In the resonance tube experiment , the first resonance is heard when length of air column is `l_(1)` and second resonance is heard when length of air column is `l_(2)`. What should be the minimum length of the tube so that third resonance can also be heard.

Options

A. `2 l_(2) - l_(1)`
B. `2 l_(1)`
C. `5 l_(1)`
D. `7 l_(1)`

Explanation

To find the minimum length of the tube so that the third resonance can also be heard, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding Resonance in the Tube:**
- In the resonance tube experiment, the first resonance occurs when the length of the air column is \( l_1 \), and the second resonance occurs when the length is \( l_2 \). The resonances correspond to specific lengths of the air column that allow for constructive interference of sound waves.

2. **Using the Resonance Formula:**
- The length of the air column at the first resonance can be expressed as:
\[ l_1 = \frac{v}{4f} + \epsilon \]
- The length of the air column at the second resonance is:
\[ l_2 = \frac{3v}{4f} + \epsilon \]
- Here, \( v \) is the speed of sound, \( f \) is the frequency of the tuning fork, and \( \epsilon \) is the end correction.

3. **Finding the Third Resonance:**
- The third resonance occurs at a length \( l_3 \):
\[ l_3 = \frac{5v}{4f} + \epsilon \]

4. **Relating the Lengths:**
- From the equations for \( l_1 \) and \( l_2 \), we can express \( l_3 \) in terms of \( l_1 \) and \( l_2 \):
- Rearranging the equations gives:
\[ l_1 = \frac{v}{4f} + \epsilon \quad \text{(1)} \]
\[ l_2 = \frac{3v}{4f} + \epsilon \quad \text{(2)} \]

5. **Expressing \( l_3 \) in terms of \( l_1 \) and \( l_2 \):**
- From the second equation, we can express \( \epsilon \):
\[ \epsilon = l_2 - \frac{3v}{4f} \]
- Substituting \( \epsilon \) back into the equation for \( l_1 \):
\[ l_1 = \frac{v}{4f} + (l_2 - \frac{3v}{4f}) = l_2 - \frac{2v}{4f} \]
- This leads to:
\[ l_2 = l_1 + \frac{2v}{4f} \]

6. **Finding the Minimum Length for Third Resonance:**
- We can express \( l_3 \) as:
\[ l_3 = 2l_2 - l_1 \]
- Substituting the expression for \( l_2 \):
\[ l_3 = 2(l_1 + \frac{2v}{4f}) - l_1 = l_1 + \frac{4v}{4f} = l_1 + \frac{v}{f} \]

7. **Final Expression for Minimum Length:**
- Therefore, the minimum length of the tube for the third resonance to be heard is:
\[ l_3 = 2l_2 - l_1 \]

### Conclusion:
The minimum length of the tube so that the third resonance can also be heard is given by:
\[ l_3 = 2l_2 - l_1 \]