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Question

A fluid is flowing through a horizontal pipe of varying cross-section, with speedvms-1 at a point where the pressure is Ppascal. At another point where pressure is P/2pascal its speed is Vms-1. If the density of the fluid is ρ in kgm-3 and the flow is streamlined, then V is equal to:


A

[(P/2ρ)+v2]

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B

[(P/ρ)+v2]

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C

[(2P/ρ)+v2]

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D

[(P/ρ)+v]

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Solution

The correct option is B

[(P/ρ)+v2]


Step 1: Given Data

Speed=vms-1

Pressure =Ppascal

Density of the fluid =ρkgm-3

At another point

Pressure =P/2pascal

Speed=Vms-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,

P1+12ρv12+ρgh1=P2+12ρv22+ρgh2

Here,

ρ=density of fluid

g=acceleration due to gravity

P1=pressure at one point

P2=pressure at other point

h1=elevation of one point

h2=elevation of other point

v1=velocity at one point

v2=velocity at other point

Since, 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+(1/2)ρv2=[P/2]+[1/2]ρV2V=[(P/ρ)+v2]

Hence, option B is correct.


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