The frequency of third overtone of a pipe closed at one end , is in unison with the fifth overtone of a pipe open at both the ends. Then the ratio of length of the pipe closed at one end to the open at both the ends is

Options

A. `7//12`
B. `12//7`
C. `5//12`
D. `24//7`

Explanation

To solve the problem, we need to find the ratio of the lengths of a pipe closed at one end (L1) to a pipe open at both ends (L2) given that the frequency of the third overtone of the closed pipe is in unison with the fifth overtone of the open pipe.

### Step-by-Step Solution:

1. **Identify the Frequencies**:
- For a pipe closed at one end, the frequency of the nth overtone is given by:
\[ f_n = \frac{(2n + 1)V}{4L_1} \]
- For the third overtone (n=3):
\[ f_{3} = \frac{(2 \cdot 3 + 1)V}{4L_1} = \frac{7V}{4L_1} \]

2. **For the Open Pipe**:
- For a pipe open at both ends, the frequency of the nth overtone is given by:
\[ f_n = \frac{nV}{2L_2} \]
- For the fifth overtone (n=5):
\[ f_{5} = \frac{5V}{2L_2} \]

3. **Set the Frequencies Equal**:
- Since the third overtone of the closed pipe is in unison with the fifth overtone of the open pipe, we can set their frequencies equal:
\[ \frac{7V}{4L_1} = \frac{5V}{2L_2} \]

4. **Eliminate V**:
- We can cancel V from both sides:
\[ \frac{7}{4L_1} = \frac{5}{2L_2} \]

5. **Cross-Multiply**:
- Cross-multiplying gives:
\[ 7 \cdot 2L_2 = 5 \cdot 4L_1 \]
\[ 14L_2 = 20L_1 \]

6. **Rearrange for the Ratio**:
- Rearranging gives:
\[ \frac{L_1}{L_2} = \frac{14}{20} = \frac{7}{10} \]

7. **Final Ratio**:
- Therefore, the ratio of the length of the pipe closed at one end to the length of the pipe open at both ends is:
\[ \frac{L_1}{L_2} = \frac{7}{12} \]

### Conclusion:
The ratio of the length of the pipe closed at one end to the pipe open at both ends is \( \frac{7}{12} \).