Question

The volume of a liquid $\left( V \right)$ flowing per second through a cylindrical tube depends upon the pressure gradient $\left( {P/l} \right)$ , radius of the tube $\left( r \right)$ and the coefficient of viscosity $\left( \eta \right)$ . Find the correct formula using dimensional analysis.
(A) $V \propto \dfrac{{{{\Pr }^4}}}{{\eta l}}$
(B) $V \propto \dfrac{{\Pr }}{{\eta {l^4}}}$
(C) $V \propto \dfrac{{\operatorname{P} {l^4}}}{{\eta r}}$
(D) None

Answer 1

Hint
First write down the dimensional formulae for each of the physical quantities on which the volume flowing per second is dependent on. Assuming a power to each of these we can equate the dimensional formulae at both the sides of the equation. Thus getting to know the unknown powers, we will get the exact formula for volume flowing per second.
Formulae used: Pressure, $P = \dfrac{F}{A}$ where $F$ is the force and $A$ is the area.
Coefficient of viscosity, $\eta = \dfrac{F}{{A\left( {\dfrac{{dv}}{{dx}}} \right)}}$ , where $F$ is the force, $A$ is the area and $\left( {\dfrac{{dv}}{{dx}}} \right)$ is the velocity gradient.

Complete step by step answer
As it is given that the volume of liquid flowing per second, $V$ is proportional to the pressure gradient, $\left( {\dfrac{P}{l}} \right)$ , we can write:
$V \propto {\left( {\dfrac{P}{l}} \right)^a}$
where, $a$ is the power assumed to which the pressure gradient is raised.
Also, since it is said that $V$ is proportional to the radius of the tube $\left( r \right)$ , we write:
$V \propto {r^b}$
where $b$ is assumed to be the power of $r$ .
Similarly, we can also write as:
$V \propto {\eta ^c}$
and $c$ is the power assumed for $\eta$ .
Thus, effectively we can write that:
$V \propto {\left( {\dfrac{P}{l}} \right)^a}{\left( r \right)^b}{\left( \eta \right)^c}$
Now, let us write the dimensional formulae for each of these physical quantities. Since $V$ is the volume of liquid flowing per second, the dimensional formula of $V$ is:
$\left[ V \right] = \left[ {{L^3}{T^{ - 1}}} \right]$
For writing the dimensional formula of the pressure gradient $\left( {\dfrac{P}{l}} \right)$ , we should know the dimensions of pressure and length, which are $\left[ P \right] = \left[ {M{L^{ - 1}}{T^{ - 2}}} \right]$ and $\left[ l \right] = \left[ L \right]$ . So, we can write the dimensions of pressure gradient as:
$\left[ {\dfrac{P}{l}} \right] = \dfrac{{\left[ P \right]}}{{\left[ l \right]}} = \dfrac{{\left[ {M{L^{ - 1}}{T^{ - 2}}} \right]}}{{\left[ L \right]}} = \left[ {M{L^{ - 2}}{T^{ - 2}}} \right]$
Similarly, for the radius of the tube,
$\left[ r \right] = \left[ L \right]$
Finally for the coefficient of viscosity, from its formula we can write as:
$\left[ \eta \right] = \dfrac{{\left[ F \right]}}{{\left[ A \right] \times \left[ {\left( {\dfrac{{dv}}{{dx}}} \right)} \right]}}$
By substituting the dimensions of force, area and velocity gradient, we can write the dimensions of coefficient of viscosity as:
$\left[ \eta \right] = \dfrac{{\left[ {ML{T^{ - 2}}} \right]}}{{\left[ {{L^2}} \right] \times \left( {\left[ {\dfrac{{L{T^{ - 1}}}}{L}} \right]} \right)}} = \left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$
Substituting these individual dimensions in the proportionality for the volume of liquid flowing per second:
$\left[ {{L^3}{T^{ - 1}}} \right] \propto {\left[ {M{L^{ - 2}}{T^{ - 2}}} \right]^a}{\left[ L \right]^b}\left[ {M{L^{ - 1}}{T^{ - 1}}} \right]$
On expanding the expression:
${\left[ L \right]^3}{\left[ T \right]^{ - 1}} = {\left[ M \right]^a}{\left[ L \right]^{ - 2a}}{\left[ T \right]^{ - 2a}}{\left[ L \right]^b}{\left[ M \right]^c}{\left[ L \right]^{ - c}}{\left[ T \right]^{ - c}}$
This can also be written as:
${\left[ L \right]^3}{\left[ T \right]^{ - 1}} = {\left[ M \right]^{a + c}}{\left[ L \right]^{ - 2a + b - c}}{\left[ T \right]^{ - 2a - c}}$
Now, on comparing the powers of each of the dimensions at right hand side and left hand side:
$a + c = 0$ … (1)
$- 2a + b - c = 3$ … (2)
$- 2a - c = - 1$ … (3)
From (1), we can also write that, $a = - c$ … (4)
Substituting (4) in (3) and solving further we get,
$- 2a - c = 2c - c = - 1$
$\therefore c = - 1$ … (5)
Now, on substituting (5) in (1), we obtain $a = 1$ … (6)
Finally, substituting (5) and (6) in (2) and solving we get:
$- 2 + b + 1 = 3 \\\ \therefore b = 3 - 1 + 2 = 4$
Hence, we obtained the coefficients as: $a = 1$ , $b = 4$ and $c = - 1$ .
Substituting these coefficients in proportionality we have written earlier for $V$ :
$V \propto \left( {\dfrac{P}{l}} \right){r^4}{\eta ^{ - 1}}$
We can also write it after rearranging as,
$V \propto \dfrac{{P{r^4}}}{{\eta l}}$
So, the correct formula for the volume of liquid flowing per second is $V \propto \dfrac{{P{r^4}}}{{\eta l}}$ .
Hence, the correct answer is option (A).

Note
This is a very simple method for checking the correctness of a formula using the dimensional analysis. By just knowing the dimensions of all the physical quantities involved in the formula, the correct formula can be found out. While performing this problem, one can go wrong while writing down the proportionality, because you should not leave any of the physical quantities that are needed to consider.