JEE MAIN - Physics (2024 - 27th January Evening Shift - No. 29)
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 $$\sqrt{V} \mathrm{~m} / \mathrm{s}$$. The value of $$V$$ is _________.
Answer
50
Explanation
$$\begin{aligned}
& \text { Change in pressure }=\frac{1}{2} \rho \mathrm{v}^2 \\
& 4.5 \times 10^4-2.0 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\
& 2.5 \times 10^4=\frac{1}{2} \times 10^3 \times \mathrm{v}^2 \\
& \mathrm{v}^2=50 \\
& \mathrm{v}=\sqrt{50} \\
& \text { Velocity of water }=\sqrt{\mathrm{V}}=\sqrt{50} \\
& =\mathrm{V}=50
\end{aligned}$$
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