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. a.`sqrt(gH)`
B. b.`1.414 sqrt(gH)`
C. c.`(sqrt(3gH))/(2)`
D. d.`sqrt((gh)/(2))`

Explanation

To solve the problem, we need to determine the speed of water flowing out of the orifice after a heavy solid cylinder is placed inside the tank. We'll follow these steps:

### Step-by-Step Solution:

1. **Understanding the Initial Conditions**:
- The tank has a height \( H \) and base area \( A \).
- Initially, the tank is half-filled with water, so the height of the water column \( h = \frac{H}{2} \).

2. **Volume of Water Before the Cylinder is Added**:
- The volume of water in the tank before the cylinder is added can be calculated as:
\[ V_{\text{initial}} = A \cdot \left(\frac{H}{2}\right) = \frac{AH}{2} \]

3. **Volume of the Cylinder**:
- The cylinder has a base area of \( \frac{A}{3} \) and height \( H \). Thus, the volume of the cylinder is:
\[ V_{\text{cylinder}} = \frac{A}{3} \cdot H \]

4. **Volume of Water After the Cylinder is Added**:
- When the cylinder is submerged, it displaces some water. The new effective height of the water column can be denoted as \( h' \).
- The remaining volume of water in the tank after the cylinder is submerged can be expressed as:
\[ V_{\text{final}} = A \cdot h' = A \cdot \left(H - \frac{A}{3} \cdot H\right) \]
- The volume of water must remain constant, so:
\[ V_{\text{initial}} = V_{\text{final}} + V_{\text{cylinder}} \]

5. **Setting Up the Equation**:
- We can set up the equation:
\[ \frac{AH}{2} = A \cdot h' + \frac{A}{3}H \]
- Simplifying this, we can cancel \( A \) from both sides:
\[ \frac{H}{2} = h' + \frac{H}{3} \]

6. **Solving for \( h' \)**:
- Rearranging gives:
\[ h' = \frac{H}{2} - \frac{H}{3} \]
- To combine these fractions, we find a common denominator (which is 6):
\[ h' = \frac{3H}{6} - \frac{2H}{6} = \frac{H}{6} \]

7. **Using Torricelli's Law**:
- According to Torricelli's law, the speed of fluid flowing out of an orifice under the influence of gravity is given by:
\[ v = \sqrt{2gh} \]
- Here, \( h \) is the height of the water column above the orifice, which is now \( h' = \frac{H}{6} \):
\[ v = \sqrt{2g \cdot \frac{H}{6}} = \sqrt{\frac{gH}{3}} \]

### Final Answer:
The speed of water flowing out of the orifice after the cylinder is placed inside the tank is:
\[ v = \sqrt{\frac{gH}{3}} \]