Question

The velocity of water in a river is $18{\text{kmh}}^{-1}$ near the surface. If the river is $5\text{m}$ deep, then the shearing stress between the surface layer and the bottom layer is:
(Given - coefficient of viscosity of water, $\eta ={10}^{-3}\text{Pa.s}$)

• A. ${10}^{-2}{\text{Nm}}^{-2}$
• B. ${10}^{-3}{\text{Nm}}^{-2}$
• C. ${10}^{-4}{\text{Nm}}^{-2}$
• D. ${10}^{-5}{\text{Nm}}^{-2}$

Answer 1

The correct option is B ${10}^{-3}{\text{Nm}}^{-2}$

Since the velocity of water in contact with the bottom of the river is zero,
$\Delta v=18{\text{kmh}}^{-1}=18\times \frac{5}{18}=5{\text{ms}}^{-1}$
Also, $\Delta z=5\text{m},\eta ={10}^{-3}\text{Pa.s}$
Velocity gradient is given as,
$\frac{\Delta v}{\Delta z}=\frac{5}{5}=1{\text{s}}^{-1}$
Viscous force is,
$F_v=\eta A\frac{dv}{dz}$
Shearing stress will be,
${\sigma }_s=\frac{F_v}{A}=\eta \frac{dv}{dz}=\eta \frac{\Delta v}{\Delta z}$
$\Rightarrow {\sigma }_s={10}^{-3}\times 1={10}^{-3}{\text{Nm}}^{-2}$