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

Explanation

By stoke.s law the terminal velocity.
`v=(2)/(9)*(r^(2)(rho-sigma)g)/(eta)`
where, `rho` is density of oil, `sigma` is density of air and `eta` the coefficient of viscosity of air. As buoyancy of the drop due to air is neglected i.e. `sigma=0`.
`:." "v=(2)/(9)*(r^(2)rho g)/(eta)`
Putting the given value, we get
`v=(2)/(9)xx((2xx10^(-5))^(2)xx(1.2xx10^(3))xx9.8)/(1.8xx10^(-5))`
`=5.8xx10^(-2)" m/s"`
The viscous force on the drop at this terminal speed is :
`F=6 pi eta rv`
`F=6xx3.14xx1.8xx10^(-5)xx2xx10^(-5)xx5.8xx10^(-2)`
`F.93xx10^(-10)N`.