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

Here,
radius of drop ( r) = 2 × 10^-5 m
density of oil ( d) = 1.2 × 10³ kg/m³
Viscosity of air ( n) = 1.8 × 10^-5 Pa-s

Use formula,
Terminal velocity (v) = {2/9} {r²dg/n}
= {2/9} { (2×10^-5)² × 1.2×10³×9.8/1.8×10^-5}
= 5.8 × 10^-2 m/s

We also know,
Viscous force acting on the drop = 6πrnv
Where , n is viscosity of air

F = 6 × 3.14 × 2 × 10^-5 × 1.8×10^-5 × 5.8 × 10^-2 N
= 3.93 × 10^-10 N