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 1

Terminal speed = 5.8 cm/s; Viscous force = 3.9 × 10^{–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, `eta = 1.8 xx 10^(-5) Pa s`

Density of air (`rho_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 xx (rho - rho_0)g)/(9eta)`

`= (2xx(2.0xx10^(-5))^2 (1.2xx10^3 - 0)xx 9.8)/(9xx1.8xx10^(-5))`

`= 5.807 xx10^(-2) ms^(-1)`

`= 5.8 cm^1`

Hence, the terminal speed of the drop is 5.8 cm s^{–1}.

The viscous force on the drop is given by:

`F = 6pietarv`

`:. F = 6xx3.14xx1.8xx10^(-5)xx2.0xx^(-5)xx 5.8xx10^(-2)`

`= 3.9 xx 10^(-10) N`

Hence, the viscous force on the drop is 3.9 × 10^{–10 }N.

#### Solution 2

Here radius of drop, r = 2.0 x 10^{-5} m, density of drop, p = 1.2 x 10^{3} kg/m^{3}, viscosity of air TI = 1.8 x 10^{-5}Pa-s.

Neglecting upward thrust due to air, we find that terminal speed is

`v_r = 2/9 (r62rhog)/eta = (2xx(2.0xx10^(-5))^2xx(1.2xx10^3))xx9.8)/(9xx(1.8xx10^(-5)))`

`= 5.81 xx 10^(-2) ms^(-1) " or " 5.81 cm s^(-1)`

Viscous force at this speed

`F = 6pietarv = 6 xx 3.14 xx (1.8xx10^(-5))xx (2.0xx10^(-5))xx (5.81xx10^(-2))`

`= 3.94 xx 10^(10) N`