state stoke's law.derive the equation.
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When a solid body moves through a fluid,the fluid in contact with the solid is dragged with it. Relative velocities are established between the layers of the fluids near the solid so that the viscous forces start operating. The fluid exerts viscous force on the solid to oppose the motion of the solid. The magnitude of the viscous force depends on the shape and size of the solid body, Its speed and the coefficient of viscosity of the fluid.
Suppose a spherical body of radius r moves at speed v through a fluid of viscosity ƞ .the viscous force F acting on the body depends on r,v and ƞ . assuming that the force is proportional to various powers of these quantities we can obtain the dependence through dimensional analysis.
Let F = kravb ƞc
MLT-2 = kLa(LT-1)b(ML-1 T-1)c
Equating the coefficients of M,L,T on both sides we get :
a,b,c each = 1
hence F = krvƞ, value of k =6π,
hence stokes law becomes,
F = 6πƞrv…… which is the required stokes law
Terminal velocity:
When a solid is dropped in a fluid, the forces acting on it are.
- Weight W acting downwards
- Viscous force F acting upwards
- Buoyant force B acting upwards
We know from Stokes law that F is proportional to velocity v.
Initially , F=0 and the solid is accelerated due to force W-B
Because of this acceleration, v increases. At a certain instant
F becomes equal to W-B. net force is 0 and solid falls with constant velocity. This is known as terminal velocity. which is given by the following formula:
V0 = (2r2(p-σ)g)/9ƞ
Wherev0 is terminal velocity,p is the density of the body, σ is density of the liquid and g is acceleration of the body.