A large open tank has two holes in the wall. One is a 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 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(2 pi))`
B. `2piL`
C. L
D. `(L)/(2 pi)`

Explanation

To solve the problem, we need to find the relationship between the side length \( L \) of the square hole and the radius \( R \) of the circular 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 of the square hole \( A_s \) is given by:
\[ A_s = L^2 \]
- The area of the circular hole \( A_c \) is given by:
\[ A_c = \pi R^2 \]

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

3. **Set Up the Flow Rate Equation:**
- Since the quantities of water flowing out per second from both holes are the same, we can equate the flow rates:
\[ A_s \cdot V_s = A_c \cdot V_c \]
- Substituting the areas and velocities:
\[ L^2 \cdot \sqrt{2gy} = \pi R^2 \cdot 2\sqrt{2gy} \]

4. **Cancel Common Terms:**
- We can cancel \( \sqrt{2gy} \) from both sides (assuming \( y \neq 0 \)):
\[ L^2 = \pi R^2 \cdot 2 \]

5. **Rearranging the Equation:**
- Rearranging gives:
\[ R^2 = \frac{L^2}{2\pi} \]

6. **Taking the Square Root:**
- Taking the square root of both sides yields:
\[ R = \frac{L}{\sqrt{2\pi}} \]

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