Question

Water flows through a horizontal pipe. At one point, the velocity is $2~m/s$ and the pressure is $2~m$ of water column. What is the pressure at another point where the velocity is $3~m/s$?
(Density of water $=10^3~kg/m^3$)

• A. $1.75~m$ of water
• B. $1.2~m$ of water
• C. $1~m$ of water
• D. $1.5~m$ of water

Hint:

In a horizontal pipe: \[ \text{Higher velocity} \Rightarrow \text{Lower pressure} \] according to Bernoulli’s principle.

Answer 1

The Correct Option is D

Concept:
For flow through a horizontal pipe, Bernoulli’s equation is: \[ \frac{P}{\rho g} + \frac{v^2}{2g} = \text{constant} \] Since the pipe is horizontal: \[ h_1 = h_2 \] therefore height terms cancel.

Step 1: Write Bernoulli’s equation between two points. \[ \frac{P_1}{\rho g} + \frac{v_1^2}{2g} = \frac{P_2}{\rho g} + \frac{v_2^2}{2g} \] Given: \[ v_1 = 2~m/s \] \[ v_2 = 3~m/s \] \[ \frac{P_1}{\rho g} = 2~m \]

Step 2: Substitute values. \[ 2 + \frac{(2)^2}{2g} = \frac{P_2}{\rho g} + \frac{(3)^2}{2g} \] Take: \[ g \approx 10~m/s^2 \] \[ 2 + \frac{4}{20} = \frac{P_2}{\rho g} + \frac{9}{20} \] \[ 2 + 0.2 = \frac{P_2}{\rho g} + 0.45 \] \[ 2.2 = \frac{P_2}{\rho g} + 0.45 \]

Step 3: Calculate pressure head. \[ \frac{P_2}{\rho g} = 2.2 - 0.45 \] \[ \frac{P_2}{\rho g} = 1.75~m \]

Step 4: Conclusion. Hence, the pressure at the second point is: \[ \boxed{1.75~m\text{ of water}} \] Therefore, the correct option is: \[ \boxed{(A)} \]