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Question: In Millikan’s oil drop experiment, what is the terminal speed of an unchanged drop of radius \[2.0 \...

In Millikan’s oil drop experiment, what is the terminal speed of an unchanged drop of radius 2.0×10−5m2.0 \times {10^{ - 5}}mand density1.2×103kgm−31.2 \times {10^3}kg{m^{ - 3}}. Take the viscosity of the air at the temperature of the experiment to be1.8×10−5Pas1.8 \times {10^{ - 5}}Pas. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Explanation

Solution

In this question, radius and density of the oil drop are given, and it is said that surrounding air opposes the motion of oil drop whose viscosity is given so we will use these data to find the terminal speed and the using this terminal speed we will find the viscous force on the oil drop.

Complete step by step answer:
The radius of the oil dropr=2.0×10−5mr = 2.0 \times {10^{ - 5}}m
The density of the oil dropρ=1.2×103kgm−3\rho = 1.2 \times {10^3}kg{m^{ - 3}}
The viscosity of the air η=1.8×10−5Pas\eta = 1.8 \times {10^{ - 5}}Pas
We know when oil is dropped with a velocity v from some height the molecules of the surrounding air opposes its motion, then the oil drop comes to a point when this vicious force becomes equal to the force on the drop due to gravity and velocity becomes constant, so this velocity is known as terminal velocity which is given as
v=2r2(ρ−σ)g9η−−(i)v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }} - - (i)
Hence by substituting the values, we get

⇒v=2r2(ρ−σ)g9η ⇒v=2×(2×10−5)2(1.2×103−0)×9.89×1.8×10−5 ⇒v=5.8×10−2ms−1 ⇒v=5.8cms−1 \Rightarrow v = \dfrac{{2{r^2}\left( {\rho - \sigma } \right)g}}{{9\eta }} \\\ \Rightarrow v = \dfrac{{2 \times {{\left( {2 \times {{10}^{ - 5}}} \right)}^2}\left( {1.2 \times {{10}^3} - 0} \right) \times 9.8}}{{9 \times 1.8 \times {{10}^{ - 5}}}} \\\ \Rightarrow v = 5.8 \times {10^{ - 2}}m{s^{ - 1}} \\\ \Rightarrow v = 5.8cm{s^{ - 1}} \\\

Therefore the terminal velocity of oil drop is=5.8cms−1 = 5.8cm{s^{ - 1}}
Now we find the viscous force on the oil drop which is given by the formula
F=6πηrv−−(ii)F = 6\pi \eta rv - - (ii)
Now we substitute the values in the equation

⇒F=6πηrv ⇒F=6×227×(1.8×10−5)×(2.0×10−5)×(5.8×10−2) ∴F=3.93×10−10N \Rightarrow F = 6\pi \eta rv \\\ \Rightarrow F = 6 \times \dfrac{{22}}{7} \times \left( {1.8 \times {{10}^{ - 5}}} \right) \times \left( {2.0 \times {{10}^{ - 5}}} \right) \times \left( {5.8 \times {{10}^{ - 2}}} \right) \\\ \therefore F = 3.93 \times {10^{ - 10}}N \\\

Hence the viscous force on the oil drop will be 3.93×10−10N3.93 \times {10^{ - 10}}N.

Note: Terminal speed is the maximum speed attained by an object when it falls through a fluid, and this occurs when the sum of the drag force and the buoyancy is equal to the downward force of gravity acting on the object.