A fluid is flowing through a horizontal pipe of varying cross-section, with speed $\mathrm{v} \mathrm{ms}^{-1}$ a ta point where the pressure is P Pascal. 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 V is equal to:

(A) $\sqrt{\frac{\mathrm{P}}{\mathrm{p}}+\mathrm{v}}$

(B) $\sqrt{\frac{2 P}{p}+v^{2}}$

(C) $\sqrt{\frac{P}{2 p}+v^{2}}$

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Fluid Mechanics

A fluid is flowing through a horizontal pipe of varying cross-section, with speed

v ms^{- 1}

a ta 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:

(A) $\frac{P}{p} + v$
(B) $\frac{2 P}{p} + v^{2}$
(C) $\frac{P}{2 p} + v^{2}$
(D) $\frac{P}{p} + v^{2}$ ✓ Correct

Answer: (D)

Solution

For a horizontal pipe Bernoulli's equation reads

P + \frac{1}{2} ρ v^{2} = \frac{P}{2} + \frac{1}{2} ρ V^{2}

, with no height term since the pipe is level. Rearranging,

\frac{1}{2} ρ V^{2} = \frac{P}{2} + \frac{1}{2} ρ v^{2}

, so

V^{2} = \frac{P}{ρ} + v^{2}

and

V = \frac{P}{ρ} + v^{2}

JEE Main · 2020 · 06 Sep 2020 · Shift 2 · Single correct option

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