Question

The velocity of water in the river is $9km/hr$ of the upper surface. The river is 10 m deep. If the coefficient of viscosity of water is ${{10}^{-2}}$ poise then the shearing stress between horizontal layers of water is.
(A). $0.25\times {{10}^{-2}}N/{{m}^{2}}$
(B). $0.25\times {{10}^{-3}}N/{{m}^{2}}$
(C). $0.5\times {{10}^{-3}}N/{{m}^{2}}$
(D). $0.75\times {{10}^{-3}}N/{{m}^{2}}$

Answer 1

Hint: First convert the units of all the given physical quantities to the same unit system. Then try to add them to the shearing stress equation. Use the equation of viscous force using the coefficient of viscosity. Put the values in the equation and we will find the answer.

Complete step by step answer:
Viscosity can be defined as the resistance given by a fluid layer to its neighbour layers to the flow of the fluid.
We know that for a river the velocity of the water will be highest at the top and the velocity will be lowest at the bottom.
Given, the velocity of the water the at top of river is, $V=9km/h$
Velocity of the water in the bottom of the river will be zero.
Depth of the river is 10 m.
Coefficient of viscosity is given as, $\eta ={{10}^{-2}}poise$
The unit of coefficient of viscosity is in the CGS unit system. Converting the unit to the SI unit system,
Coefficient of viscosity is, $\eta ={{10}^{-2}}\times {{10}^{-1}}kg{{m}^{-1}}{{s}^{-1}}={{10}^{-3}}kg{{m}^{-1}}{{s}^{-1}}$
Again, $V=9km/h=9\times \dfrac{{{10}^{3}}}{3600}=2.5m/s$
Now, shearing stress between the horizontal layers of water is given by,
$\text{stress = }\dfrac{force}{area}=\dfrac{F}{A}$
Where, F is the force and A is the area.
Now, the force is given by the equation,
$F=\eta A\dfrac{dv}{dx}$
Where, $\dfrac{dv}{dx}$ is the velocity gradient or the change in velocity per unit distance.
Pitting the value of force on the equation of shearing stress, we get,
$\text{stress = }\dfrac{F}{A}=\dfrac{\eta A\dfrac{dv}{dx}}{A}=\eta \dfrac{dv}{dx}$
Putting the values on the above equation, we get,
$\text{stress = 1}{{\text{0}}^{-3}}\times \dfrac{2.5}{10}=0.25\times {{10}^{-3}}N/{{m}^{2}}$
The correct option is (B).

Note: Shearing stress can be defined as the deformation caused by the force in the direction of the plane in which direction the force is applied. This is also called the tangential stress. i.e. it acts along the surface or parallel to the surface.