A tank of height `H` and base area `A` is half filled with water and there is a small orifice at the bottom and there is a heavy solid cylinder having base area `(A)/(3)` and height of the cylinder is same as that of the tak. The water is flowing out of the orifice. Here cylinder is put into the tank to increase the speed of water flowing out.
The speed of water flowing out of the orifice after the cylinder is kept inside it is

Options

A. `sqrt(3gH)`
B. `sqrt(2gH)`
C. `sqrt((3gH)/(2))`
D. `sqrt((gH)/(2))`

Explanation

To solve the problem step by step, we need to analyze the situation involving the tank, the water, and the cylinder.

### Step 1: Understand the initial conditions
The tank has a height \( H \) and a base area \( A \). It is half-filled with water, so the initial height of the water \( h \) is:
\[ h = \frac{H}{2} \]

### Step 2: Determine the volume of water before the cylinder is added
The volume of water in the tank before the cylinder is added can be calculated as:
\[ \text{Volume of water} = A \cdot h = A \cdot \frac{H}{2} \]

### Step 3: Analyze the cylinder's dimensions
The cylinder has a base area of \( \frac{A}{3} \) and a height equal to that of the tank, which is \( H \).

### Step 4: Calculate the volume of the cylinder
The volume of the cylinder can be calculated as:
\[ \text{Volume of cylinder} = \text{Base Area} \times \text{Height} = \frac{A}{3} \cdot H \]

### Step 5: Find the new height of water after the cylinder is submerged
When the cylinder is submerged in the tank, it displaces some water. The volume of water after the cylinder is submerged must equal the initial volume of water minus the volume of the cylinder. Therefore, we set up the equation:
\[ A \cdot \frac{H}{2} = (A - \frac{A}{3}) \cdot h' \]
Here, \( h' \) is the new height of water in the tank.

### Step 6: Simplify the equation
The base area of the remaining water is:
\[ A - \frac{A}{3} = \frac{2A}{3} \]
Thus, we can rewrite the equation as:
\[ A \cdot \frac{H}{2} = \frac{2A}{3} \cdot h' \]

### Step 7: Cancel out \( A \) from both sides
Dividing both sides by \( A \):
\[ \frac{H}{2} = \frac{2}{3} h' \]

### Step 8: Solve for \( h' \)
To find \( h' \), we rearrange the equation:
\[ h' = \frac{H}{2} \cdot \frac{3}{2} = \frac{3H}{4} \]

### Step 9: Use Torricelli’s Law to find the speed of water flowing out
According to Torricelli’s Law, the speed \( v \) of fluid flowing out of an orifice under the influence of gravity is given by:
\[ v = \sqrt{2gh'} \]
Substituting \( h' = \frac{3H}{4} \):
\[ v = \sqrt{2g \cdot \frac{3H}{4}} = \sqrt{\frac{3gH}{2}} \]

### Conclusion
Thus, the speed of water flowing out of the orifice after the cylinder is kept inside is:
\[ v = \sqrt{\frac{3gH}{2}} \]