Question

The terminal velocity of a spherical ball of radius r falling through a viscous liquid is proportional to

Answer 1

According to the Stokes’ law of viscosity the frictional force exerted on spherical objects in a

viscous fluid is given by

F 6 R fr   

Here  is the fluid viscosity,  is the particle’s velocity and R is the particle’s radius

There are three forces acting on a sphere falling in a fluid:

1. Ffr (acting upward)

2. the gravity force mg (acting downward)

3. buoyancy force F g V b fluid     (acting upward)

According to the Newton’s second law of motion and taking in account that the velocity of the

particle remains constant (terminal velocity), we can write:

fr b

particle

3

3 3

fluid particle

2

particle fluid

2 particle fluid

F F mg 0

m V

4

V R

3

4 4 6 R g R g R 0

3 3

4

R g( )

3 2

R

6 9

  

  

 

            

     

  

 

Hence the terminal viscosity is proportional to the particle’s radius squared.

Answer

The terminal viscosity is proportional to the particle’s radius squared.

Answer 2

Answer:

The terminal velocity of a spherical ball of radius r falling through a viscous liquid is proportional to particle 'Radius squared'

Explanation:

As per Stokes’ law of viscosity the frictional force exerted on spherical objects in viscous fluid is  F(fr) = 6 Π μ R υ  ... (1)

μ - Fluid viscosity

R - Particle's radius

υ - Particle's velocity

now, F(fr) + F(b) - mg = 0  ..... (2)

m = ρ(particle) x V

V = 4/3 Π R³

therefore, mg = ρ(particle) x 4/3 Π R³x g  ... (3)

F(b) = ρ . g. 4/3 Π R³ ---- (4)

Substituting (1) , (3) , (4) in (2) we get,

v = 2.9. R²  ρ (particle - fluid)

viscous liquid is proportional to square of radius.