In Millikan's oil drop experiment,what is the terminal speed of an uncharged drop of radius $2.0 \times 10^{-5} \; m$ and density $1.2 \times 10^{3} \; kg \; m^{-3}$? Take the viscosity of air at the temperature of the experiment to be $1.8 \times 10^{-5} \; Pa \; s$. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

(N/A) Given:
Radius $r = 2.0 \times 10^{-5} \; m$
Density $\rho = 1.2 \times 10^{3} \; kg \; m^{-3}$
Viscosity $\eta = 1.8 \times 10^{-5} \; Pa \; s$
Acceleration due to gravity $g = 9.8 \; m \; s^{-2}$
Terminal velocity $v$ is given by the formula:
$v = \frac{2r^2(\rho - \rho_0)g}{9\eta}$
Since we neglect buoyancy,$\rho_0 = 0$.
$v = \frac{2 \times (2.0 \times 10^{-5})^2 \times (1.2 \times 10^3) \times 9.8}{9 \times 1.8 \times 10^{-5}}$
$v = \frac{2 \times 4.0 \times 10^{-10} \times 1.2 \times 10^3 \times 9.8}{16.2 \times 10^{-5}}$
$v = \frac{94.08 \times 10^{-7}}{16.2 \times 10^{-5}} \approx 5.8 \times 10^{-2} \; m \; s^{-1} = 5.8 \; cm \; s^{-1}$
Viscous force $F$ is given by Stokes' Law:
$F = 6 \pi \eta r v$
$F = 6 \times 3.14 \times 1.8 \times 10^{-5} \times 2.0 \times 10^{-5} \times 5.8 \times 10^{-2}$
$F \approx 3.9 \times 10^{-10} \; N$