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In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×105 m and density 1.2×103kgm3? Take the viscosity of air at the temperature of the experiment to be 1.8×105 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.


Solution

Terminal speed = 5.8 cm/s
Viscous force =3.9×1010N
Radius of the given uncharged drop, r=2.0×105m
Density of the uncharged drop, ρ=1.2×103kgm3
Viscosity of air, η=1.8×105Pas
Density of air (ρ0) can be taken as zero in order to neglect buoyancy of air.
Acceleration due to gravity, g=9.8m/s2
Terminal velocity (v) is given by the relation:
v=2r2×(ρρ0)g9η=2×(2.0×105)2(1.2×1030)×9.89×1.8×105=5.807×102ms1=5.8cms1
Hence, the terminal speed of the drop is 5.8cms1.
The viscous force on the drop is given by:
F=6πηrv
F=6×3.14×1.8×105×2.0×105×5.8×102=3.9×1010N
Hence, the viscous force on the drop is 3.9×1010N.

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