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 water flowing out per second from both holes is the same.

### Step-by-Step Solution:

1. **Identify the Areas of the Holes:**
- The area of hole \( A \) (square) is given by:
\[ A_A = L^2 \]
- The area of hole \( B \) (circle) is given by:
\[ A_B = \pi r^2 \]

2. **Determine the Velocities of Efflux:**
- The velocity of efflux from hole \( A \) at depth \( h \) is:
\[ V_A = \sqrt{2gh} \]
- The velocity of efflux from hole \( B \) at depth \( 3h \) is:
\[ V_B = \sqrt{2g \cdot 3h} = \sqrt{6gh} \]

3. **Use the Discharge Equation:**
- The discharge \( Q \) from each hole can be expressed as:
\[ Q_A = A_A \cdot V_A = L^2 \cdot \sqrt{2gh} \]
\[ Q_B = A_B \cdot V_B = \pi r^2 \cdot \sqrt{6gh} \]
- Since the discharges are equal, we set \( Q_A = Q_B \):
\[ L^2 \cdot \sqrt{2gh} = \pi r^2 \cdot \sqrt{6gh} \]

4. **Cancel Common Terms:**
- We can cancel \( \sqrt{gh} \) from both sides:
\[ L^2 \cdot \sqrt{2} = \pi r^2 \cdot \sqrt{6} \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ L^2 = \frac{\pi r^2 \cdot \sqrt{6}}{\sqrt{2}} = \pi r^2 \cdot \sqrt{3} \]

6. **Taking the Square Root:**
- Taking the square root of both sides:
\[ L = r \cdot \sqrt{\pi \cdot \sqrt{3}} \]

### Final Expression:
Thus, we find that:
\[ L = r \cdot \sqrt{\frac{\pi}{\sqrt{3}}} \]