In Millikan's oil experiment, what is the terminal speed of a speed of a drop of radius `2.0 xx 10^(5)m` and density `1.2xx10^(3)m^(-3)`? Take the viscosity of air at the temperature of the expermental to be `1.8xx10^(-5)Nsm^(2)`. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Explanation

Here, `r=2.0xx10^(-5)m , rho = 1.2 xx10^(3)kgm^(-3) , eta = 1.8xx10^(-5)Nsm^(-2), sigma = 0, upsilon = ? , F = ?`
Terminal velocity `upsilon = (2r^(2)(rho-sigma)g)/(9eta) = (2xx(2.0 xx 10^(-5))^(2)(1.2xx10^(3)-0)xx9.8)/(9xx1.8xx10^(-5))`
`=5.8xx10^(-2)ms^(-1)=5.8cms^(-1)`
Viscous force on the drop, `F = 6 pi eta r upsilon`
`=6xx22/7 xx (1.8xx10^(-5))xx(2.0xx10^(-5)) xx (5.8xx10^(-2)) = 3.93xx10^(-10)N`.