mathrm{\pi \eta rv} ...\left(\mathrm{i}\right)$

Here, v is terminal velocity.

The terminal velocity of drop can be given as

$\begin{array}{l} \mathrm{v}=\frac{2{\mathrm{r}}^2\left(\mathrm{\rho }-{\mathrm{\rho }}_{\mathrm{a}}\right)\mathrm{g}}{9\mathrm{\eta }}\\ \Rightarrow \mathrm{v}=\frac{2\times {\left(2.0\times {10}^{-5}\right)}^2\left(1.2\times {10}^3-0\right)\times 9.8}{9\times 1.8\times {10}^{-5}}\\ =5.807\times {10}^{-2}{\mathrm{ms}}^{-1}\end{array}$

Substituting all values in Eq. (i), we get

           $$\begin{gathered} \mathrm{F}=6\mathrm{\pi }\times 1.8\times {10}^{-5}\times 2.0\times {10}^{-5}\times 5.807\times {10}^{-2} \\ \begin{array}{l} =3.94\times {10}^{-10}\mathrm{N}\\ \simeq 3.9\times {10}^{-10}\mathrm{N}\end{array} \end{gathered}$$