Establish stoke formula of viscosity.

The viscous force acting on a sphere is directly proportional to the following parameters:

• the radius of the sphere
• coefficient of viscosity
• the velocity of the object

Mathematically, this is represented as

F∝ηarbvc

Now let us evaluate the values of a, b and c.

Substituting the proportionality sign with an equality sign, we get

F=kηarbvc (1)

Here, k is the constant of proportionality which is a numerical value and has no dimensions.

Writing the dimensions of parameters on either side of equation (1), we get

[MLT^–2] = [ML^–1T^–1]^a [L]^b [LT^-1]^c

Simplifying the above equation, we get

[MLT^–2] = M^a ⋅ L^–a+b+c ⋅ T^–a–c (2)

According to <a href="https://byjus.com/physics/mechanics/">classical mechanics</a>, mass, length and time are independent entities.

Equating the superscripts of mass, length and time respectively from equation (2), we get

a = 1 (3)

–a + b + c = 1 (4)

–a –c = 2 or a + c = 2 (5)

Substituting (3) in (5), we get

1 + c = 2

c = 1 (6)

Substituting the value of (3) & (6) in (4), we get

–1 + b + 1 = 1

b = 1 (7)

Substituting the value of (3), (6) and (7) in (1), we get

F=kηrv

The value of k for a spherical body was experimentally obtained as 6π

Therefore, the viscous force on a spherical body falling through a liquid is given by the equation

F=6πηrv