A cylindrical tank of height H is completely filled with water. On its vertical side there are two tiny holes, one above the middle at a height ` h_1 ` and the other below the middle at a depth ` h_2 ` If the jets of water from the holes meet at the same point at the horizontal plane through the bottom of the tank then the ratio ` h_1//h_2 ` is :

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the problem, we need to analyze the situation involving the cylindrical tank with two holes at different heights. The goal is to find the ratio \( \frac{h_1}{h_2} \) given that the jets of water from the holes meet at the same point on the horizontal plane through the bottom of the tank.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The cylindrical tank has a height \( H \) and is filled with water.
- There are two holes: one at height \( h_1 \) above the middle of the tank and the other at depth \( h_2 \) below the middle.

2. **Identifying Heights**:
- The middle of the tank is at height \( \frac{H}{2} \).
- The height of the first hole from the bottom is \( \frac{H}{2} + h_1 \).
- The height of the second hole from the bottom is \( \frac{H}{2} - h_2 \).

3. **Using Torricelli’s Law**:
- The velocity of water flowing out of a hole can be derived from Torricelli's theorem, which states that the speed \( v \) of efflux of a fluid under gravity through an orifice is given by:
\[ v = \sqrt{2gh} \]
- For the first hole (at height \( h_1 \)):
\[ v_1 = \sqrt{2g\left(\frac{H}{2} + h_1\right)} \]
- For the second hole (at depth \( h_2 \)):
\[ v_2 = \sqrt{2g\left(\frac{H}{2} - h_2\right)} \]

4. **Calculating the Range**:
- The horizontal range \( R \) for a projectile is given by:
\[ R = v \cdot t \]
- The time \( t \) taken for the water to fall from height \( h \) is given by:
\[ t = \sqrt{\frac{2h}{g}} \]

5. **Setting Up the Equations**:
- For the first hole, the range \( R_1 \) can be expressed as:
\[ R_1 = v_1 \cdot t_1 = \sqrt{2g\left(\frac{H}{2} + h_1\right)} \cdot \sqrt{\frac{2\left(\frac{H}{2} + h_1\right)}{g}} = 2\left(\frac{H}{2} + h_1\right) \]
- For the second hole, the range \( R_2 \) is:
\[ R_2 = v_2 \cdot t_2 = \sqrt{2g\left(\frac{H}{2} - h_2\right)} \cdot \sqrt{\frac{2\left(\frac{H}{2} - h_2\right)}{g}} = 2\left(\frac{H}{2} - h_2\right) \]

6. **Equating the Ranges**:
- Since the jets meet at the same point:
\[ R_1 = R_2 \]
- This gives us:
\[ 2\left(\frac{H}{2} + h_1\right) = 2\left(\frac{H}{2} - h_2\right) \]
- Simplifying this equation:
\[ \frac{H}{2} + h_1 = \frac{H}{2} - h_2 \]
- Rearranging gives:
\[ h_1 + h_2 = 0 \quad \Rightarrow \quad h_1 = h_2 \]

7. **Finding the Ratio**:
- Therefore, the ratio \( \frac{h_1}{h_2} \) is:
\[ \frac{h_1}{h_2} = 1 \]

### Final Answer:
The ratio \( \frac{h_1}{h_2} \) is \( 1 \).