Question

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 × 103 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.

Answer 1

Given that

Radius of the drop is r=2.0× 10 −5  m

Density of the drop is ρ=1.2× 10 3   kg/ m 3

Viscosity of the air is η=1.8× 10 −5  Pa-s

Neglect upward thrust due to air at drop, the terminal speed of the uncharged drop is,

υ T = 2 9 r 2 ρg η

Substitute the values.

υ T = 2× ( 2× 10 −5 ) 2 ×1.2× 10 3 ×9.8 9×1.8× 10 −5 υ T =0.0581  ms −1 υ T =5.81  cms −1

Thus, the terminal speed of an unchanged drop is 5.81  cms −1 .

Viscous force on the drop at this terminal speed is given by,

F=6πηr υ T F=6× 22 7 ×1.8× 10 −5 ×2× 10 −5 ×0.0581 F=3 .94×10 −10  N

Thus, the viscous force on the drop at this terminal speed is 3 .94×10 −10  N.