A closed organ pipe of length `L ` is in resonance with a tuning fork. If a hole is made in the pipe at a distance `L/4` from closed - end, it will be in resonance again, when:

Options

A. Tuning fork is replaced by another of high frequency
B. Tuning fork is replaced by another of lower frequency.
C. It will be resonance with same tuning fork
D. Now the pipe will never resonate with any tuning fork.

Answer

To solve the problem, we need to analyze the situation of a closed organ pipe and the effect of making a hole in it.

### Step-by-Step Solution:

1. **Understanding the Closed Organ Pipe**:
- A closed organ pipe is closed at one end and open at the other. The fundamental frequency (first harmonic) of a closed pipe is given by the formula:
\[ f = \frac{v}{4L} \]
where \( v \) is the speed of sound in air and \( L \) is the length of the pipe.

2. **Initial Condition**:
- The pipe of length \( L \) is in resonance with a tuning fork, meaning it is vibrating at its fundamental frequency.

3. **Making a Hole**:
- A hole is made at a distance of \( \frac{L}{4} \) from the closed end. This effectively changes the characteristics of the pipe. The pipe now behaves as a combination of a closed pipe (from the closed end to the hole) and an open pipe (from the hole to the open end).

4. **New Effective Lengths**:
- After making the hole, the section of the pipe from the closed end to the hole (\( \frac{L}{4} \)) behaves like a closed pipe, and the section from the hole to the open end (\( \frac{3L}{4} \)) behaves like an open pipe.

5. **Resonance Conditions**:
- The closed pipe (from closed end to hole) will have its fundamental frequency:
\[ f_{closed} = \frac{v}{4 \cdot \frac{L}{4}} = \frac{v}{L} \]
- The open pipe (from hole to open end) will have its fundamental frequency:
\[ f_{open} = \frac{v}{2 \cdot \frac{3L}{4}} = \frac{2v}{3L} \]

6. **Finding Resonance**:
- For the pipe to be in resonance again after making the hole, the frequencies of the two sections must match. However, the frequencies calculated above (\( \frac{v}{L} \) and \( \frac{2v}{3L} \)) are not equal, indicating that they cannot resonate together.

7. **Conclusion**:
- Since the two sections cannot resonate together, the pipe will not resonate with any tuning fork after the hole is made. Therefore, the correct answer is that the pipe will never resonate with any tuning fork.

### Final Answer:
The pipe will never resonate with any tuning fork after the hole is made.