If water falls from a tap, then the volume rate of flow of water at depth of h, (`A_0` is the area of cross-section of the mouth and `A_0/2` is the corresponding area at depth h)

Options

A. `A_0sqrt(2gh)`
B. `A_0sqrt(2/3gh)`
C. `2A_0sqrt(gh)`
D. `A_0sqrt(3gh)`

Answer

To solve the problem of finding the volume rate of flow of water at a depth \( h \) when water falls from a tap, we can follow these steps:

### Step 1: Understand the Areas
We have two cross-sectional areas:
- At the mouth of the tap: \( A_0 \)
- At depth \( h \): \( A = \frac{A_0}{2} \)

### Step 2: Apply the Equation of Continuity
According to the principle of continuity, the product of the area and velocity at two points must be constant. Thus, we can write:
\[ A_0 \cdot v_1 = \frac{A_0}{2} \cdot v_2 \]
From this equation, we can express \( v_1 \) in terms of \( v_2 \):
\[ v_1 = \frac{v_2}{2} \]

### Step 3: Apply Bernoulli’s Theorem
Bernoulli's equation states that for an incompressible fluid, the total mechanical energy along a streamline is constant. We can apply it at two points:
1. At the top (point 1):
- Pressure = \( P_0 \)
- Height = \( h \)
- Velocity = \( v_1 \)

2. At the bottom (point 2):
- Pressure = \( P \) (which we can assume to be atmospheric pressure, \( P_0 \))
- Height = 0
- Velocity = \( v_2 \)

Using Bernoulli's equation:
\[ P_0 + \rho g h + \frac{1}{2} \rho v_1^2 = P + \frac{1}{2} \rho v_2^2 \]
Since \( P = P_0 \), we can simplify this to:
\[ \rho g h + \frac{1}{2} \rho v_1^2 = \frac{1}{2} \rho v_2^2 \]

### Step 4: Substitute \( v_1 \) into Bernoulli's Equation
Substituting \( v_1 = \frac{v_2}{2} \) into the equation:
\[ \rho g h + \frac{1}{2} \rho \left(\frac{v_2}{2}\right)^2 = \frac{1}{2} \rho v_2^2 \]
This simplifies to:
\[ \rho g h + \frac{1}{8} \rho v_2^2 = \frac{1}{2} \rho v_2^2 \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ \rho g h = \frac{1}{2} \rho v_2^2 - \frac{1}{8} \rho v_2^2 \]
\[ \rho g h = \left(\frac{4}{8} - \frac{1}{8}\right) \rho v_2^2 \]
\[ \rho g h = \frac{3}{8} \rho v_2^2 \]

### Step 6: Solve for \( v_2 \)
Cancelling \( \rho \) from both sides:
\[ g h = \frac{3}{8} v_2^2 \]
\[ v_2^2 = \frac{8}{3} g h \]
Taking the square root:
\[ v_2 = \sqrt{\frac{8}{3} g h} \]

### Step 7: Calculate the Volume Rate of Flow
The volume rate of flow \( Q \) at depth \( h \) can be expressed as:
\[ Q = A \cdot v_2 = \frac{A_0}{2} \cdot v_2 \]
Substituting \( v_2 \):
\[ Q = \frac{A_0}{2} \cdot \sqrt{\frac{8}{3} g h} \]
This simplifies to:
\[ Q = \frac{A_0 \sqrt{8 g h}}{2 \sqrt{3}} \]

### Final Answer
Thus, the volume rate of flow of water at a depth \( h \) is:
\[ Q = \frac{A_0 \sqrt{8 g h}}{2 \sqrt{3}} \]