Derive Stokes’ Law For The Viscosity Of A Small Sphere Falling Through A Viscous Liquid Using Method Of Dimension.

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Stokes’ Law:

Let us consider, a spherical body of radius \( r \) falls through a liquid of infinite extent having a coefficient of viscosity \( \eta \). After some time it attains a terminal velocity \( v \), then the resistive force on the spherical body due to viscous liquid becomes \( F=6\pi\eta{rv} \). This is known as Stokes’ law.

Derivation of Stokes’ law from dimensional analysis:

If a small sphere is falling through a viscous medium, then due to viscosity the resistive force \( (F) \) on the sphere depends upon the radius \( r \) of the sphere, coefficient of viscosity \( \eta \) of the medium and the velocity \( v \) of the sphere.

Therefore \( F=Kv^ar^b{\eta}^c \),

where, K is a dimensionless constant and a,b,c are dimensionless numbers to which v,r, \( \eta \) are raised respectiely.

Now, using dimensions of different terms, we have

\( [MLT^{-2}]={[LT^{-1}]}^a{[L]}^b{[ML^{-1}T^{-1}]}^c\\={[M]}^c{[L]}^{a+b+c}{[T]}^{-a-c} \)

Equating the power of M,L and T respectively, we get

\( C=1\tag{1} \) \( a+b-c=1\tag{2} \) \( -a-c=-2\tag{3} \)

or, \( a+c=2\\or, \ a=2-c\\or, \ a=2-1\\or,\ a=1 \)

\( b=1+c-a=1+1-1=1 \)

Therefore, \( F=Kvr\eta \).

The value of K was found by Stokes to be \( 6\pi \).

Therefore, \( F=6\pi\eta{rv} \).

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