An organ pipe `A` closed at one end vibrating in its fundamental frequency and another pipe `B` open at both ends is vibrating in its second overtone are in resonance with a given tuning fork . The ratiio of length of pipe `A` to that of `B` is

Options

A. `1 : 2`
B. `3 : 8`
C. `2 : 3`
D. `1 : 6`

Explanation

To find the ratio of the lengths of the two pipes, we will use the fundamental frequency formula for a closed pipe and the formula for the second overtone of an open pipe.

### Step-by-Step Solution:

1. **Identify the frequencies of the pipes:**
- Pipe A (closed at one end) is vibrating in its fundamental frequency.
- Pipe B (open at both ends) is vibrating in its second overtone (which is the third harmonic).

2. **Write the frequency formula for Pipe A:**
- The fundamental frequency \( f_A \) of a closed pipe is given by:
\[ f_A = \frac{V}{4L_A} \]
where \( V \) is the speed of sound in air, and \( L_A \) is the length of pipe A.

3. **Write the frequency formula for Pipe B:**
- The second overtone (third harmonic) frequency \( f_B \) of an open pipe is given by:
\[ f_B = \frac{3V}{2L_B} \]
where \( L_B \) is the length of pipe B.

4. **Set the frequencies equal to each other:**
- Since both pipes are in resonance with the same tuning fork, we have:
\[ f_A = f_B \]
Therefore:
\[ \frac{V}{4L_A} = \frac{3V}{2L_B} \]

5. **Cancel \( V \) from both sides:**
- This simplifies to:
\[ \frac{1}{4L_A} = \frac{3}{2L_B} \]

6. **Cross-multiply to find the relationship between \( L_A \) and \( L_B \):**
- Cross-multiplying gives:
\[ 2L_B = 12L_A \]

7. **Rearranging gives the ratio of lengths:**
- Dividing both sides by \( 2L_A \):
\[ \frac{L_A}{L_B} = \frac{1}{6} \]

### Final Answer:
The ratio of the length of pipe A to that of pipe B is:
\[ \frac{L_A}{L_B} = \frac{1}{6} \]