What is Stokes Law?
Stokes law discusses the active force applied on a body when it is dropped in a liquid. Initially, because of low viscous force, this velocity of the falling body remains low. But, as the spherical body falls with its effective weight, it gains acceleration, and this velocity of the body increases gradually.
This also makes the liquid in contact move with a velocity same as that of this body. The movement of an object in this fluid with increasing velocity causes motion in liquid layers resulting in the development of viscous force. This force increases with the increasing velocity until the point of time when it matches this effective force with which this body moves.
In such a situation, the net force on the body becomes zero, and it attains a constant velocity named terminal velocity (Vt). This idea helped state Stokes law and this derivation of frictional force or Stokes’ drag applies to the interface between the body and fluid.
According to Sir George Stokes, “the force acting between the liquid and falling body interface is proportional to velocity and radius of the spherical object and viscosity of fluid”.
Stoke's law explains the main reason why raindrops falling from the sky do not harm us on the ground. Stokes law is an extremely important concept that is a part of physics
Stokes Law Formula
Stokes came up with this formula in 1851 to calculate this drag force or frictional force of spherical objects immersed in viscous fluids. Here, look at the formula mentioned below.
F = 6 * πηrv
Where,
F is the drag force or frictional force at the interface
η is the viscosity of a liquid
r is the radius of the spherical body
V is the velocity of flow
Stokes Law Derivation
Stokes’ proposition regarding this immersion of the spherical body in a viscous fluid can be mathematically represented as,
F ∝ ηa rb vc
By solving this proportional expression, we can get the Stokes law equation. To change this proportionality sign into equality, we must add a constant to the equation. Let us consider that constant as ‘K’, so the transformed equation becomes
F = K * ηa rb vc ……….. (1)
Now, Let us write down the dimensions of this equation on both sides. Please note, K is a constant which has no dimension.
[MLT-2] = [[ML-1T-1]a . Lb . [LT-1]c ]
We need to simplify this expression by separating each variable as follows
[MLT-2] = [Ma ] [L-a+b+c] [T-a-c]
Comparing the value of length, mass and time, following equations can be found.
a = +1 ……….(2)
-a+b+c = +1 ……....(3)
-a-c= -2 ……………..(4)
On solving equations 2,3, and 4, we get a =1, b=1, c=1. On substituting these values in equation 1, we have
F = K * η1 r1 v1 = K ηrv
Further, the value of K was found to be 6π for spheres through experimental observation. The above calculations helped in Stokes equation derivation along with its fundamental formula.
Terminal Velocity Formula
As explained earlier, terminal velocity is attained at an equilibrium position when the net force acting upon the spherical body and acceleration becomes zero. Here is the formula for terminal velocity derived from Stokes law definition.
Vt = 2a2 (ρ−σ) g / 9η
Where ρ is the mass density of a spherical object and σ is the mass density of a fluid.
Assumptions made in the Stokes Law
Some assumptions were made in the Stokes law. The assumptions made in the Stokes law include the following:
Laminar flow
Particles are spherical
The material is homogeneous or uniform in composition
Surfaces are smooth
Particles do not interfere with each other
Applications of Stoke Law
Stokes law is applicable in many areas. Some applications of stokes law include the following:
It helps in finding the settling of sediment in freshwater
It is also used to measure the viscosity of fluids.
FAQs on Stokes Law Derivation
1. What is Stokes law and how does it describe the drag force on a sphere moving through a fluid?
Stokes law states that the frictional force (drag force) experienced by a small spherical object moving at low velocity through a viscous fluid is directly proportional to the sphere's radius, the viscosity of the fluid, and the object's velocity. Mathematically, it is written as F = 6πηrv, where η is the fluid's viscosity, r is radius, and v is velocity.
2. How is Stokes law derived using dimensional analysis for Class 11 and 12 Physics?
The derivation starts by assuming the drag force depends on fluid viscosity (η), sphere radius (r), and velocity (v): F ∝ ηa rb vc. Assign dimensional forms to each variable and balance them to solve for a, b, and c. This gives F = kηrv, and experiments show k = 6π for spheres, resulting in F = 6πηrv as per CBSE 2025–26.
3. Under what conditions is Stokes law valid for moving objects in a liquid?
Stokes law applies when:
- The particle is spherical and small enough so the flow remains laminar (low Reynolds number)
- The fluid is homogeneous and of uniform viscosity
- The object's velocity is low
- No interaction occurs between nearby particles
- The surface of the sphere is smooth
4. Why does a raindrop reach a constant speed when falling through air, according to Stokes law?
As a raindrop falls, it accelerates until the upward viscous drag force equals its effective weight. At this point, net force is zero, and the drop attains a constant terminal velocity as described by Stokes law. This explains why falling raindrops do not cause harm upon impact with the ground.
5. What is the formula for terminal velocity derived from Stokes law, and what do its terms mean?
The terminal velocity (Vt) for a sphere falling through a fluid is given by: Vt = (2r2(ρ − σ)g)/(9η), where r = radius of the sphere, ρ = density of sphere, σ = density of fluid, η = viscosity of fluid, and g = acceleration due to gravity.
6. How does the viscosity of a fluid influence the movement of objects as explained in Stokes law?
A higher viscosity (η) increases the drag force on the object, thereby reducing its speed through the fluid. For example, a sphere moves slower through glycerin than water because glycerin is more viscous.
7. What role did Stokes law play in determining the electronic charge in Millikan's oil drop experiment?
Stokes law was crucial in Millikan's experiment because it enabled precise calculation of the viscous drag on falling oil drops. This allowed accurate measurement of their terminal velocity, which is essential in determining the value of the electron's charge.
8. How does temperature affect the viscosity of liquids and the application of Stokes law?
Increasing temperature generally decreases the viscosity of liquids, so the drag force experienced by moving objects also decreases. However, for gases, viscosity increases with temperature. This affects the rate at which objects settle in different fluids under varying temperature conditions.
9. What are common misconceptions students have about the applicability of Stokes law?
Common misconceptions include assuming Stokes law is valid for high-speed motions or non-spherical objects. In reality, the law only holds for small, smooth, spherical particles in laminar flow at low velocities. It does not apply when turbulent flow occurs or particles are irregularly shaped.
10. In what ways is Stokes law applied in real-life and industrial contexts?
Applications include:
- Measuring viscosity of unknown fluids using settling velocities of spheres
- Predicting sedimentation rates in water bodies
- Designing settling tanks and centrifuges
- Understanding aerosol and pollutant behavior in the atmosphere
