Mechanical Properties Of Fluids Question 4
Question 4 - 2024 (27 Jan Shift 2)
The reading of pressure metre attached with a closed pipe is $4.5 \times 10^{4} \mathrm{~N} / \mathrm{m}^{2}$. 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
Show Answer
Answer: (50)
Solution:
Change in pressure $=\frac{1}{2} \rho \mathrm{v}^{2}$
$4.5 \times 10^{4}-2.0 \times 10^{4}=\frac{1}{2} \times 10^{3} \times \mathrm{v}^{2}$
$2.5 \times 10^{4}=\frac{1}{2} \times 10^{3} \times \mathrm{v}^{2}$
$\mathrm{v}^{2}=50$
$\mathrm{v}=\sqrt{50}$
Velocity of water $=\sqrt{\mathrm{V}}=\sqrt{50}$
$=\mathrm{V}=50$
