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.
After long time, when the height of water inside the tank again becomes equal to `(H)/(2)`, the solid cylinder is taken out. then the velocity of liquid flowing out of the orifice (just after removing the cylinder) will be

Options

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

Explanation

To solve the problem, we need to determine the velocity of the water flowing out of the orifice just after the solid cylinder is removed from the tank. We will use the principles of fluid dynamics and the concept of hydrostatic pressure.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The tank has a height \( H \) and a base area \( A \).
- The tank is half filled with water, so the initial height of water \( h = \frac{H}{2} \).
- A solid cylinder with a base area \( \frac{A}{3} \) is submerged in the tank, and its height is equal to the height of the tank \( H \).

2. **Volume of Water Displaced by the Cylinder**:
- The volume of the cylinder submerged in the water is given by:
\[ V_{cylinder} = \text{Base Area} \times \text{Height} = \frac{A}{3} \times H \]

3. **Effective Height of Water**:
- When the cylinder is submerged, it displaces some water. The effective height of water above the orifice can be calculated by considering the volume of water that remains after accounting for the volume of the cylinder.
- The volume of water in the tank when the cylinder is submerged is:
\[ V_{water} = A \times \frac{H}{2} \]
- The effective height of water \( h' \) when the cylinder is submerged can be found using the equation:
\[ A \times h' = A \times \frac{H}{2} - \frac{A}{3} \times H \]
- Simplifying this gives:
\[ h' = \frac{H}{2} - \frac{H}{3} = \frac{3H - 2H}{6} = \frac{H}{6} \]

4. **Velocity of Water Flowing Out**:
- The velocity of efflux \( v \) from the orifice can be calculated using Torricelli's theorem, which states that the velocity of fluid flowing out of an orifice is given by:
\[ v = \sqrt{2gh'} \]
- Substituting \( h' = \frac{H}{6} \):
\[ v = \sqrt{2g \cdot \frac{H}{6}} = \sqrt{\frac{2gH}{6}} = \sqrt{\frac{gH}{3}} \]

5. **Final Result**:
- Therefore, the velocity of the liquid flowing out of the orifice just after removing the cylinder is:
\[ v = \sqrt{\frac{gH}{3}} \]