Question

The velocity of water in a river is $18\, km\,h ^{-1}$ near the surface. If the river is $5\, m$ 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
$=10^{-3} \,Pa\, s$
$v=18 \,km h ^{-1}=\frac{18000}{3600}=5\, m \, s ^{-1}$
$d=5\, m$
We know that viscous force, $F=\eta A \frac{ d v}{ d x}$
where,
$\frac{ d v}{ d x}$ 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{ d v}{ d x} $
$=10^{-3} \times \frac{5}{5} $
$=10^{-3} \,N m$