Question

The volume of a liquid (v) flowing per second through a cylindrical tube depends upon the pressure gradient ($\dfrac{p}{l}$) radius of the tube (r) coefficient of viscosity ($\eta$)of the liquid by dimensional method of the correct formula is
(A) $V\alpha \dfrac{{{\Pr }^{4}}}{\eta l}$
(B) $V\alpha \dfrac{\Pr }{\eta {{l}^{4}}}$
(C) $V\alpha \dfrac{\operatorname{P}{{l}^{4}}}{\eta r}$
(D) None

Answer 1

Hint: We know that viscosity is defined as the measure of the fluid’s resistance to the flow of the fluid. This gives us an idea about the internal friction of the moving fluid that we have taken into consideration. So, we can say that a fluid with large viscosity gives a resistance to the motion because of the molecular structure which results in the formation of a large amount of internal friction. Based on this concept we have to solve this question.

Complete step by step answer:
We should know that the resistance that is developed in case of viscosity is basically a frictional force that is acting between the parts of the fluid that are travelling at different speeds.
We know that the formula to relate the volume and pressure is given below:
$V\propto {{R}^{a}}{{\left( \dfrac{P}{L} \right)}^{b}}{{\eta }^{c}}$
The values of variables are given to us in the question as:
$V={{L}^{3}}{{T}^{-1}}$
$R=L$
The dimension is given as:
$\dfrac{P}{L}=\left[ M \right]{{\left[ T \right]}^{-2}}{{\left[ L \right]}^{-2}}$
So, now the coefficient of viscosity is given as:
$\eta =\left[ M \right]{{\left[ T \right]}^{-1}}{{\left[ L \right]}^{-1}}$
Now we have to further evaluate to get the expression as:
$\Rightarrow {{\left[ L \right]}^{3}}{{\left[ T \right]}^{-1}}={{\left[ L \right]}^{a}}{{\left[ M \right]}^{b}}{{\left[ T \right]}^{-2b}}{{\left[ L \right]}^{-2b}}{{\left[ M \right]}^{c}}{{\left[ L \right]}^{-c}}{{\left[ T \right]}^{-c}}$
Now we get the algebraic equations as:
$a-2b-c=3$
$b+c=0$
$2b-c=1$
After solving the above expressions, we get the value of the variables as:
$a=4,\ b=1,\ c=-1$
So, the dimensional method of the correct formula is given to us by:
$V\propto \dfrac{P{{R}^{4}}}{\eta L}$

Hence, the correct answer is C.

Note : We know that there are two types of viscosity which are commonly considered as kinematic and dynamic. So, let us begin with dynamic viscosity is the relationship between the shear stress and the shear rate in case of a fluid. Now let us define the kinematic viscosity. The Kinetic viscosity is defined as the relationship between viscous and the inertial forces in case of a fluid.
The viscosity gives us an idea about how a fluid is basically a measure of how sticky it is. In case of the fluids flowing through the pipes, the viscosity will produce a force which is resistive in nature.