The first overtone frequency of a closed organ pipe `P_1` is equal to the fundamental frequency of an open organ pipe `P_2`. If the length of the pipe `P_1` is 30 cm, what will be the length of` P_2`?

Explanation

To solve the problem, we need to find the length of the open organ pipe \( P_2 \) given that the first overtone frequency of the closed organ pipe \( P_1 \) is equal to the fundamental frequency of the open organ pipe \( P_2 \). The length of the closed organ pipe \( P_1 \) is given as 30 cm.

### Step-by-Step Solution:

1. **Understand the Frequencies**:
- The first overtone frequency of a closed organ pipe is given by the formula:
\[ f_1 = \frac{3}{4} \cdot \frac{V}{L_1} \]
where \( V \) is the speed of sound in air and \( L_1 \) is the length of the closed organ pipe.
- The fundamental frequency of an open organ pipe is given by the formula:
\[ f_2 = \frac{V}{2L_2} \]
where \( L_2 \) is the length of the open organ pipe.

2. **Set the Frequencies Equal**:
- According to the problem, the first overtone frequency of \( P_1 \) is equal to the fundamental frequency of \( P_2 \):
\[ f_1 = f_2 \]
- Therefore, we can write:
\[ \frac{3}{4} \cdot \frac{V}{L_1} = \frac{V}{2L_2} \]

3. **Cancel \( V \) from Both Sides**:
- Since \( V \) is the same for both pipes, we can cancel it out:
\[ \frac{3}{4L_1} = \frac{1}{2L_2} \]

4. **Cross-Multiply to Solve for \( L_2 \)**:
- Cross-multiplying gives:
\[ 3 \cdot 2L_2 = 4L_1 \]
- This simplifies to:
\[ 6L_2 = 4L_1 \]

5. **Rearrange to Find \( L_2 \)**:
- Now, solve for \( L_2 \):
\[ L_2 = \frac{4L_1}{6} = \frac{2L_1}{3} \]

6. **Substitute the Value of \( L_1 \)**:
- Given that \( L_1 = 30 \) cm, substitute this value into the equation:
\[ L_2 = \frac{2 \times 30 \text{ cm}}{3} = \frac{60 \text{ cm}}{3} = 20 \text{ cm} \]

7. **Conclusion**:
- The length of the open organ pipe \( P_2 \) is \( 20 \) cm.

### Final Answer:
The length of the open organ pipe \( P_2 \) is **20 cm**.