Question

A manometer connected to a closed tap reads $3.5\times {10}^5{\text{N/m}}^2$. When the valve is opened, the reading of the manometer falls to $3.0\times {10}^5{\text{N/m}}^2$, then the velocity of flow of water is:
(consider the water pipe to be horizontal)

• A. $100\text{m/s}$
• B. $10\text{m/s}$
• C. $1\text{m/s}$
• D. $10\sqrt{10}\text{m/s}$

Answer 1

The correct option is B $10\text{m/s}$

Applying Bernoulli's theorem per unit mass of liquid,
$\frac{P_1}{\rho }+\frac{v_1^2}{2}=\frac{P_2}{\rho }+\frac{v_2^2}{2}$
Since the pipe carrying the water is horizontal, the elevation is same everywhere.
$v_1=0$ as initially, tap was closed
& $v_2=v$ when tap opens and water starts to flow.
As the liquid starts flowing, its kinetic energy increases and pressure will decrease.
$\Rightarrow \frac{1}{2}{v}^2=\frac{P_1-P_2}{\rho }$
$\Rightarrow \frac{1}{2}{v}^2=\frac{(3.5\times {10}^5)-(3\times {10}^5)}{{10}^3}$
$\Rightarrow v^2=\frac{2\times 0.5\times {10}^5}{{10}^3}$
$\Rightarrow v=\sqrt{100}=10\text{m/s}$