Question

The velocity of water in a river is $18km{h}^{-1}$ near the surface. If the river is $5m$ deep, find the shearing stress between the horizontal layers of water. The coefficient of viscosity of water $=10^{-2}$ poise.

• A. $10^{-1}N{m}^{-2}$
• B. $10^{-2}N{m}^{-2}$
• C. $10^{-3}N{m}^{-2}$
• D. $10^{-4}N{m}^{-2}$

Answer 1

Answer: (C)

Given,
$\eta =10^{-2}$ poise
$=10^{-3}Pas$
$v=18km{h}^{-1}=\frac{18000}{3600}=5m{s}^{-1}$
$d=5m$
We know that viscous force, $F=\eta A\frac{dv}{dx}$
where,
$\frac{dv}{dx}$ is the velocity gradient,
$A$ is the area under consideration,
$\eta$ is the coefficient of viscosity of water.
We know that,
shearing stress $=\frac{F}{A}$
$\frac{F}{A}=\eta \frac{dv}{dx}$
$=10^{-3}\times \frac{5}{5}$
$=10^{-3}Nm$