A fluid is flowing through a horizontal pipe of varying cross-section, with speed $v {ms}^{- 1}$ at a point where the pressure is $P$ Pascal. At another point where pressure is $\frac{P}{2}$ Pascal its speed is $V {ms}^{- 1}$ . If the density of the fluid is ${ρ kg m}^{- 3}$ and the flow is streamline, then V is equal to :

see full answer

Your Exam Success, Personally Taken Care Of

1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.

An Intiative by Sri Chaitanya

$\sqrt{\frac{P}{ρ} + v}$

$\sqrt{\frac{2 P}{ρ} + v^{2}}$

$\sqrt{\frac{P}{ρ} + v^{2}}$

$\sqrt{\frac{P}{2 ρ} + v^{2}}$

answer is D.

(Unlock A.I Detailed Solution for FREE)

Best Courses for You

JEE

NEET

Foundation JEE

Foundation NEET

CBSE

Detailed Solution

Using Bernoulli’s equation
$P_{1} + \frac{1}{2} ρ υ_{1}^{2} + ρ g h_{1} = P_{2} + \frac{1}{2} ρ υ_{2}^{2} + ρ g h_{2}$
For horizontal pipe, $h_{1} = 0$ and $h_{2} = 0$ and taking $P_{1} = P, P_{2} = \frac{P}{2}$
,we get $\Rightarrow P + \frac{1}{2} ρ υ^{2} = \frac{P}{2} + \frac{1}{2} ρ V^{2}$ $\Rightarrow \frac{p}{2} + \frac{1}{2} ρ υ^{2} = \frac{1}{2} ρ V^{2} \Rightarrow V = \sqrt{υ^{2} + \frac{P}{ρ}}$