Question

state Stokes law derive an expression for the terminal velocity of a sphere falling through a viscous fluid​

Answer 1

According to Stokes’ law, the backward viscous force acting on a small spherical body of radius r moving with uniform velocity v through fluid of viscosity η is given by

                          F=6 πη rv

  Where, r = Radius of the spherical body

v = velocity of the spherical body

It gives the relationship between retarding force and velocity. When viscous force plus buoyant force becomes equal to force due to gravity, the net force becomes zero. The sphere then descends with a constant terminal velocity (vt).

Now,

                           6 π η rv =4/3 πr^3( p-σ )

 Where, ρ = Density of the liquid

σ = Density of the spherical body

Terminal velocity of a spherical body of density ρ and radius r moving through a liquid of density ρ’ is

where,

ρ = density of body,

sigma = density of liquid,

η = coefficient of viscosity of liquid and,

g = acceleration due to gravity

1)  If ρ > ρ0, the body falls downwards.

2) If ρ < ρ0, the body moves upwards with the constant velocity.

3) If po << ρ, v = (2r2ρg/9η)

Answer 2

Stokes law states that the force that retards a sphere moving through a viscous fluid is directly proportional to the velocity and radius of the sphere and the viscosity of the fluid. Sir George G Stokes gave the formula as :

F = 6$\pi$∩rv

where F is retarding force

∩ denotes viscosity of liquid

r is radius of sphere

v is velocity through which it falls

Terminal velocity: It is maximum constant velocity  of the body while falling freely in viscous liquid or medium.

When a spherical body falls through a  liquid, three types of forces act on it:

(i) Weight of the body that acts downwards (W)

(ii) Buoyant force that acts upwards and is equal to the weight of water displaced ($F_{B}$)

(iii) Force acting upwards as calculated from Stokes Law ($F_{s}$)

A diagram is attached for clear understanding

Let p be the density of spherical ball

and $p_{0}$  be density of liquid

So to balance out all forces we have equation as:

W = $F_{B}$ + $F_{s}$       ....(i)

Weight = Volume x density x g(acceleration due to gravity)

W = $\frac{4\pi r^{3} }{3}$ x p x g

$F_{B}$ = Volume of ball x density of liquid x g

$F_{B}$ =  $\frac{4\pi r^{3} }{3}$ x  $p_{0}$ x g

If v is the terminal velocity then as oer Stokes Law upward force $F_{s}$ is

$F_{s}$ = 6$\pi$∩rv

Putting values of all in (i) we get

$\frac{4\pi r^{3} }{3}$ x p x g =   $\frac{4\pi r^{3} }{3}$ x  $p_{0}$ x g + 6$\pi$∩rv

6$\pi$∩rv = $\frac{4\pi r^{3} }{3}$ x p x g - $\frac{4\pi r^{3} }{3}$ x  $p_{0}$ x g

6$\pi$∩rv = $\frac{4\pi r^{3} }{3}$ x(p- $p_{0}$) x g

v = $\frac{2r^{2} (p- p_{0} g )}{9n}$

Terminal velocity depends on the inverse relation of viscosity of medium and in direct relation with densities of body and medium and square of radius.