Physics · Fluid Mechanics

JEE Main 2024 — 27 January, Shift 2 — Question 54

The reading of pressure metre attached with a closed pipe is $4.5 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ . On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$ . The velocity of water is found to be $\sqrt{\mathrm{V}} \mathrm{m} / \mathrm{s}$ . The value of V is $\qquad$

Answer: 50

Numerical answer — enter this value.

Step-by-step solution

Change in pressure $=\frac{1}{2} \rho v^{2}$

$4.5 \times 10^{4}-2.0 \times 10^{4}=\frac{1}{2} \times 10^{3} \times \mathrm{v}^{2}$

$2.5 \times 10^{4}=\frac{1}{2} \times 10^{3} \times \mathrm{v}^{2}$

$\mathrm{v}^{2}=50$

$\mathrm{v}=\sqrt{50}$

Velocity of water $=\sqrt{V}=\sqrt{50}$

$=\mathrm{V}=50$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 27 January, Shift 2
Subject: Physics
Chapter: Fluid Mechanics
Topic: Bernoulli's Equation and its Applications