Q.Two horizontal layers of a liquid, separated by a perpendicular distance of $2.0\ \text{mm}$, are set into steady streamline flow such that the upper layer moves with a velocity of $0.50\ \text{m/s}$ relative to the lower layer, which is at rest. If the coefficient of viscosity of the liquid is $1.2\times10^{-3}\ \text{Pa·s}$ and the two layers are in contact over an area of $0.05\ \text{m}^2$, find

Find [η] from its defining relation, then assume F depends on η, r, v as a power law and match dimensions to derive Stokes' law up to the constant.

Streamline (laminar) flow: flow in which every fluid particle passing a given point follows the same smooth path (the path of the particle before it), with layers of fluid sliding over one another without mixing.

Turbulent flow: flow at high velocity in which the motion becomes irregular and chaotic, with eddies and mixing between layers — streamlines are no longer well defined.

Dimension of coefficient of viscosity ($\eta$): from $F=\eta A\dfrac{dv}{dx}$:

$[\eta]=\frac{[MLT^{-2}]}{[L^2][T^{-1}][L^{-1}]}=[ML^{-1}T^{-1}]$

Stokes' law (by dimensional analysis): assume the viscous drag force on a sphere depends on $\eta$, radius $r$, and velocity $v$ as $F=k\,\eta^a r^b v^c$ ($k$ dimensionless).

$[MLT^{-2}]=[ML^{-1}T^{-1}]^a[L]^b[LT^{-1}]^c$

Matching powers of M: $a=1$. Matching T: $-a-c=-2 \implies c=1$. Matching L: $-a+b+c=1 \implies b=1$.

So $F=k\eta rv$. The constant $k$ cannot be found from dimensional analysis alone; a full derivation (Stokes, 1851) gives $k=6\pi$:

$F=6\pi\eta rv$