JEE Mains · Physics · STD 11 - Fluid mechanics
- JEE Mains 2024
- Hard
- Numerical
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 \(\mathrm{V}\) is__________
Pick an option to check your answer.
Answer & Solution
Correct Answer
(A) \(50\)
Step-by-step Solution
Detailed explanation
\( \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 \)
\( 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \)
\( 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \)
\( \mathrm{v}^2=50 \)
\( \mathrm{v}=\sqrt{50} \)
\( \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} \)
\( =\mathrm{V}=50\)
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