A vessel completely filled with water has holes 'A' and 'B' at depths 'h' and '3h' from the top respectively. Hole 'A' is a square of side 'L' and 'B' is circle of radius 'r'. The water flowing out per second from both the holes is same. Then 'L' is equal to

Options

A. `r^(1/2)(pi)^(1/2)(3)^(1/2)`
B. `r(pi)^(1/4)(3)^(1/4)`
C. `r(pi)^(1/2)(3)^(1/4)`
D. `r^(1/2)(pi)^(1/3)(3)^(1/2)`

Answer

To solve the problem, we need to find the relationship between the side length \( L \) of the square hole \( A \) and the radius \( r \) of the circular hole \( B \) given that the volume flow rates from both holes are the same.

### Step-by-Step Solution:

1. **Identify the given parameters**:
- Hole \( A \) is a square with side length \( L \) and is located at depth \( h \).
- Hole \( B \) is a circle with radius \( r \) and is located at depth \( 3h \).

2. **Determine the efflux speed**:
- The efflux speed \( V_A \) from hole \( A \) at depth \( h \) can be calculated using Torricelli's theorem:
\[ V_A = \sqrt{2gh} \]
- The efflux speed \( V_B \) from hole \( B \) at depth \( 3h \) is:
\[ V_B = \sqrt{2g(3h)} = \sqrt{6gh} \]

3. **Calculate the cross-sectional areas**:
- The area \( A_A \) of hole \( A \) (square) is:
\[ A_A = L^2 \]
- The area \( A_B \) of hole \( B \) (circle) is:
\[ A_B = \pi r^2 \]

4. **Set up the equation for volume flow rates**:
- Since the volume flow rates are equal, we can write:
\[ A_A \cdot V_A = A_B \cdot V_B \]
- Substituting the areas and velocities:
\[ L^2 \cdot \sqrt{2gh} = \pi r^2 \cdot \sqrt{6gh} \]

5. **Cancel common terms**:
- We can cancel \( \sqrt{2gh} \) from both sides:
\[ L^2 = \pi r^2 \cdot \frac{\sqrt{6gh}}{\sqrt{2gh}} = \pi r^2 \cdot \sqrt{3} \]

6. **Solve for \( L \)**:
- Rearranging gives:
\[ L^2 = \pi r^2 \sqrt{3} \]
- Taking the square root of both sides:
\[ L = r \sqrt{\pi \sqrt{3}} \]

7. **Final expression**:
- The final expression for \( L \) in terms of \( r \) is:
\[ L = r \sqrt{\pi \sqrt{3}} \]