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Question

# A fluid is flowing through a horizontal pipe of varying cross-section, with speed$vm{s}^{-1}$ at a point where the pressure is $Ppascal$. At another point where pressure is $P/2pascal$ its speed is $Vm{s}^{-1}$. If the density of the fluid is $\rho$ in $kg{m}^{-3}$ and the flow is streamlined, then $V$ is equal to:

A

$\sqrt{\left[\left(P/2\rho \right)+{v}^{2}\right]}$

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B

$\sqrt{\left[\left(P/\rho \right)+{v}^{2}\right]}$

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C

$\sqrt{\left[\left(2P/\rho \right)+{v}^{2}\right]}$

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D

$\sqrt{\left[\left(P/\rho \right)+v\right]}$

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Solution

## The correct option is B $\sqrt{\left[\left(P/\rho \right)+{v}^{2}\right]}$Step 1: Given DataSpeed$=vm{s}^{-1}$Pressure $=Ppascal$Density of the fluid $=\rho kg{m}^{-3}$At another pointPressure $=P/2pascal$Speed$=Vm{s}^{-1}$Step 2: Find $V$Bernoulli's equation formula is a relation between pressure, kinetic energy, and gravitational potential energy of a fluid in a container.By Bernoulli’s equation,${P}_{1}+\frac{1}{2}\rho {v}_{1}^{2}+\rho g{h}_{1}={P}_{2}+\frac{1}{2}\rho {v}_{2}^{2}+\rho g{h}_{2}$Here,$\rho =$density of fluid$g=$acceleration due to gravity${P}_{1}=$pressure at one point${P}_{2}=$pressure at other point${h}_{1}=$elevation of one point${h}_{2}=$elevation of other point${v}_{1}=$velocity at one point${v}_{2}=$velocity at other pointSince, pipe is horizontal, elevation of both point is same and potential energy will be same at both point and it will cancel out upon equating.$P+\left(1/2\right)\rho {v}^{2}=\left[P/2\right]+\left[1/2\right]\rho {V}^{2}\phantom{\rule{0ex}{0ex}}V=\sqrt{\left[\left(P/\rho \right)+{v}^{2}\right]}$Hence, option B is correct.

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