Hey students, we know that the viscous force acting on any object depends upon the area of layer of liquid parallel to direction of streamline flow and the velocity gradient.
Coefficient of viscosity is given as,
Viscosity plays important in explanation of properties of fluid. Shape of rain drop, honey flowing on plane and smooth surface are some important applications of viscous force. In this article we are going to discuss the viscous force acting on spherical body when it is falling in viscous fluid.
Let’s discuss the concept in details……………!
Consider the spherical body of radius ‘r’ is falling freely in viscous fluid as shown in fig. below,
Let, η = coefficient of viscosity of fluid
r = radius of spherical body
v = velocity of spherical body
When the spherical body is falling in viscous fluid, the velocity of body is decreases due to opposition by fluid. This means that the fluid has exerted the force on spherical body.
Sir G. Stoke’s has given as law to find the magnitude of viscous force acting on the spherical body. The Stoke’s law stated as,
“Viscous force acting on the spherical body falling freely in viscous fluid is directly proportional to,
- Velocity of body(v).
- Radius of spherical body(r).
- Coefficient of viscosity of fluid (η).
∴F α r v η
∴ F = k r v η
Where ‘k’ is constant of proportionality.
The magnitude of k was obtained through some experiments and it is given by,
k = 6π
Then the formula for viscous force acting on the spherical body falling freely in viscous fluid is,
∴ F = 6π r v η
This formula can be verified dimensionally for checking its correctness. Can you try it students?
Let’s solve some numerical for more illustration…..!
Ex:1) Viscous force acting on metal sphere of radius 2.4 cm falling with velocity of 34 cm/s is 52 dyne. Find the coefficient of viscosity of fluid.
Solution:
Here, r = 2.4 cm = 2.4 × 10-2 m,
v = 34 cm/s =34 × 10-2 m/s
F = 52 dyne
We have,
∴ F = 6π r v η