In Millikan's oil drop 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 experimental 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 radius of drop `r=2.0xx10^(-5)m` density of drop `p=1.2xx10^(3)kg//m^(3)`, viscosity of air `Tl=1.8xx10^(-5)Pa-s`
Neglecting upward thrust due to air, we find the terminal speed is
`v_(T)=(2)/(9)(r^(2)rhog)/(eta)=(2xx(2.0xx10^(-5))^(2)xx(1.2xx10^(3))xx9.8)/(9xx(1.8xx10^(-5)))`
`=5.81xx10^(-2)ms^(-1)` or `5.81cms^(-1)`
viscous force at this speed
`F=6pietarv=6xx3.14xx(1.8xx10^(-5))xx(2.0xx10^(-5))xx(5.81xx10^(-2))`
`=3.94xx10^(-10)N`