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I was reading the R.A. Millikan's oil drop experiment and the following part confused me:

By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser).

In Newtonian physics, the rate of fall (or velocity) of a freely falling body does not depend on the mass of the object. So, how did Millikan measure the mass of these oil droplets? Does anyone know the equations he used?

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By measuring the rate of fall, Millikan was able to measure the fall of the oil droplets (which were dropped through a tiny hole in the upper electrical condenser)

The rate of fall here means terminal velocity of oil, $v_\mathrm{t}.$ Oil was made to drop at constant rate so no acceleration. Drag force on oil drop is:

$$F_\mathrm{d} = 6\pi r\eta v_\mathrm{t},\tag{1}$$

where $r$ is the radius of oil drop and $\eta$ is the viscosity of air.

Assume oil droplet is spherical. Apparent weight of oil droplet in air is the true weight minus the upthrust (which equals the weight of air displaced by the oil drop). So, it can be written as:

$$w = \frac{4\pi r^3(\rho - \rho_\mathrm{air})g}{3}.\tag{2}$$

No acceleration is taking place so the total force acting on it must be zero and thus the two forces must cancel each other (i.e $F_\mathrm{d} = w).$ Solving for $r$ gives

$$r = \sqrt{\frac{9\eta v_\mathrm{t}}{2g(\rho - \rho_\mathrm{air})}}.\tag{3}$$

The density of oil is known so by known so by putting $r,$ you can calculate the mass of oil droplet as

$$m = \rho V = \rho\frac{4\pi r^3}{3}.\tag{4}$$

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