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JEE Main 20244 AprilMorning Shift
Exam
JEE Main
Year
2024
Date
4 April
Shift
Morning Shift
Subject
Physics
Chapter
Mechanical Properties of Fluids

JEE Main 2024 · 4 April · Morning Shift · Physics · Mechanical Properties of Fluids

Given below are two statements: Statement I : When speed of liquid...

Given below are two statements: Statement I : When speed of liquid is zero everywhere, pressure difference at any two points depends on equation P1−P2=ρg(h2−h1)P_1-P_2=\rho g\left(h_2-h_1\right). Statement II : In ventury tube shown 2gh=v12−v222 \mathrm{gh}=v_1^2-v_2^2 Given below are two statements: Statement I : When speed of liquid is zero everywhere, pressure difference at any two
In the light of the above statements, choose the most appropriate answer from the options given below.

Options

  1. ABoth Statement I and Statement II are correct.
  2. BStatement I is correct but Statement II is incorrect.
  3. CStatement I is incorrect but Statement II is correct.
  4. DBoth Statement I and Statement II are incorrect.

Correct answer

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

Step-by-step solution

Given below are two statements: Statement I : When speed of liquid is zero everywhere, pressure difference at any two
Applying Bernoulli's equation P1+ρgh1+12ρv12=P2+ρgh2+12ρv22[h1&h2\mathrm{P}_1+\rho \mathrm{gh}_1+\frac{1}{2} \rho \mathrm{v}_1^2=\mathrm{P}_2+\rho \mathrm{gh}_2+\frac{1}{2} \rho \mathrm{v}_2^2 \left[h_1 \& h_2\right. are height of point from any reference level] Given V1=V2=0\mathrm{V}_1=\mathrm{V}_2=0 (for statement-1) ∴P1−P2=g(h2−h2)\therefore \mathrm{P}_1-\mathrm{P}_2=\mathrm{g}\left(\mathrm{h}_2-\mathrm{h}_2\right)
For statement-2 P1+12ρv12=P2+12ρv22\mathrm{P}_1+\frac{1}{2} \rho \mathrm{v}_1^2=\mathrm{P}_2+\frac{1}{2} \rho \mathrm{v}_2^2P1−P2=ρghP1−P2=12ρv22−12ρv12ρgh=12ρv22−12ρv122gh=v22−v12\begin{aligned} & P_1-P_2=\rho g h \\ & P_1-P_2=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2 \\ & \rho g h=\frac{1}{2} \rho v_2^2-\frac{1}{2} \rho v_1^2 \\ & 2 g h=v_2^2-v_1^2 \end{aligned}$$
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