Question
When a small sphere moves at low speed through a fluid, the viscous force $$F,$$ opposing the motion is experimentally found to depend upon the radius $$r,$$ the velocity $$v$$ of the sphere and the viscosity $$\eta $$ of the fluid. Expression for force is
A.
$$4\pi \eta r{v^2}$$
B.
$$4\pi \eta {r^2}v$$
C.
$$2\pi \eta {r^2}v$$
D.
$$6\pi \eta rv$$
Answer :
$$6\pi \eta rv$$
Solution :
We can thus say that the viscous force $$\left( F \right)$$ is the function of radius $$\left( r \right),$$ velocity $$\left( v \right)$$ and viscosity $$\left( \eta \right).$$
$${\text{or}}\,\,F = f\left( {\eta ,r,v} \right)\,\,{\text{or}}\,\,F = k{\eta ^x}{r^y}{v^z}\,......\left( {\text{1}} \right)$$
Where $$k$$ is a constant.
Now, dimensions of the constituents are
$$\eqalign{
& \therefore \left[ {ML{T^{ - 2}}} \right] = {\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]^x}{\left[ L \right]^y}{\left[ {L{T^{ - 1}}} \right]^z} \cr
& = \left[ {{M^x}{L^{ - x + y + z}}{T^{ - x - z}}} \right] \cr} $$
Equating the exponents of similar quantities of both sides we get, $$x = 1; - x + y + z = 1$$ and $$ - x - z = - 2$$
Solving for $$x,y\,\& \,z,$$ we get $$x = y = z = 1$$
Equation (1) becomes $$F = k\eta rv$$
Experimentally, it was found that
$$k = 6\pi \,\,{\text{or}}\,\,F = 6\pi \eta rv,$$ which is the famous Stokes' law.