An underwater sonar source operating at a frequency of 60 kHz directs its beam towards the surface. If velocity of sound in air is 330 m/s, wavelength and frequency of the waves in air are :-

Options

A. 5.5 mm,60 kHz
B. 3.30 m,60 kHz
C. 5.5 mm, 30 kHz
D. 5.5 mm,80 kHz

Explanation

To solve the problem, we need to find the wavelength and frequency of the waves in air given that the underwater sonar source operates at a frequency of 60 kHz and the velocity of sound in air is 330 m/s.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Frequency of the sonar source (in water): \( f = 60 \text{ kHz} = 60 \times 10^3 \text{ Hz} \)
- Velocity of sound in air: \( v = 330 \text{ m/s} \)

2. **Understand the Relationship Between Velocity, Frequency, and Wavelength:**
- The relationship is given by the formula:
\[ v = f \lambda \]
where \( v \) is the velocity of sound, \( f \) is the frequency, and \( \lambda \) is the wavelength.

3. **Determine the Frequency in Air:**
- The frequency of the wave does not change when it travels from one medium to another. Therefore, the frequency of the wave in air is the same as in water:
\[ f_{\text{air}} = f_{\text{water}} = 60 \text{ kHz} = 60 \times 10^3 \text{ Hz} \]

4. **Calculate the Wavelength in Air:**
- Rearranging the formula \( v = f \lambda \) to find the wavelength:
\[ \lambda = \frac{v}{f} \]
- Substitute the values of \( v \) and \( f \):
\[ \lambda = \frac{330 \text{ m/s}}{60 \times 10^3 \text{ Hz}} \]

5. **Perform the Calculation:**
- Calculate \( \lambda \):
\[ \lambda = \frac{330}{60000} = 0.0055 \text{ m} \]
- Convert meters to millimeters (1 m = 1000 mm):
\[ \lambda = 0.0055 \text{ m} \times 1000 = 5.5 \text{ mm} \]

6. **Final Results:**
- Frequency in air: \( 60 \text{ kHz} \)
- Wavelength in air: \( 5.5 \text{ mm} \)

### Conclusion:
The frequency of the waves in air is \( 60 \text{ kHz} \) and the wavelength is \( 5.5 \text{ mm} \).

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