__Question 10.28__:
In
Millikan’s oil drop experiment, what is the terminal speed of
an uncharged drop of radius 2.0 × 10

^{–5}m and density 1.2 × 10^{3}kg m^{–3}? Take the viscosity of air at the temperature of the experiment to be 1.8 × 10^{–5}Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

__Solution__:^{–10 }N

Radius of the given uncharged drop,

*r*= 2.0 × 10

^{–5}m

Density of the uncharged drop,

*ρ*= 1.2 × 10

^{3}kg m

^{–3}

Viscosity of air, η = 1.8 X 10

^{-5}Pa s

Density of air (ρ

_{0}) can be taken as zero in order to neglect buoyancy of air.

Acceleration due to gravity, g = 9.8 m/s

^{2}

Terminal velocity (

*v*) is given by the relation:

v = 2r

^{2}X (ρ - ρ

_{0}) g / 9η

= 2 X (2 X 10

^{-5})

^{2}(1.2 X 10

^{3}- 0 ) X 9.8 / (9 X 1.8 X 10

^{-5})

= 5.8 X 10

^{-2}m/s

= 5.8 cm s

^{-1}

Hence, the terminal speed of the drop is 5.8 cm s

^{–1}.

The viscous force on the drop is given by:

F = 6πηrv

∴ F = 6 X 3.14 X 1.8 X 10

^{-5}X 2 X 10

^{-5}X 5.8 X 10

^{-2}

= 3.9 X 10

^{-10}N

Hence, the viscous force on the drop is 3.9 × 10

^{–10 }N.

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