A fluid is flowing through a horizontal pipe

Question:

A fluid is flowing through a horizontal pipe of varying cross-section, with speed $\mathrm{v} \mathrm{ms}^{-1}$ at a point where the pressure is P Pascal. P At another point where pressure is $\frac{\mathrm{P}}{2}$ 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. $\sqrt{\frac{\mathrm{P}}{2 \rho}+\mathrm{v}^{2}}$

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

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

  4. $\sqrt{\frac{\mathrm{P}}{\rho}+\mathrm{v}}$


Correct Option: , 2

Solution:

Applying Bernoulli's Equation

$\mathrm{P}_{1}+\frac{1}{2} \rho \mathrm{v}_{1}^{2}+\rho \mathrm{g} \mathrm{y}_{1}=\mathrm{P}_{2}+\frac{1}{2} \rho \mathrm{v}_{2}^{2}+\rho \mathrm{g} \mathrm{y}_{2}$

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

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

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

Ans. (2)

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