A sound wave travels with a velocity of 300 m `s^(-1)` through a gas. 9 beats are produced in 3 s when two waves pass through it simultaneously. lf one of the waves has 2 m wavelength, the wavelength of the other wave is a) 1.98 m b) 2.04 m c) 2.06 m d) 1.99 m

Options

A. 1.98 m
B. 2.04 m
C. 2.06 m
D. 1.99 m

Explanation

To solve the problem, we need to find the wavelength of the second wave given that a sound wave travels with a velocity of 300 m/s, and 9 beats are produced in 3 seconds when two waves pass through the gas simultaneously. One of the waves has a wavelength of 2 m.

### Step-by-Step Solution:

1. **Calculate the Number of Beats per Second:**
\[ \text{Number of beats per second} = \frac{\text{Total beats}}{\text{Total time}} = \frac{9 \text{ beats}}{3 \text{ s}} = 3 \text{ beats/s} \]

2. **Use the Beat Frequency Formula:**
The beat frequency (number of beats per second) is given by the difference in frequencies of the two waves:
\[ \text{Number of beats per second} = |f_1 - f_2| \]
where \( f_1 \) and \( f_2 \) are the frequencies of the two waves.

3. **Relate Frequency to Wavelength:**
The frequency of a wave can be calculated using the formula:
\[ f = \frac{v}{\lambda} \]
where \( v \) is the velocity of the wave and \( \lambda \) is the wavelength. For the two waves:
\[ f_1 = \frac{v}{\lambda_1} \quad \text{and} \quad f_2 = \frac{v}{\lambda_2} \]

4. **Substitute Frequencies into the Beat Frequency Equation:**
Using the relationship for frequencies:
\[ 3 = \left| \frac{v}{\lambda_1} - \frac{v}{\lambda_2} \right| \]
Since \( v = 300 \, \text{m/s} \) and \( \lambda_1 = 2 \, \text{m} \):
\[ 3 = \left| \frac{300}{2} - \frac{300}{\lambda_2} \right| \]
This simplifies to:
\[ 3 = \left| 150 - \frac{300}{\lambda_2} \right| \]

5. **Solve for \( \lambda_2 \):**
We can set up two equations based on the absolute value:
- Case 1: \( 150 - \frac{300}{\lambda_2} = 3 \)
- Case 2: \( 150 - \frac{300}{\lambda_2} = -3 \)

**Case 1:**
\[ 150 - 3 = \frac{300}{\lambda_2} \implies 147 = \frac{300}{\lambda_2} \implies \lambda_2 = \frac{300}{147} \approx 2.04 \, \text{m} \]

**Case 2:**
\[ 150 + 3 = \frac{300}{\lambda_2} \implies 153 = \frac{300}{\lambda_2} \implies \lambda_2 = \frac{300}{153} \approx 1.96 \, \text{m} \]
Since we are looking for a wavelength close to 2 m, we will consider the first case.

6. **Final Answer:**
The wavelength of the second wave is approximately \( 2.04 \, \text{m} \).

### Conclusion:
The correct option is (b) 2.04 m.