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.

Answer 2

The rate of flow of water through a hole in the tank is given by Torricelli’s law, which states that the velocity of the water emerging from the hole is given by: \[ v = \sqrt{2gh} \] where \( h \) is the depth of the hole from the water surface and \( g \) is the acceleration due to gravity. The rate of flow (or volume flux) is the product of the velocity and the area of the hole. The area of the hole is \( A = \pi r^2 \), where \( r \) is the radius of the hole. Therefore, the rate of flow \( Q \) is: \[ Q = A \cdot v = \pi r^2 \cdot \sqrt{2gh} \] However, there is an additional term due to the difference in pressure between the inside of the tank and atmospheric pressure. The pressure inside the tank at a depth \( h \) is \( P_0 \), and the pressure at the hole is \( P_a \) (the atmospheric pressure). The pressure difference contributes to the velocity and thus the rate of flow. The modified formula for the rate of flow is: \[ Q = \pi r^2 \cdot \sqrt{2gh + \frac{2(P_0 - P_a)}{\rho}} \] where \( \rho \) is the density of water. This equation accounts for both the gravitational and pressure-driven components of the flow.

Thus, the correct answer is: (A) \( \pi r^2 \sqrt{2gh + \frac{2(P_0 - P_a)}{\rho}} \)