A squre hole of side length l is made at a depth of h and a circular hole is made at a depht of 4 h from the surface of water in a water tank kept on a horizontal surface (where l ltlt h). Find the radius r of the circular hole if equal amount of water comes out of the vessel through the holes per second.

Options

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

Answer

To solve the problem of finding the radius \( r \) of the circular hole such that equal amounts of water flow out of both the square and circular holes per second, we can follow these steps:

### Step 1: Identify the parameters
- Let the side length of the square hole be \( l \).
- The depth of the square hole from the surface of the water is \( h \).
- The depth of the circular hole from the surface of the water is \( 4h \).
- The area of the square hole \( A_1 = l^2 \).
- The area of the circular hole \( A_2 = \pi r^2 \).

### Step 2: Determine the efflux speed
According to Torricelli's theorem, the speed of efflux \( v \) of a fluid under gravity through a hole is given by:
\[ v = \sqrt{2gh} \]
where \( g \) is the acceleration due to gravity and \( h \) is the depth of the hole from the surface of the fluid.

For the square hole at depth \( h \):
\[ v_1 = \sqrt{2gh} \]

For the circular hole at depth \( 4h \):
\[ v_2 = \sqrt{2g(4h)} = \sqrt{8gh} = 2\sqrt{2gh} \]

### Step 3: Set up the flow rate equation
The volume flow rate through each hole is given by:
\[ Q_1 = A_1 v_1 = l^2 \sqrt{2gh} \]
\[ Q_2 = A_2 v_2 = \pi r^2 (2\sqrt{2gh}) \]

### Step 4: Equate the flow rates
Since the amount of water flowing out through both holes is the same, we can set \( Q_1 = Q_2 \):
\[ l^2 \sqrt{2gh} = \pi r^2 (2\sqrt{2gh}) \]

### Step 5: Simplify the equation
We can cancel \( \sqrt{2gh} \) from both sides (assuming \( h \neq 0 \)):
\[ l^2 = 2\pi r^2 \]

### Step 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 radius \( r \) of the circular hole is:
\[ r = \frac{l}{\sqrt{2\pi}} \]

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