10.28 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.
The radius of the uncharged drop, r = 2.0 × 10–5 m
The density of the uncharged drop, ρ = 1.2 × 103 kg m–3
The viscosity of air, η=1.8×10-5 Pa s
The density of air (ρ0) can be taken as zero in order to neglect buoyancy of air.
Terminal velocity (v) is:
v=2r2×(ρ-ρ0)g9η=2×(2.0×10-5)2×(1.2×103-0)9.89×1.8×10-5=5.807×10-2 ms-1=5.8 cm s-1
The viscous force on the drop=6πηrv=6×3.14×1.8×10-5×2.0×10-5×5.8×10-2=3.9×10-10 N
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