 # A fluid is flowing through a horizontal pipe of varying cross-section Question:

A fluid is flowing through a horizontal pipe of varying cross-section, with speed $v \mathrm{~ms}^{-1}$ at a point where the pressure is $P$ Pascal. At another point where pressure is

$\frac{P}{2}$ Pascal its speed is $\mathrm{V} \mathrm{ms}^{-1}$. If the density of the fluid is

$\rho \mathrm{kg} \mathrm{m}^{-3}$ and the flow is streamline, then $\mathrm{V}$ is equal to:

1. (1) $\sqrt{\frac{P}{\rho}+v}$

2. (2) $\sqrt{\frac{2 P}{\rho}+v^{2}}$

3. (3) $\sqrt{\frac{P}{2 \rho}+v^{2}}$

4. (4) $\sqrt{\frac{P}{\rho}+v^{2}}$

Correct Option: , 4

Solution:

(4) Using Bernoulli's equation

$P_{1}+\frac{1}{2} \rho v_{1}^{2}+\rho g h_{1}=P_{2}+\frac{1}{2} \rho v_{2}^{2}+\rho g h_{2}$

For horizontal pipe, $h_{1}=0$ and $h_{2}=0$ and taking

$P_{1}=P, P_{2}=\frac{P}{2}$, we get

$\Rightarrow P+\frac{1}{2} \rho v^{2}=\frac{P}{2}+\frac{1}{2} \rho V^{2}$

$\Rightarrow \frac{P}{2}+\frac{1}{2} \rho v^{2}=\frac{1}{2} \rho V^{2} \Rightarrow V=\sqrt{v^{2}+\frac{P}{\rho}}$  