__Viscosity__ : When a fluid flows such that a velocity gradient is set up within it, forces act within the fluid so as to prevent the velocity gradient from existing. This force is due to a property called viscosity.

Suppose that a glass plate in contact with a water column of height h is moved with constant velocity v. Forces of viscosity appear between the solid surface and the layer in contact.

$ \displaystyle F = -\eta A \frac{dv}{dz} $

where η is a constant called co-efficient of viscosity, its cgs unit is poise. Dimension is ML^{-1}T^{-1} . The SI unit of viscosity equals 10 poise.

__Stoke’s Law & Terminal Velocity__

When a sphere of radius r moves with a velocity v through a fluid of viscosity η, the viscous force opposing the motion of the sphere is

F = 6πη r v

If for a sphere viscous force become equal to the net weight acting downward, the velocity of the body become constant and is known as terminal velocity.

$ \displaystyle 6\pi \eta r v = \frac{4}{3}\pi r^3 (\rho -\sigma)g $

$ \displaystyle v = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $

Exercise : Raindrops fall from a very great height. As we see the speed with which it falls is not that large. Why is this so ?

Illustration : A spherical ball of radius 1×10^{-4} m and density 10^{4} kg/m^{3} falls freely under gravity through a distance h before entering a tank of water. If after entering the water the velocity of the ball does not change, find h. The viscosity of water is 9.8 × 10^{-6}N -m^{2}

Solution : After falling through a height h, the velocity of the ball becomes

$ \displaystyle v = \sqrt{2 g h} $

After entering water, this velocity does not change, this velocity is equal to the terminal velocity,

$ \displaystyle \sqrt{2 g h} = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $

On putting given values , we get

or, h = 20.41 m

Illustration : Two identical drops of water are falling through air with a steady velocity v. If the drops coalesced, what will be the new velocity?

Solution : Let r be the radius of each drop. The terminal velocity v_{T} of a drop of radius r is given by

$ \displaystyle v = \frac{2 r^2 (\rho – \sigma) g}{9 \eta} $ . . . (i)

Now when two drops each of radius r coalesce to form a new drop, the volume of coalesced drop will be given by

$ \displaystyle \frac{4}{3}\pi R^3 = 2\times \frac{}{}\pi r^3 $

The radius of the coalesced drop will be

R = (2)^{1/3} r

$ \displaystyle \frac{v’}{v} = (\frac{R}{r})^2 = 2^{2/3} $

Hence, the new terminal velocity of the coalesced drop is

or v’ = (2)^{2/3} v [as v_{T} = v]

### Also Read:

→ Surface Tension → Surface Energy → Excess Pressure inside a soap bubble → Angle of contact → Capillarity & Ascent Formula → Poiseuille’s formula |