Question

Derivation of Stoke’s law.

Answer 1

Hint:Stoke’s Law is defined as the settling velocities of the small spherical particles in a fluid medium. The viscous force \[F\]is acting on a small sphere of radius \[r\]moving with velocity \[\nu \]through the liquid is given by\[F = 6\pi \eta r\nu \]. Calculate the dimensions of \[n\]the coefficient of viscosity.

Complete step-by-step solution:
Stoke’s Law Derivation:
The following parameters are directly proportional to the viscous force acting on a sphere.
the radius of the sphere
coefficient of viscosity
the velocity of the object
Mathematically, this is represented as,
\[F \propto {\eta ^a}{r^b}{\nu ^c}\]
Let us evaluate the values of\[a,b,c\].
Substitute the proportionality sign with an equality sign, we get
\[F = k{\eta ^a}{r^b}{\nu ^c}.............(1)\]
Here, \[k\] is the constant value which is a numerical value and has no dimensions.
Now writing the dimensions of parameters on either side of the equation\[(1)\], we get
\[\left[ {ML{T^{ - 2}}} \right] = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^a}{\left[ L \right]^b}{\left[ {L{T^{ - 1}}} \right]^c}\]
Simplifying the above equation, we get
\[\left[ {ML{T^{ - 2}}} \right] = {M^a} \cdot {L^{ - a + b + c}} \cdot {T^{ - a - c}}.........\left( 2 \right)\]
The independent entities are classical mechanics, mass, length, and time.
Equating the superscripts of mass, length, and time respectively from the equation\[\left( 2 \right)\], we get
\[a = 1..........\left( 3 \right)\]
\[ - a + b + c = 1...........\left( 4 \right)\]
\[ - a - c = 2\]
\[a + c = 2\left( 5 \right)\]
Using the eq. \[\left( 3 \right)\]in the eq.\[\left( 5 \right)\], we get
\[1 + c = 2\]
\[ \Rightarrow c = 1..........\left( 6 \right)\]
By putting the values of the eq.\[\left( 3 \right)\]&\[\left( 6 \right)\] in the eq.\[\left( 4 \right)\], we get
\[ \Rightarrow - 1 + b + 1 = 1\]
\[ \Rightarrow b = 1.........\left( 7 \right)\]
Substituting the values of the eq.\[\left( 3 \right)\], \[\left( 6 \right)\] and \[\left( 7 \right)\]in the eq.\[\left( 1 \right)\], we get
\[F = k\eta r\nu \]
The value of \[k\] for a spherical body was experimentally obtained as\[6\pi \].
Hence, the viscous force on a spherical body falling through a liquid is given by the equation
\[F = 6\pi \eta r\nu \]

Note:Stoke’s law is derived from the forces acting on a small particle as it sinks through the liquid column under the influence of gravity.
A viscous fluid is directly proportional to the velocity and the radius of the sphere, and the viscosity of the fluid when the force that retards a sphere is moving.