In Millikan's oil drop experiment what is the therminal sped of an uncharged drop of radius `2.0xx10^(-5)` m and density `1.2xx10^(3)kg m^(-3)`. Take the viscosity of air at the temperature of the experiment to be `1.8xx10^(5)` Pa s.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=1.2xx10^(3)kgm-3,eta=1.8xx10^(-5)Nsm^(-2)`
`P_(O)=0, V=?, F=?`
Terminal velocity V= `(2r^(2)(rho-rho_(o))g)/(9eta)`
`=(2xx(2.0xx10^(-5))^(2)xx(1.2xx10^(3)-0)xx9.8)/(9xx1.8xx10^(-5))`
`=5.8xx10^(-2)ms^(-1)=5.8cms^(-1)`
Viscous force on the drop, `F=6pietarv`
`=6xx(22)/(7)xx(1.8xx10^(-5))xx(2.0xx10^(-5))xx(5.8xx10^(-2))`
`=3.93xx10^(-10)N`.