Question

The viscous force acting on a body falling under gravity in a viscous fluid will be
(A) $\dfrac{6\pi V}{\eta r}$
(B) $\dfrac{6\pi \eta r}{V}$
(C) $6\pi \eta rV$
(D) $\dfrac{6\pi rV}{6\pi \eta }$

Answer 1

In general, the viscous force between a body and a fluid (either a liquid or a gas) moving past it, in a direction to oppose the flow of the fluid past the object. As such, the viscous force can be viewed as the force of friction in a fluid. Viscous force does not necessarily act between a fluid and a solid, it can act between two layers of flowing fluids as well.

Complete step by step answer:
Stokes law is named after George Gabriel Stokes and it describes the relationship between the frictional force on a sphere moving in a liquid and other quantities such as particle radius and velocity of the particle.
The assumption made for Stokes law is that the Reynolds number of the fluid is less than one, which means that the flow of the fluid is laminar. If the above condition is fulfilled, the viscous force on a body falling freely under gravity will be $F=6\pi \eta rV$ where $F$ is the viscous force to be overcome, $V$ is the speed of the falling body relative to the fluid velocity, $r$ is the particle radius or radius of the body (since the body is generally considered to be spherical) and $\eta$ is the dynamic viscosity.
Hence we can say that option (C) gives the correct expression for the required viscous force and is hence the correct option.

Note: The expression for Stokes law is valid only if certain ideal conditions are fulfilled. Some of these conditions are that the flow should be streamlined, the falling particle must be spherical, the particle and the fluid must have homogeneous compositions; the surface of the falling particle must be smooth and that the particles do not interfere with each other. Hence Stokes law is not valid for practical purposes in real life and changes must be made in it appropriately.