A closed organ pipe of radius `r_(1)` and an open organ pipe of radius `r_(2)` and having same length 'L' resonate when excited with a given tuning fork. Closed organ-pipe resonates in its fundamental mode where as open organ pipe resonates in its first overtone, then :-

Options

A. `r_(2)-r_(1) = L`
B. `r_(2)-r_(1) = L//2`
C. `r_(2)-2r_(1) = 2.5L`
D. `2r_(2)-2r_(1) = 2.5L`

Answer

To solve the problem, we need to analyze the conditions of resonance for both the closed organ pipe and the open organ pipe.

### Step-by-Step Solution:

1. **Understanding the Closed Organ Pipe:**
- A closed organ pipe resonates in its fundamental mode. The effective length of the closed organ pipe is given by:
\[ L_{\text{closed}} = L + 0.6 R_1 \]
- In the fundamental mode, the wavelength (\(\lambda\)) is related to the effective length by:
\[ L_{\text{closed}} = \frac{\lambda}{4} \]
- Therefore, we can write:
\[ L + 0.6 R_1 = \frac{\lambda}{4} \]

2. **Understanding the Open Organ Pipe:**
- An open organ pipe resonates in its first overtone. The effective length of the open organ pipe is given by:
\[ L_{\text{open}} = L + 1.2 R_2 \]
- In the first overtone, the wavelength is related to the effective length by:
\[ L_{\text{open}} = \frac{\lambda}{2} \]
- Therefore, we can write:
\[ L + 1.2 R_2 = \frac{\lambda}{2} \]

3. **Equating the Two Expressions for Wavelength:**
- From the two equations derived, we can express \(\lambda\) for both pipes:
- From the closed pipe:
\[ \lambda = 4(L + 0.6 R_1) \]
- From the open pipe:
\[ \lambda = 2(L + 1.2 R_2) \]
- Setting these two expressions for \(\lambda\) equal gives:
\[ 4(L + 0.6 R_1) = 2(L + 1.2 R_2) \]

4. **Simplifying the Equation:**
- Expanding both sides:
\[ 4L + 2.4 R_1 = 2L + 2.4 R_2 \]
- Rearranging the equation:
\[ 4L - 2L = 2.4 R_2 - 2.4 R_1 \]
\[ 2L = 2.4 R_2 - 2.4 R_1 \]
- Dividing through by 2.4:
\[ \frac{2L}{2.4} = R_2 - R_1 \]
- Simplifying gives:
\[ R_2 - R_1 = \frac{5}{6}L \]

5. **Final Relationship:**
- Rearranging gives:
\[ R_2 - 2R_1 = \frac{5}{6}L + R_1 \]
- This leads to:
\[ R_2 - 2R_1 = 3L \]

### Conclusion:
The final relationship derived is:
\[ R_2 - 2R_1 = 3L \]