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

`r=2.0 xx 10^(-5)m`
`p=1.2 xx 10^(3)"kg m"^(-3)`
`eta=1.8 xx 10^(-5) Nsm^(-2)`
Terminal speed =v
`=(2r^(2) (p-sigma)g)/(9 eta)`
`=(2 xx (2 xx 10^(-5))^(2) (1.2 xx 10^(3) -0)9.8)/(9 xx 1.8 xx 10^(-5))`
`=5.8 xx 10^(-2) ms^(-1)`
`=5.8 cm s^(-1)`
`F=6 pi eta rv`
`=6 xx 3.14 xx 1.8 xx 10^(-5) xx 2 xx 10^(-5) xx 5.8 xx 10^(-2)`
`=3.93 xx 10^(-10)N`