A large open tank has two holes in the wall. One is square hole of side L at a depth y from the top and the other is a circular hole of radius R at a depth of 4y from the top. When the tank is completely filled with water, the quantities of water flowing out per second from both holes are the same. Then R is equal to

Options

A. `L//sqrt((2pi))`
B. `2piL`
C. L
D. `L//2pi`

Explanation

To solve the problem, we need to find the relationship between the radius \( R \) of the circular hole and the side length \( L \) of the square hole, given that the quantities of water flowing out per second from both holes are the same.

### Step-by-Step Solution:

1. **Identify the Areas of the Holes:**
- The area \( A_1 \) of the square hole is given by:
\[ A_1 = L^2 \]
- The area \( A_2 \) of the circular hole is given by:
\[ A_2 = \pi R^2 \]

2. **Determine the Velocities of Water Flowing Out:**
- The velocity \( V_1 \) of water flowing out of the square hole at depth \( y \) is given by Torricelli's theorem:
\[ V_1 = \sqrt{2gy} \]
- The velocity \( V_2 \) of water flowing out of the circular hole at depth \( 4y \) is:
\[ V_2 = \sqrt{2g(4y)} = \sqrt{8gy} \]

3. **Use the Continuity Equation:**
- According to the continuity equation, the flow rates from both holes must be equal:
\[ A_1 V_1 = A_2 V_2 \]
- Substituting the areas and velocities into the equation:
\[ L^2 \cdot \sqrt{2gy} = \pi R^2 \cdot \sqrt{8gy} \]

4. **Simplify the Equation:**
- Cancel \( \sqrt{gy} \) from both sides:
\[ L^2 \cdot \sqrt{2} = \pi R^2 \cdot \sqrt{8} \]
- This simplifies to:
\[ L^2 \cdot \sqrt{2} = \pi R^2 \cdot 2\sqrt{2} \]

5. **Further Simplification:**
- Dividing both sides by \( \sqrt{2} \):
\[ L^2 = 2\pi R^2 \]

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

### Final Answer:
Thus, the value of \( R \) is:
\[ R = \frac{L}{\sqrt{2\pi}} \]