A vertical pipe open at both ends is partially submerged in water . A tuning fork is unknown frequency is placed near the top of the pipe and made to vibrate . The pipe can be moved up and down and thus length of air column the pipe can be adjusted. For definite lengths of air column in the pipe, standing waves will be set up as a result of superposition of sound waves traveling in opposite directions. Smallest value of length of air column , for which sound intensity is maximum is `10 cm` [ take speed of sound , `v = 344 m//s`].
Answer the following questions.
The air column here is closed at one end because the surface of water acts as a wall. Which of the following is correct ?

Options

A. a. At the closed end of air column , there is a displacement node and also a pressure node
B. b. At the closed end of the air column , there is a displacement node and also a pressure antinode
C. c. At the closed end of the air column , there is a displacement antinode and a pressure node
D. d. At the closed end of the air column , there is a displacement antinode and also a pressure antinode

Explanation

To solve the problem, we need to analyze the behavior of standing waves in a vertical pipe that is partially submerged in water. The pipe is open at both ends, but the water surface acts as a closed end for the air column.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- The pipe is open at the top and closed at the bottom (due to the water).
- The length of the air column can be adjusted, and we are interested in the conditions for standing waves.

2. **Identifying the Characteristics of Standing Waves**:
- In a closed pipe (closed at one end), the standing wave pattern has a displacement node at the closed end and an antinode at the open end.
- The closed end (water surface) will have minimum displacement (node) and maximum pressure (antinode).

3. **Analyzing the Given Length**:
- The smallest value of the length of the air column for maximum sound intensity is given as `10 cm` (0.1 m).
- This corresponds to the fundamental frequency (first harmonic) of the pipe.

4. **Wave Properties**:
- For a closed pipe, the fundamental frequency (first harmonic) can be described by the equation:
\[ L = \frac{\lambda}{4} \]
where \(L\) is the length of the air column and \(\lambda\) is the wavelength.

5. **Calculating the Wavelength**:
- Rearranging the formula gives:
\[ \lambda = 4L = 4 \times 0.1 \, \text{m} = 0.4 \, \text{m} \]

6. **Finding the Frequency**:
- The speed of sound in air is given as \(v = 344 \, \text{m/s}\).
- The frequency \(f\) can be calculated using the wave equation:
\[ f = \frac{v}{\lambda} = \frac{344 \, \text{m/s}}{0.4 \, \text{m}} = 860 \, \text{Hz} \]

7. **Identifying Nodes and Antinodes**:
- At the closed end (water surface), there is a displacement node and a pressure antinode.
- At the open end (top of the pipe), there is a displacement antinode and a pressure node.

8. **Conclusion**:
- The correct statements regarding the closed end of the air column are:
- There is a displacement node at the closed end.
- There is a pressure antinode at the closed end.

### Final Answer:
The correct option is that at the closed end of the air column, there is a displacement node and a pressure antinode.