`{:(,"Column I",,"Column II"),((A),"A stretched string",(p),"n antinodes"),((B),"Open organ pipes",(q),"n nodes"),((C),"Closed organ pipe",(r),"(n + 1) antinodes"),((D),"A stationary wave produced",(s),"(n + 1) nodes"),(,"in organ pipes may have",,):}`

Explanation

To solve the problem of matching the items in Column I with those in Column II, we will analyze each item in Column I and determine the corresponding item in Column II based on the properties of standing waves in different mediums.

### Step-by-Step Solution:

1. **A Stretched String**:
- A stretched string fixed at both ends will have nodes at the ends and antinodes in between.
- The number of nodes (N) is given by N = n + 1, where n is the number of loops or antinodes.
- Therefore, for a stretched string, there are **n antinodes** and **n + 1 nodes**.
- **Match**: A → S (n + 1 nodes).

2. **Open Organ Pipes**:
- Open organ pipes are open at both ends, which means there are antinodes at both ends.
- The number of nodes (N) is given by N = n, where n is the number of loops or antinodes.
- Therefore, for open organ pipes, there are **n + 1 antinodes** and **n nodes**.
- **Match**: B → R (n + 1 antinodes) and B → Q (n nodes).

3. **Closed Organ Pipe**:
- A closed organ pipe has one end closed, which means there is a node at the closed end and an antinode at the open end.
- The number of nodes (N) is given by N = n, where n is the number of loops or antinodes.
- Therefore, for a closed organ pipe, there are **n antinodes** and **n nodes**.
- **Match**: C → P (n nodes) and C → Q (n nodes).

4. **A Stationary Wave Produced in Organ Pipes**:
- Stationary waves in organ pipes can have various configurations depending on whether the pipe is open or closed.
- The combinations can include both n nodes and n + 1 antinodes.
- Therefore, the possible matches are **n + 1 antinodes** and **n nodes**.
- **Match**: D → R (n + 1 antinodes) and D → Q (n nodes).

### Final Matches:
- A → S (n + 1 nodes)
- B → R (n + 1 antinodes) and B → Q (n nodes)
- C → P (n nodes) and C → Q (n nodes)
- D → R (n + 1 antinodes) and D → Q (n nodes)