In Millikan's oil experiement, what is the terminal velocity of a liquid droplet of radius 0.02 mm and density `1.2 g//c c`. Take the viscosity of air the temperature of the experiment to be `1.8 xx 10^(-5)Nm^(-2)s`. Find the viscous force on the broplet at the given velocity. Neglect the buoyancy of the droplet due to air.

Explanation

Here, `r=0.02 mm = 0.02 xx 10^(-3)m`
`= 2 xx 10^(-5) m`,
`rho = 1.2 g//c c = 1.2 xx 10^(3) kg m^(-3)`
`eta = 1.8 xx 10^(-5) Nm^(-2) s, sigma =0`
Terminal velocity , `upsilon = 2/9 (r^(2) (rho-sigma)g)/(eta)`
`2/9 xx ((2xx10^(-5))^(2) xx (1.2 xx 10^(3)-0)xx9.8)/(1.8 xx 10^(-5))`
`= 5.81 xx 10^(-2) ms^(-1)`
Viscous force on the droplet is
`F= 6 pi eta r upsilon = 6 xx 3.142 xx (1.8 xx 10^(-5))`
` xx (2.0 xx 10^(-5)) xx (5.81 xx 10^(-2))`
`= 3.94 xx 10^(-10)N`.