Question

A dimensionally consistent relation for the volume V of a liquid of coefficient of viscosity $'\eta '$ flowing per second, through a tube of radius r and length l and having a pressure difference P across its ends, is:
A. $V=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}$
B. $V=\dfrac{\pi \eta }{8{{\Pr }^{4}}}$
C. $V=\dfrac{8\text{P}\eta }{\pi {{r}^{4}}}$
D. $V=\dfrac{\pi \text{P}\eta }{8{{r}^{4}}}$

Answer 1

Hint: We need to find the dimensionally consistent relation of volume, ‘V’ from the given relations. ‘V’ is the volume of the liquid flowing per given time. We can find the dimension of volume flowing per unit time. Now check the dimensions of the given relations and thus we can find the correct relation.

Formula used:
Pressure, $P=\dfrac{F}{A}$
Coefficient of viscosity, $\eta =\left( \dfrac{\tau }{\dfrac{du}{dt}} \right)$

Complete answer:
The equation for volume, ‘V’ of a liquid that has a coefficient of viscosity $'\eta '$ flowing per second through a tube is given.
The tube has a radius ‘r’ and length ‘l’, the pressure difference across the pipe’s ends is given as ‘P’.
Since, here we have volume per second; the unit of volume per second will be,
$V=\dfrac{V}{\sec }=\dfrac{{{m}^{3}}}{\sec }$
Hence its dimension will be,
$\left[ V \right]=\left( {{L}^{3}}{{T}^{-1}} \right)$
We need to check for the relation with the dimension same as above.
We are given four relations; we have to check for the dimensionally consistent relation for volume.
In the 1st case we have the equation for V as,
\[V=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}\]
Here $'\pi '$ and $'8'$ are constants, therefore it has no dimension.
Pressure, P is the ratio of force and area.
$P=\dfrac{F}{A}$, where ‘F’ is force and ‘A’ is area.
Therefore unit of pressure,
$P=\dfrac{N}{{{m}^{2}}}$
We know, $N=\dfrac{kgm}{{{\sec }^{2}}}$
Therefore,
$\begin{align}
  & P=\dfrac{kgm}{{{\sec }^{2}}{{m}^{2}}} \\
 & P=\dfrac{kg}{m{{\sec }^{2}}} \\
\end{align}$
Therefore the dimension of ‘P’ is,
$\left[ P \right]=\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-2}} \right]$
In the equation for ‘V’, ‘r’ is radius.
Dimension for radius is,
$\left[ r \right]=\left[ {{L}^{1}} \right]$
And we know, ‘l’ is length.
Dimension of length is,
$\left[ l \right]=\left[ {{L}^{1}} \right]$
Coefficient of viscosity,
$\eta =\dfrac{\tau }{\left( \dfrac{du}{dt} \right)}$, Were $'\tau '$ is shear stress and $'\dfrac{du}{dt}'$ is rate of change of velocity with respect to time.
Unit of $'\eta '$ is,
$\begin{align}
  & \eta =\dfrac{kgm\sec }{{{\sec }^{2}}\times {{m}^{2}}} \\
 & \eta =\dfrac{kg}{m\sec } \\
\end{align}$
Therefore, the dimension of $'\eta '$ is,
$\left[ \eta \right]=\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-1}} \right]$
Now that we have dimensions of all variables, dimension of volume in the 1st relation is,
$\begin{align}
  & V=\dfrac{\pi {{\Pr }^{4}}}{8\eta l} \\
 & \implies \left[ V \right]=\dfrac{\left[ {{M}^{1}}{{L}^{-1}}T{{-}^{2}} \right]\left[ {{L}^{4}} \right]}{\left[ {{M}^{1}}{{L}^{-1}}{{T}^{-1}} \right]\left[ L \right]} \\
 & \implies \left[ V \right]=\left[ {{L}^{3}}{{T}^{-1}} \right] \\
\end{align}$
This is equal to the dimension of V.

So, the correct answer is “Option A”.

Note:
The powers to which the fundamentals quantities are raised to obtain other physical quantities are known as dimensions.
The representation of a physical quantity in its dimensions as the fundamental quantities is called a dimensional formula.
When two quantities are dimensionally consistent, their dimensional formula will be equal.