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In Millikan’s oil drop experiment, what ...

In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius `2.0xx 10^5` m and density `1.2xx 10^3 kg m^-3`. Take the viscosity of air at the temperature of the experiment to be `1.8xx 1 0^-5` Pa s. How much is the viscous force on the drop at that speed ? Neglect buoyancy of the drop due to air

Text Solution

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`r=2xx10^(-5)m,rho=1.2xx10^(3)kgm^(-3)`
`eta=1.8xx10^(-5)` Pa S
desity of air, `d=1.293kgm^(-3)`
Using relation : `v=(2)/(9)(r^(2)(rho-d)g)/(eta)`
`v=(2xx(2xx10^(-5))^(2)(1.2xx10^(3)-1.293)xx9.8)/(9xx1.8xx10^(-5))`
`v=0.058ms^(-1)` Also, viscous force `F=6pietarv`
`F=6xx3.14xx2xx10^(-5)xx0.058xx1.8xx10^(-5)=3.93xx10^(-10)N`.
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