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 \( h \), we can follow these steps:

### Step 1: Understand the Problem
We have a tap with a cross-sectional area \( A_0 \) at the top and a reduced area \( A_0/2 \) at a depth \( h \). We need to find the volume flow rate of water at this depth.

### Step 2: Apply the Continuity Equation
According to the principle of continuity, the product of the cross-sectional area and the velocity of the fluid must remain constant along a streamline. Thus, we can write:

\[ A_0 V_1 = \frac{A_0}{2} V_2 \]

Where:
- \( V_1 \) is the velocity of water at the top (cross-section \( A_0 \))
- \( V_2 \) is the velocity of water at depth \( h \) (cross-section \( A_0/2 \))

### Step 3: Solve for \( V_2 \)
From the continuity equation, we can rearrange it to find \( V_2 \):

\[ V_2 = 2 V_1 \]

### Step 4: Apply Bernoulli's Equation
Now, we apply Bernoulli's equation between the two points (the top of the tap and the point at depth \( h \)):

\[ \frac{P_1}{\rho g} + \frac{V_1^2}{2g} + h_1 = \frac{P_2}{\rho g} + \frac{V_2^2}{2g} + h_2 \]

Assuming atmospheric pressure acts on both points, we can simplify this to:

\[ \frac{V_1^2}{2g} + h = \frac{V_2^2}{2g} \]

### Step 5: Substitute \( V_2 \) and Solve for \( V_1 \)
Substituting \( V_2 = 2V_1 \) into the equation gives:

\[ \frac{V_1^2}{2g} + h = \frac{(2V_1)^2}{2g} \]

This simplifies to:

\[ \frac{V_1^2}{2g} + h = \frac{4V_1^2}{2g} \]

Rearranging gives:

\[ h = \frac{4V_1^2}{2g} - \frac{V_1^2}{2g} \]

\[ h = \frac{3V_1^2}{2g} \]

### Step 6: Solve for \( V_1 \)
Now, solving for \( V_1 \):

\[ V_1^2 = \frac{2gh}{3} \]

\[ V_1 = \sqrt{\frac{2gh}{3}} \]

### Step 7: Find \( V_2 \)
Now, substituting back to find \( V_2 \):

\[ V_2 = 2V_1 = 2\sqrt{\frac{2gh}{3}} = \sqrt{\frac{8gh}{3}} \]

### Step 8: Calculate Volume Flow Rate
The volume flow rate \( Q \) at depth \( h \) is given by:

\[ Q = A_2 V_2 \]

Where \( A_2 = \frac{A_0}{2} \):

\[ Q = \frac{A_0}{2} \cdot \sqrt{\frac{8gh}{3}} \]

### Step 9: Final Expression
Thus, the final expression for the volume flow rate \( Q \) is:

\[ Q = \frac{A_0 \sqrt{8gh}}{2\sqrt{3}} \]