Animated Solution for Physics - Properties of Solids and Liquids: In Millikan's oil drop experiment, what is viscous force acting on an uncharged drop of radius $2.0\times 10^{-5}$ m and density $1.2\times 10^3{\text{ kgm}}^{-3}$? Take viscosity of liquid $=1.8\times 10^{-5}{\text{ Nsm}}^{-2}$. (Neglect buoyancy due to air).

A Falling Drop

Imagine a tiny oil drop falling through the air, just like in Millikan's famous oil drop experiment. As it falls, gravity pulls it down. However, the air isn't just empty space; it has a property called viscosity. This viscosity creates a drag force that pushes back up against the drop's motion.

The problem explicitly tells us to neglect the buoyancy due to air. This makes our physical model beautifully simple: we only have two forces to worry about. The downward force of gravity ($F_g$) and the upward viscous drag force ($F_v$).

When the drop first starts falling, it accelerates because gravity is stronger than the drag force. But as it speeds up, the drag force increases. Eventually, the drop reaches a speed where the upward viscous force exactly balances the downward gravitational force. At this point, the net force is zero, and the drop falls at a constant speed known as the terminal velocity ($v_T$).

Mathematically, this dynamic equilibrium is expressed as:

We know from Stokes' Law that the viscous force on a small sphere is given by:

And the weight of the spherical drop is its volume times its density times gravity:

Finding Velocity First

One way to solve this problem is to first find the terminal velocity. By equating the two forces, we get the formula for terminal velocity:

Let's plug in the given values: - Radius, $r=2.0\times 10^{-5}\text{ m}$ - Density, $\rho =1.2\times 10^3{\text{ kgm}}^{-3}$ - Viscosity, $\eta =1.8\times 10^{-5}{\text{ Nsm}}^{-2}$ - Gravity, $g=9.8{\text{ ms}}^{-2}$

Substituting these in:

Now that we have the velocity, we can plug it back into Stokes' Law to find the viscous force:

Direct Force Balance

While the method above works perfectly, there is a much faster and more elegant way to solve this.

Remember our equilibrium condition? At terminal velocity, the viscous force is exactly equal to the weight of the drop.

This means we don't even need to calculate the terminal velocity! We can just calculate the weight of the drop directly, and that will be our answer for the viscous force.

Let's calculate the weight:

Both methods yield the exact same result, but the second method is a fantastic shortcut that saves time and reduces the chance of calculation errors during a high-pressure exam. The closest matching option is (b).