
What will be the shearing stress between horizontal layers of the river water when it is flowing at $5m{s^{ - 1}}$ and the depth of the river is $10m$? Given that viscosity of water is 1 milli - decapoise.
A. $0.1 \times {10^{ - 3}}N{m^{ - 2}}$
B. $0.5 \times {10^{ - 3}}N{m^{ - 2}}$
C. ${10^{ - 3}}N{m^{ - 2}}$
D. $2 \times {10^{ - 3}}N{m^{ - 2}}$
Answer
491.7k+ views
Hint: Convert the viscosity given to its S.I units. As we know that,
$1mdaP = {10^{ - 3}}daP$
$daP$ 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
$Strain = \dfrac{v}{d}$
where, $v$ is the speed of river water
$d$ 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 $\tau .$
Let the viscosity be $\eta $ and let the stress be $\sigma .$
According to the question
$\eta = 1mdaP = {10^{ - 3}}daP$
Now, the shearing stress can be expressed as
$\sigma = \dfrac{{Strain}}{\eta } \cdots (1)$
Now, we have to calculate the shearing strain
So, let the speed of river water be $v$ and $d$ be the depth of river water.
The formula for calculating the shearing strain is-
$Strain = \dfrac{v}{d} \cdots (2)$
It is given that, speed of water is $5m{s^{ - 1}}$ and depth of river water is $10m$ therefore, putting these values in their respective places in equation $(2)$, we get
$
Strain = \dfrac{5}{{10}} \\
\Rightarrow Strain = \dfrac{1}{2} \\
$
Now, we have the value of strain and viscosity therefore, in equation $(1)$, we get
\[
\sigma = \dfrac{1}{2}({10^{ - 3}}) \\
\sigma = 0.5 \times {10^{ - 3}}N{m^{ - 2}} \\
\]
The value of shearing stress is \[0.5 \times {10^{ - 3}}N{m^{ - 2}}\]
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 $N/{m^2}$ or Pascal $(Pa).$ Shear stress is the ratio of Force acting on an object and area of an object. Mathematically, it can be represented as-
$\sigma = \dfrac{F}{A}$.
$1mdaP = {10^{ - 3}}daP$
$daP$ 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
$Strain = \dfrac{v}{d}$
where, $v$ is the speed of river water
$d$ 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 $\tau .$
Let the viscosity be $\eta $ and let the stress be $\sigma .$
According to the question
$\eta = 1mdaP = {10^{ - 3}}daP$
Now, the shearing stress can be expressed as
$\sigma = \dfrac{{Strain}}{\eta } \cdots (1)$
Now, we have to calculate the shearing strain
So, let the speed of river water be $v$ and $d$ be the depth of river water.
The formula for calculating the shearing strain is-
$Strain = \dfrac{v}{d} \cdots (2)$
It is given that, speed of water is $5m{s^{ - 1}}$ and depth of river water is $10m$ therefore, putting these values in their respective places in equation $(2)$, we get
$
Strain = \dfrac{5}{{10}} \\
\Rightarrow Strain = \dfrac{1}{2} \\
$
Now, we have the value of strain and viscosity therefore, in equation $(1)$, we get
\[
\sigma = \dfrac{1}{2}({10^{ - 3}}) \\
\sigma = 0.5 \times {10^{ - 3}}N{m^{ - 2}} \\
\]
The value of shearing stress is \[0.5 \times {10^{ - 3}}N{m^{ - 2}}\]
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 $N/{m^2}$ or Pascal $(Pa).$ Shear stress is the ratio of Force acting on an object and area of an object. Mathematically, it can be represented as-
$\sigma = \dfrac{F}{A}$.
Recently Updated Pages
Express the following as a fraction and simplify a class 7 maths CBSE

The length and width of a rectangle are in ratio of class 7 maths CBSE

The ratio of the income to the expenditure of a family class 7 maths CBSE

How do you write 025 million in scientific notatio class 7 maths CBSE

How do you convert 295 meters per second to kilometers class 7 maths CBSE

Write the following in Roman numerals 25819 class 7 maths CBSE

Trending doubts
Give 10 examples of unisexual and bisexual flowers

Draw a labelled sketch of the human eye class 12 physics CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Differentiate between insitu conservation and exsitu class 12 biology CBSE

What are the major means of transport Explain each class 12 social science CBSE

Franz thinks Will they make them sing in German even class 12 english CBSE
