Question

A closed pipe containing liquid showed a pressure \(P_1\) by gauge. When the valve was opened, pressure was reduced to \(P_2\). The speed of water flowing out of the pipe is (\(ρ\)=density of water)

• A. \([\frac{(P_1-P_2)}{ρ}]^{\frac{1}{2}}\)
• B. \([\frac{2(P_1-P_2)}{ρ}]^{\frac{1}{2}}\)
• C. \([\frac{(P_2-P_1)}{ρ}]^{\frac{1}{2}}\)
• D. \([\frac{2(P_2-P_1)}{ρ}]^{\frac{1}{2}}\)

Hint:

Apply Bernoulli's equation: the pressure drop becomes kinetic energy.

Answer 1

The Correct Option is B

Step 1: Work-energy view:
A volume $\Delta V$ of water leaves, and the pressure does work $(P_1 - P_2)\Delta V$.

Step 2: Kinetic energy:
This becomes kinetic energy $\tfrac12\rho\Delta V\,v^2$.

Step 3: Solve:
$v^2 = \dfrac{2(P_1-P_2)}{\rho}$, option (B).

Final Answer:
The speed is root of 2(P1 - P2)/rho. \[ \boxed{\text{(B) }\left[\dfrac{2(P_1-P_2)}{\rho}\right]^{1/2}} \]