In the `5th` overtone of an open organ pipe, these are (N-stands for nodes and A- for antinodes)

Options

A. `2N,3A`
B. `3N,4A`
C. `4N,5A`
D. `5N,4A`

Explanation

To solve the problem regarding the 5th overtone of an open organ pipe, we need to understand the relationship between overtones, harmonics, nodes, and antinodes.

### Step-by-Step Solution:

1. **Understanding Overtones and Harmonics**:
- The 5th overtone corresponds to the 6th harmonic of the pipe. This is because the nth overtone is equal to (n+1)th harmonic. Therefore, for the 5th overtone:
\[ \text{Harmonic} = 5 + 1 = 6 \]

2. **Determining Nodes and Antinodes**:
- For an open organ pipe, the harmonics have a specific arrangement of nodes (N) and antinodes (A). The formula to determine the number of nodes and antinodes in an open pipe is:
- Number of nodes (N) = Harmonic number
- Number of antinodes (A) = Harmonic number + 1

3. **Calculating Nodes and Antinodes for the 6th Harmonic**:
- Since we have established that the 5th overtone corresponds to the 6th harmonic:
- Number of nodes (N) = 6
- Number of antinodes (A) = 6 + 1 = 7

4. **Final Answer**:
- In the 5th overtone of an open organ pipe, there are **6 nodes (N)** and **7 antinodes (A)**.

### Summary:
- For the 5th overtone of an open organ pipe:
- Nodes (N) = 6
- Antinodes (A) = 7