Question

A tube of length $L$ and radius $R$ is joined to another tube of length $\dfrac{L}{3}$ and radius $\dfrac{R}{2}$. A fluid is flowing through this tube. If the tube pressure difference across the first tube is $P$, then the pressure difference across the tube is
$\begin{align}
  & \text{A}\text{. }\dfrac{16P}{3} \\
 & \text{B}\text{. }\dfrac{4P}{3} \\
 & \text{C}\text{. }P \\
 & \text{D}\text{. }\dfrac{3P}{16} \\
\end{align}$

Answer 1

Hint: The rate of flow in both the pipes will remain constant. So from the Poiseuillie’s Formula calculate the rate of flow. Then equate them and by putting the changed values of the radius and length calculate the new tube pressure in terms of old tube pressure.

Formula used: Poiseuillie’s formula gives the rate of flow and is given by, volume of liquid flowing per unit time.so,
$Q=\dfrac{V}{t}=\dfrac{\pi {{\Pr }^{4}}}{8\eta l}$, where
$\begin{align}
  & P=\text{ pressure inside the tube} \\
 & \eta =\text{ coefficient of viscosity} \\
 & r=\text{ radius of tube} \\
 & l=\text{ length of tube} \\
\end{align}$

Complete step by step answer:
 As the rate of flow of liquid remains the same.
Case 1: When the radius of tube is $R$ and length of the tube is $L$ then according to Poiseuillie’s formula the rate of flow of liquid will be -
$Q=\dfrac{\pi \operatorname{P}{{R}^{4}}}{8\eta L}$, where
\[\begin{align}
  & P=\text{ pressure inside the tube} \\
 & \eta =\text{ coefficient of viscosity} \\
 & R=\text{ radius of tube} \\
 & L=\text{ length of tube} \\
\end{align}\]

Case 2: When the length of the tube is reduced to $L'=\dfrac{L}{3}$ and radius $R'=\dfrac{R}{2}$ then the rate of flow will be ,
$Q'=\dfrac{\pi \operatorname{P}'R{{'}^{4}}}{8\eta L'}$
As the tubes were connected so $Q'=Q$. Equating $Q$ and $Q'$we get.
$\begin{align}
  & Q'=Q \\
 & \Rightarrow \dfrac{\pi \operatorname{P}'R{{'}^{4}}}{8\eta L'}=\dfrac{\pi \operatorname{P}{{R}^{4}}}{8\eta L} \\
\end{align}$
But $L'=\dfrac{L}{3}$ and $R'=\dfrac{R}{2}$
Putting the value of $L'\text{ and }R'$in above equation.
$\begin{align}
  & \dfrac{\pi \operatorname{P}'{{\left( \dfrac{R}{2} \right)}^{4}}}{8\eta \dfrac{L}{3}}=\dfrac{\pi \operatorname{P}{{R}^{4}}}{8\eta L} \\
 & \Rightarrow P'=\dfrac{16P}{3} \\
\end{align}$

So, the correct answer is “Option A”.

Note: Poiseullie’s formula is only valid for fluid flowing steadily parallel to the axis of the tube. The pressure across the tube is constant over any cross-section of the tube. The liquid velocity is zero at the wall of the tube and increases towards the axis of the tube. The tube should be held horizontal so that gravity has no effect on the flow of liquid inside the tube.