Question

The velocity of water in the river is $9\,{\rm{km/hr}}$ of the upper surface. The river is
$ {\rm{10}}\,{\rm{m}} $ deep. If the coefficient of viscosity of water is $ {\rm{1}}{{\rm{0}}^{ -
2}} $poise, then the shearing stress between horizontal layers of water is
A. $ 0.25 \times {10^{ - 2}}\,{\rm{N/}}{{\rm{m}}^{\rm{2}}} $
B. $ 0.25 \times {10^{ - 3}}\,{\rm{N/}}{{\rm{m}}^{\rm{2}}} $
C. $ 0.5 \times {10^{ - 3}}\,{\rm{N/}}{{\rm{m}}^{\rm{2}}} $
D. $ 0.75 \times {10^{ - 3}}\,{\rm{N/}}{{\rm{m}}^{\rm{2}}} $

Answer 1

Hint: In this solution, using Newton’s law of viscosity $ F = \eta A\dfrac{{d\nu }}{{dx}} $, find torque $ \tau $ and put all the value and find the stress in the formula to find the correct answer.

Complete Step by Step Solution
Velocity: The significance of an object’s velocity can be defined as the rate of change in the location of the object in relation to a frame of reference and time. It may sound confusing but velocity in a particular direction is simply speeding. It is a quantity of vectors which implies that we need both magnitude (speed) and direction to define velocity. If there is a change in magnitude or direction in a body’s velocity the body is said to be moving, the SI unit of it is meter per second ($ {\rm{m}}{{\rm{s}}^{ - {\rm{2}}}} $).

Newton’s law of viscosity: Viscosity is the physical property that characterizes simple fluid flow-resistance. Newton’s viscosity law describes the relation between the shear stress and shear rate of a mechanically stressed fluid. Kinematic viscosity is the viscosity of dynamics separated by density.
Newton’s law of viscosity, we have:
$ F = \eta A\dfrac{{d\nu }}{{dx}} $
$ \tau = \dfrac{F}{A} = \eta \dfrac{{d\nu }}{{dx}} $
Also known as the strain rate is the velocity gradient. The law stipulates that shearing stress is directly proportional to the strain rate in a fluid. The proportionality constant is dubbed the viscosity coefficient. Unlike solid artifacts, a fluid has the property of yielding under infinitesimal shear stress, so stress is expressed in terms of strain rate and not strain. Note that the SI unit of viscosity is decapoise, where 1 decapoise = 10 poise.
Substituting, we have:
Shearing stress
$ \begin{array}{c}\tau = 1{0^{ - 3}} \times \dfrac{{5 \times 9}}{{18 \times 10}}\\ = 0.25
\times {10^{ - 3}}{\rm{N/}}{{\rm{m}}^{\rm{2}}}\end{array} $

Hence, the correct answer is B.

Note: In this solution, using Newton law of viscosity, find the expression of force per unit area which is designated as torque. Later put all the value on that given formula and find the correct answer.