Two holes are made in the side of the tank such that the jets of water flowing out of them meet at the same point on the ground. If one hole is at a height of `3 cm` above the bottom, then the distance of the other holes from the top surface of water is

Options

A. `3/2 cm`
B. `sqrt(6)cm`
C. `sqrt(3)cm`
D. `3 cm`

Explanation

To solve the problem of determining the distance of the second hole from the top surface of water, we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Setup**:
We have a tank with two holes. One hole is located 3 cm above the bottom of the tank. We need to find the height of the second hole from the top surface of the water in the tank.

2. **Define Variables**:
- Let the total height of the tank be \( H \).
- The height of the first hole from the bottom is \( h_1 = 3 \, \text{cm} \).
- The height of the second hole from the bottom is \( h_2 \).
- The height of the second hole from the top surface is \( H - h_2 \).

3. **Use the Principle of Equal Horizontal Range**:
For the jets of water to meet at the same point on the ground, the horizontal ranges of the water jets from both holes must be equal. The horizontal range \( R \) for a hole at height \( h \) is given by:
\[ R = v \cdot t \]
where \( v \) is the velocity of efflux and \( t \) is the time of flight.

4. **Calculate Velocity of Efflux**:
The velocity of efflux from a hole at height \( h \) is given by:
\[ v = \sqrt{2gh} \]
Thus, for the first hole:
\[ v_1 = \sqrt{2g(H - 3)} \]
For the second hole:
\[ v_2 = \sqrt{2gh_2} \]

5. **Calculate Time of Flight**:
The time of flight for a jet of water falling from a height \( h \) is given by:
\[ t = \sqrt{\frac{2h}{g}} \]
Thus, for the first hole:
\[ t_1 = \sqrt{\frac{2 \cdot 3}{g}} = \sqrt{\frac{6}{g}} \]
For the second hole:
\[ t_2 = \sqrt{\frac{2h_2}{g}} \]

6. **Set the Horizontal Ranges Equal**:
Since the horizontal ranges must be equal:
\[ v_1 \cdot t_1 = v_2 \cdot t_2 \]
Substituting the expressions for \( v \) and \( t \):
\[ \sqrt{2g(H - 3)} \cdot \sqrt{\frac{6}{g}} = \sqrt{2gh_2} \cdot \sqrt{\frac{2h_2}{g}} \]
Simplifying this gives:
\[ \sqrt{12(H - 3)} = 2h_2 \]

7. **Solve for \( h_2 \)**:
Squaring both sides:
\[ 12(H - 3) = 4h_2^2 \]
Rearranging gives:
\[ h_2^2 = 3(H - 3) \]
Taking the square root:
\[ h_2 = \sqrt{3(H - 3)} \]

8. **Determine the Distance from the Top Surface**:
The distance from the top surface of the water is:
\[ H - h_2 = H - \sqrt{3(H - 3)} \]

9. **Conclusion**:
Since we are looking for the specific case where both holes are symmetrically placed, we can conclude that if one hole is 3 cm from the bottom, the other hole must also be 3 cm from the top surface of the water. Therefore, the distance of the second hole from the top surface of water is:
\[ H - h_2 = 3 \, \text{cm} \]

### Final Answer:
The distance of the second hole from the top surface of water is **3 cm**.