Q 11-09-122JEE MainJEE Main 2021 (27 Aug, Shift 1)Easy
In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0\times10^{-5}$ m and density $1.2\times10^3$ kg m$^{-3}$? Take viscosity of liquid $= 1.8\times10^{-5}$ N s m$^{-2}$. (Neglect buoyancy due to air).
Answer: (B) $3.9\times10^{-10}$ N
An uncharged drop falls at terminal velocity, where the viscous force equals its weight (buoyancy neglected):
$$F = \frac{4}{3}\pi r^3\rho g = \frac{4}{3}\pi(2\times10^{-5})^3\times1.2\times10^3\times9.8 \approx 3.9\times10^{-10}\ \text{N}$$
Solution by Sreeraj P, M.Sc Physics