Question

A closed water tank has cross-sectional area A. It has small hole at a depth of h from the free surface of water. The radius of the hole is r so that r << \(\sqrt{\frac{A}{\pi}}\). If P_o is the pressure inside the tank above water level, and P_a is the atmospheric pressure, the rate of flow of the water coming out of the hole is [ ρ is the density of water]

• A. \(\pi r^2\sqrt{2gh+\frac{2(P_o-P_a)}{\rho}}\)
• B. \(\pi r^2\sqrt{2gH}\)
• C. \(\pi r^2\sqrt{gh+\frac{2(P_o-P_a)}{\rho}}\)
• D. \(\pi r^2\sqrt{2gh}\)

Answer 1

The Correct Option is A

Given: 

• Cross-sectional area of the tank: \( A \)
• Radius of the hole: \( r \)
• Depth of hole below water surface: \( h \)
• Pressure above water level: \( P_o \)
• Atmospheric pressure: \( P_a \)
• Density of water: \( \rho \)

Step 1: Applying Bernoulli’s Theorem

Using Bernoulli’s equation between the surface of the liquid and the hole:

\[ P_o + \rho g H = P_a + \frac{1}{2} \rho v^2 \]

Solving for \( v \), we get:

\[ v = \sqrt{2gh + \frac{2(P_o - P_a)}{\rho}} \]

Step 2: Determining the Rate of Flow

The rate of flow \( Q \) is given by:

\[ Q = A_{hole} v = \pi r^2 \sqrt{2gh + \frac{2(P_o - P_a)}{\rho}} \]

Answer: The correct option is A.