There are two holes P and Q at depths h and 4h from the top of a large vessel, completely filled with water. P is a square hole of side L and Q is a circular hole of radius r. If the same quantity of water flows out per second from both the holes, then the relation between L and r is

Options

A. `L=r/2`
B. `L=sqrt(2pir)`
C. `L-2pir`
D. `L=(sqrt(2pi))/(r )`

Explanation

To solve the problem, we need to find the relationship between the side length \( L \) of the square hole at depth \( h \) and the radius \( r \) of the circular hole at depth \( 4h \) when the same quantity of water flows out per second from both holes.

### Step-by-Step Solution:

1. **Identify the velocities at both holes**:
- For hole P at depth \( h \), the velocity of water flowing out can be derived from Torricelli’s theorem:
\[ v_P = \sqrt{2gh} \]
- For hole Q at depth \( 4h \):
\[ v_Q = \sqrt{2g(4h)} = \sqrt{8gh} = 2\sqrt{2gh} \]

2. **Calculate the areas of the holes**:
- The area of the square hole P:
\[ A_P = L^2 \]
- The area of the circular hole Q:
\[ A_Q = \pi r^2 \]

3. **Set up the volume flow rate equation**:
- The volume flow rate (Q) for both holes must be equal:
\[ Q_P = A_P \cdot v_P = L^2 \cdot \sqrt{2gh} \]
\[ Q_Q = A_Q \cdot v_Q = \pi r^2 \cdot (2\sqrt{2gh}) \]

4. **Equate the flow rates**:
\[ L^2 \cdot \sqrt{2gh} = \pi r^2 \cdot (2\sqrt{2gh}) \]

5. **Simplify the equation**:
- Cancel \( \sqrt{2gh} \) from both sides:
\[ L^2 = 2\pi r^2 \]

6. **Solve for \( L \)**:
\[ L = \sqrt{2\pi} r \]

### Final Relation:
The relationship between \( L \) and \( r \) is:
\[ L = \sqrt{2\pi} r \]