Given below are two statements : Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation \(P_{1}-P_{2}=\rho \mathrm{g}\left(\mathrm{h}_{2}-\mathrm{h}_{1}\right)\) Statement II : In ventury tube shown \(2 \mathrm{gh}=v_{1}^{2}-v_{2}^{2}\) In the light of the above statements, choose the most appropriate answer from the options given below.

(D) Statement I is correct but Statement II is incorrect.

(A) Both Statement I and Statement II are correct.

(B) Statement I is incorrect but Statement II is correct.

(C) Both Statement I and Statement II are incorrect.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Fluid Mechanics

Given below are two statements :

Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation

P_{1} - P_{2} = ρ g (h_{2} - h_{1})

Statement II : In ventury tube shown

2 gh = v_{1}^{2} - v_{2}^{2}

In the light of the above statements, choose the most appropriate answer from the options given below.

Answer: (D)

Apply Bernoulli's equation between two points of the fluid:

P_{1} + ρ g h_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + ρ g h_{2} + \frac{1}{2} ρ v_{2}^{2}

Statement I: When the liquid is at rest everywhere,

v_{1} = v_{2} = 0

, so

P_{1} - P_{2} = ρ g (h_{2} - h_{1}) .

This is the hydrostatic result — correct.

Statement II: In a Venturi tube the two points are at the same level (

h_{1} = h_{2}

), so

P_{1} + \frac{1}{2} ρ v_{1}^{2} = P_{2} + \frac{1}{2} ρ v_{2}^{2} .

The manometer reading gives

P_{1} - P_{2} = ρ g h

, hence

ρ g h = \frac{1}{2} ρ v_{2}^{2} - \frac{1}{2} ρ v_{1}^{2} \Rightarrow 2 g h = v_{2}^{2} - v_{1}^{2} .

The statement writes

2 g h = v_{1}^{2} - v_{2}^{2}

, which has the wrong sign — incorrect.

Statement I correct, Statement II incorrect

\Rightarrow

option D.

JEE Main · 2024 · 04 Apr 2024 · Shift 1 · Single correct option