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. (a) `(L)/(sqrt(2pi))`
B. (b) `2piL`
C. (c) `L`
D. (d) `(L)/(2pi)`

Explanation

To solve the problem, we need to use Torricelli's theorem, which states that the speed of efflux of a fluid under the force of gravity through an orifice is given by the equation:

\[ v = \sqrt{2gh} \]

where \( v \) is the velocity of the fluid, \( g \) is the acceleration due to gravity, and \( h \) is the height of the fluid above the hole.

### Step 1: Calculate the velocity of water flowing out of the square hole
For the square hole at depth \( y \):
- The area \( A_s \) of the square hole is given by:
\[ A_s = L^2 \]
- The height of water above the square hole is \( h = y \).
- The velocity \( v_s \) of water flowing out of the square hole is:
\[ v_s = \sqrt{2gy} \]

### Step 2: Calculate the velocity of water flowing out of the circular hole
For the circular hole at depth \( 4y \):
- The area \( A_c \) of the circular hole is given by:
\[ A_c = \pi R^2 \]
- The height of water above the circular hole is \( h = 4y \).
- The velocity \( v_c \) of water flowing out of the circular hole is:
\[ v_c = \sqrt{2g(4y)} = \sqrt{8gy} \]

### Step 3: Set the flow rates equal
The flow rate \( Q \) through each hole can be expressed as:
\[ Q_s = A_s v_s \]
\[ Q_c = A_c v_c \]

Since the quantities of water flowing out per second from both holes are the same, we have:
\[ Q_s = Q_c \]

Substituting the expressions for \( Q_s \) and \( Q_c \):
\[ L^2 \cdot \sqrt{2gy} = \pi R^2 \cdot \sqrt{8gy} \]

### Step 4: Simplify the equation
We can cancel \( \sqrt{gy} \) from both sides (assuming \( y \neq 0 \)):
\[ L^2 \cdot \sqrt{2} = \pi R^2 \cdot \sqrt{8} \]

Rearranging gives:
\[ L^2 \cdot \sqrt{2} = \pi R^2 \cdot 2\sqrt{2} \]

Dividing both sides by \( \sqrt{2} \):
\[ L^2 = 2\pi R^2 \]

### Step 5: Solve for \( R \)
Now, we can solve for \( R \):
\[ R^2 = \frac{L^2}{2\pi} \]
\[ R = \sqrt{\frac{L^2}{2\pi}} = \frac{L}{\sqrt{2\pi}} \]

Thus, the radius \( R \) is equal to:
\[ R = \frac{L}{\sqrt{2\pi}} \]

### Final Answer:
\[ R = \frac{L}{\sqrt{2\pi}} \]

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