### Q.29:- 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.

**Answer:-**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, *η* = 1.8 × 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* = 2*r*^{2} × (*ρ* – *ρ*_{0}) g / 9*η*

= 2 × (2 × 10^{-5})^{2} (1.2 × 10^{3} – 0 ) × 9.8 / (9 × 1.8 × 10^{-5})

= 5.8 × 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* = 6*πηrv*

∴ *F* = 6 × 3.14 × 1.8 × 10^{-5} × 2 × 10^{-5} × 5.8 × 10^{-2}

= 3.9 × 10^{-10} N

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