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Question: What will be the shearing stress between horizontal layers of the river water when it is flowing at ...

What will be the shearing stress between horizontal layers of the river water when it is flowing at 5ms−15m{s^{ - 1}} and the depth of the river is 10m10m? Given that viscosity of water is 1 milli - decapoise.
A. 0.1×10−3Nm−20.1 \times {10^{ - 3}}N{m^{ - 2}}
B. 0.5×10−3Nm−20.5 \times {10^{ - 3}}N{m^{ - 2}}
C. 10−3Nm−2{10^{ - 3}}N{m^{ - 2}}
D. 2×10−3Nm−22 \times {10^{ - 3}}N{m^{ - 2}}

Explanation

Solution

Convert the viscosity given to its S.I units. As we know that,
1mdaP=10−3daP1mdaP = {10^{ - 3}}daP
daPdaP is used to denote decapoise.
Now, use the expression of shearing stress which is-
Shearing Stress = Shearing Strain ×\times Viscosity
And shearing strain can be expressed as
Strain=vdStrain = \dfrac{v}{d}
where, vv is the speed of river water
dd is the depth of river water

Complete step-by-step solution:
Shearing is the process in which the parallel layers slide over each other.
Shearing stress is the stress which is experienced by the object when it undergoes deformation if the external force is applied on it. When the direction of this external force is parallel to the plane of the object. The deformation of an object is also along that plane. It is denoted by τ.\tau .
Let the viscosity be η\eta and let the stress be σ.\sigma .
According to the question
η=1mdaP=10−3daP\eta = 1mdaP = {10^{ - 3}}daP
Now, the shearing stress can be expressed as
σ=Strainη⋯(1)\sigma = \dfrac{{Strain}}{\eta } \cdots (1)
Now, we have to calculate the shearing strain
So, let the speed of river water be vv and dd be the depth of river water.
The formula for calculating the shearing strain is-
Strain=vd⋯(2)Strain = \dfrac{v}{d} \cdots (2)
It is given that, speed of water is 5ms−15m{s^{ - 1}} and depth of river water is 10m10m therefore, putting these values in their respective places in equation (2)(2), we get
Strain=510 ⇒Strain=12  Strain = \dfrac{5}{{10}} \\\ \Rightarrow Strain = \dfrac{1}{2} \\\
Now, we have the value of strain and viscosity therefore, in equation (1)(1), we get

σ=12(10−3) σ=0.5×10−3Nm−2  \sigma = \dfrac{1}{2}({10^{ - 3}}) \\\ \sigma = 0.5 \times {10^{ - 3}}N{m^{ - 2}} \\\

The value of shearing stress is 0.5×10−3Nm−20.5 \times {10^{ - 3}}N{m^{ - 2}}
Hence, the correct answer is option (B).

Note:- Shearing stress has both magnitude and direction. Therefore, it is a vector quantity. The S.I unit for shearing stress is N/m2N/{m^2} or Pascal (Pa).(Pa). Shear stress is the ratio of Force acting on an object and area of an object. Mathematically, it can be represented as-
σ=FA\sigma = \dfrac{F}{A}.