An organ pipe open at one end is vibrating in first overtone and is in resonance with another pipe open at both ends and vibrating in third harmonic. The ratio of length of two pipes is–

Options

A. `1:2`
B. `4:1`
C. `8:3`
D. `3:8`

Explanation

To solve the problem, we need to find the ratio of the lengths of two pipes: one pipe open at one end (closed pipe) vibrating in its first overtone and another pipe open at both ends vibrating in its third harmonic.

### Step-by-Step Solution:

1. **Understand the Frequencies of the Pipes**:
- For a closed pipe (open at one end), the fundamental frequency \( f \) is given by:
\[ f = \frac{v_s}{4L} \]
where \( v_s \) is the speed of sound and \( L \) is the length of the pipe.
- The first overtone of a closed pipe corresponds to the third harmonic, which means:
\[ f_{1st \, overtone} = \frac{3v_s}{4L_1} \]
where \( L_1 \) is the length of the closed pipe.

2. **Open Pipe Frequency**:
- For an open pipe (open at both ends), the fundamental frequency is given by:
\[ f = \frac{v_s}{2L} \]
- The third harmonic of an open pipe is given by:
\[ f_{3rd \, harmonic} = \frac{3v_s}{2L_2} \]
where \( L_2 \) is the length of the open pipe.

3. **Setting the Frequencies Equal**:
- Since the two pipes are in resonance, their frequencies must be equal:
\[ \frac{3v_s}{4L_1} = \frac{3v_s}{2L_2} \]

4. **Canceling Common Terms**:
- We can cancel \( 3v_s \) from both sides:
\[ \frac{1}{4L_1} = \frac{1}{2L_2} \]

5. **Cross Multiplying**:
- Cross multiplying gives:
\[ 2L_2 = 4L_1 \]

6. **Finding the Ratio**:
- Rearranging the equation gives us:
\[ \frac{L_1}{L_2} = \frac{2}{4} = \frac{1}{2} \]

### Final Answer:
The ratio of the lengths of the two pipes \( \frac{L_1}{L_2} \) is \( \frac{1}{2} \).