An organ pipe `P_(1)` open at one end vibrating in its first harmonic and another pipe `P_(2)` open at ends vibrating in its third harmonic are in resonance with a given tuning fork. The ratio of the length of `P_(1)` to that `P_(2)` is

Explanation

To find the ratio of the lengths of the two organ pipes \( P_1 \) and \( P_2 \), we need to use the principles of harmonics for open and closed pipes.

### Step-by-Step Solution:

1. **Understanding the Harmonics**:
- For an organ pipe open at one end (like \( P_1 \)), the fundamental frequency (first harmonic) is given by:
\[ f_1 = \frac{V}{4L_1} \]
- For an organ pipe open at both ends (like \( P_2 \)), the frequency for the third harmonic is given by:
\[ f_2 = \frac{3V}{2L_2} \]

2. **Setting Up the Resonance Condition**:
- Since both pipes are in resonance with the same tuning fork, we can set their frequencies equal:
\[ f_1 = f_2 \]
- Substituting the expressions for \( f_1 \) and \( f_2 \):
\[ \frac{V}{4L_1} = \frac{3V}{2L_2} \]

3. **Eliminating \( V \)**:
- We can cancel \( V \) from both sides (assuming \( V \neq 0 \)):
\[ \frac{1}{4L_1} = \frac{3}{2L_2} \]

4. **Cross-Multiplying**:
- Cross-multiply to solve for the relationship between \( L_1 \) and \( L_2 \):
\[ 2L_2 = 12L_1 \]

5. **Finding the Ratio**:
- Rearranging gives:
\[ \frac{L_1}{L_2} = \frac{2}{12} = \frac{1}{6} \]

### Final Answer:
The ratio of the length of \( P_1 \) to that of \( P_2 \) is:
\[ \frac{L_1}{L_2} = \frac{1}{6} \]