In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×10–5 m and density 1.2×103kgm–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.
In Millikan’s oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0×10–5 m and density 1.2×103kgm–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 = 5.8 cm/s
Viscous force =3.9×10–10N
Radius of the given uncharged drop, r=2.0×10–5m
Density of the uncharged drop, ρ=1.2×103kgm–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)g9η=2×(2.0×10−5)2(1.2×103−0)×9.89×1.8×10−5=5.807×10−2ms−1=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=6×3.14×1.8×10−5×2.0×10−5×5.8×10−2=3.9×10−10N
Hence, the viscous force on the drop is 3.9×10−10N.
Terminal speed = 5.8 cm/s
Viscous force =3.9×10–10N
Radius of the given uncharged drop, r=2.0×10–5m
Density of the uncharged drop, ρ=1.2×103kgm–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)g9η=2×(2.0×10−5)2(1.2×103−0)×9.89×1.8×10−5=5.807×10−2ms−1=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=6×3.14×1.8×10−5×2.0×10−5×5.8×10−2=3.9×10−10N
Hence, the viscous force on the drop is 3.9×10−10N.
