In Millikan's oil drop experiment , what is the terminal speed of an uncharged drop of radius `2.0xx10^(-5)m` and density `1.2xx10^(3)kgm^(-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 radius of drop `r=2.0xx10^(-5)m`
Density of oil `rho=1.2xx10^(3)kgm^(-3)`
Density of air (drop is of air ) `rho_(o)=0`
The coefficient of viscosity of oil,
`eta=1.8xx10^(-5)` Pas
Terminal velocity from Stock.s law ,
`v_(t)=(2)/(9)(r^(2)g)/(eta)(rho-rho_(o))`
`=(2r^(2)gQ)/(9eta)`
`=(2xx4xx10^(-10)xx9.8xx1.2xx10^(3))/(9xx1.8xx10^(-5))`
`=5.807xx10^(-2)`
`v_(t)=5.8cms^(-1)`
Viscous force on drop from Stoke.s law
`F_(v)=6pietarv_(t)`
`=6xx3.14xx1.8xx10^(-5)xx2xx10^(-5)xx5.8xx10^(-2)`
`=393.34xx10^(-12)`
`=3.93xx10^(-10)N`