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.
Terminal speed =5.8cm/s Viscous force =3.9×10−10N Radius of the given uncharged drop, r=2.0×10−5m Density of the uncharged drop, ρ=1.2×10−3kgm−3 Viscosity of air, η=1.8×10−5Pas Density of air (ρ0) can be taken as zero in order to neglect buoyancy of air. Acceleration due to gravity, g=9.8m/s2 Terminal velocity (v) is given by the relation: V=2r2×(ρ−ρ0)g/9η =2×(2×10−5)2(1.2×103−0)×9.8/(9×1.8×10−5) =5.8×10−2m/s =5.8cms−1 Hence, the terminal speed of the drop is 5.8cms−1. The viscous force on the drop is given by: F=6πηrv ∴F=6×3.14×1.8×10−5×2×10−5×5.8×10−2 =3.9×10−10N Hence, the viscous force on the drop is 3.9×10−10N.
