Mechanical Properties Of Fluids Question 4
Question 4 - 2024 (27 Jan Shift 2)
The reading of pressure metre attached with a closed pipe is $4.5 \times 10^{4} N / m^{2}$. On opening the valve, water starts flowing and the reading of pressure metre falls to $2.0 \times 10^{4} N / m^{2}$. The velocity of water is found to be $\sqrt{V} m / s$. The value of $V$ is
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Answer: (50)
Solution:
Change in pressure $=\frac{1}{2} \rho v^{2}$
$4.5 \times 10^{4}-2.0 \times 10^{4}=\frac{1}{2} \times 10^{3} \times v^{2}$
$2.5 \times 10^{4}=\frac{1}{2} \times 10^{3} \times v^{2}$
$v^{2}=50$
$v=\sqrt{50}$
Velocity of water $=\sqrt{V}=\sqrt{50}$
$=V=50$
