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: Understanding the Concept:
Bernoulli's equation for a streamline at the same level: \(P + \tfrac12\rho v^2 = \text{constant}\).

Step 2: Apply it:
Inside the closed pipe the water is at rest at pressure \(P_1\). At the outlet the pressure is \(P_2\) and the speed is \(v\):
\[ P_1 = P_2 + \tfrac12\rho v^2 \]

Step 3: Solve for v:
\[ v = \sqrt{\frac{2(P_1 - P_2)}{\rho}} \]

Step 4: Check the options:
Option (B) matches. Option (A) misses the factor 2. Options (C) and (D) have \(P_2 - P_1\), which is negative because \(P_2 < P_1\).

Final Answer:
Pressure difference turns into 1/2 rho v squared. \[ \boxed{\text{(B) }\left[\dfrac{2(P_1-P_2)}{\rho}\right]^{1/2}} \]