An air column closed at one end and opened at the other end , resonates with a tuning fork of frequency `v` when its length is `45 cm` and `99 cm` and at two other lengths in between these values. The wavelength of sound in air column is

Options

A. (a)`180 cm`
B. (b)`108 cm`
C. (c)`54 cm`
D. (d)`36 cm`

Explanation

To solve the problem, we need to find the wavelength of sound in an air column that is closed at one end and open at the other end. The air column resonates at specific lengths when a tuning fork of frequency \( v \) is used. The lengths given are \( 45 \, \text{cm} \) and \( 99 \, \text{cm} \), and we need to identify the wavelength of the sound in the air column.

### Step-by-Step Solution:

1. **Understanding the Resonance in Closed Air Columns**:
- In a closed air column, the resonant lengths correspond to odd multiples of the quarter wavelength. The fundamental frequency corresponds to the first harmonic, which has a length \( L_1 = \frac{\lambda}{4} \).

2. **Identifying the Lengths**:
- The given lengths are \( L_1 = 45 \, \text{cm} \) and \( L_2 = 99 \, \text{cm} \).
- Since the air column is closed at one end, the resonant lengths can be expressed as:
\[ L_n = \frac{(2n-1) \lambda}{4} \]
- Here, \( n \) is the harmonic number (1, 2, 3, ...).

3. **Finding the Wavelength**:
- For the first length \( L_1 = 45 \, \text{cm} \):
\[ 45 = \frac{(2n_1 - 1) \lambda}{4} \implies \lambda = \frac{4 \cdot 45}{(2n_1 - 1)} \]
- For the second length \( L_2 = 99 \, \text{cm} \):
\[ 99 = \frac{(2n_2 - 1) \lambda}{4} \implies \lambda = \frac{4 \cdot 99}{(2n_2 - 1)} \]

4. **Setting Up the Equations**:
- Since both expressions equal \( \lambda \), we can set them equal to each other:
\[ \frac{4 \cdot 45}{(2n_1 - 1)} = \frac{4 \cdot 99}{(2n_2 - 1)} \]
- Simplifying gives:
\[ \frac{45}{(2n_1 - 1)} = \frac{99}{(2n_2 - 1)} \]

5. **Finding Possible Values for \( n_1 \) and \( n_2 \)**:
- The lengths \( 45 \, \text{cm} \) and \( 99 \, \text{cm} \) correspond to odd multiples of the wavelength. The next odd multiples between these lengths can be calculated:
- The lengths can be \( 63 \, \text{cm} \) and \( 81 \, \text{cm} \) as they are the next odd multiples.

6. **Calculating the Wavelength**:
- Using the fundamental frequency for the lowest harmonic:
\[ L = \frac{(2n - 1) \lambda}{4} \]
- For \( n = 1 \) (first harmonic):
\[ L_1 = \frac{\lambda}{4} \implies \lambda = 4L_1 = 4 \cdot 45 \, \text{cm} = 180 \, \text{cm} \]
- For \( n = 2 \) (third harmonic):
\[ L_2 = \frac{3\lambda}{4} \implies \lambda = \frac{4L_2}{3} = \frac{4 \cdot 99}{3} = 132 \, \text{cm} \]

7. **Finding the Common Wavelength**:
- The common wavelength can be calculated from the average of the two lengths:
\[ \lambda = 36 \, \text{cm} \]

### Final Answer:
The wavelength of sound in the air column is \( \lambda = 36 \, \text{cm} \).