A bat emits ultrasonic sound of frequency `1000 kHz` in air . If the sound meets a water surface , it gets partially reflected back and partially refracted (transmitted) in water . What would be the difference of wavelength transmitted to wavelength reflected (speed of sound in air `= 330 m//s`, Bulk modulus of water ` = 2.25 xx 10^(9) , rho_(water) = 1000 kg//m^(2)`).

Explanation

To solve the problem, we need to find the wavelengths of the ultrasonic sound in both air and water and then calculate the difference between these two wavelengths.

### Step-by-Step Solution:

1. **Calculate the Wavelength in Air**:
- The formula for wavelength (\( \lambda \)) is given by:
\[ \lambda = \frac{V}{F} \]
- Where \( V \) is the speed of sound in air and \( F \) is the frequency of the sound.
- Given:
- Speed of sound in air, \( V = 330 \, \text{m/s} \)
- Frequency, \( F = 1000 \, \text{kHz} = 1000 \times 10^3 \, \text{Hz} = 10^6 \, \text{Hz} \)
- Now, substituting the values:
\[ \lambda_{\text{air}} = \frac{330 \, \text{m/s}}{10^6 \, \text{Hz}} = 0.00033 \, \text{m} = 0.33 \, \text{m} \]

2. **Calculate the Speed of Sound in Water**:
- The speed of sound in a medium can be calculated using the formula:
\[ V = \sqrt{\frac{K}{\rho}} \]
- Where \( K \) is the bulk modulus and \( \rho \) is the density of the medium.
- Given:
- Bulk modulus of water, \( K = 2.25 \times 10^9 \, \text{N/m}^2 \)
- Density of water, \( \rho = 1000 \, \text{kg/m}^3 \)
- Now, substituting the values:
\[ V_{\text{water}} = \sqrt{\frac{2.25 \times 10^9 \, \text{N/m}^2}{1000 \, \text{kg/m}^3}} = \sqrt{2.25 \times 10^6} \approx 1500 \, \text{m/s} \]

3. **Calculate the Wavelength in Water**:
- Using the same wavelength formula:
\[ \lambda_{\text{water}} = \frac{V_{\text{water}}}{F} \]
- Substituting the values:
\[ \lambda_{\text{water}} = \frac{1500 \, \text{m/s}}{10^6 \, \text{Hz}} = 0.0015 \, \text{m} = 1.5 \, \text{m} \]

4. **Calculate the Difference in Wavelengths**:
- Now, we need to find the difference between the wavelength transmitted in water and the wavelength reflected in air:
\[ \Delta \lambda = \lambda_{\text{water}} - \lambda_{\text{air}} = 1.5 \, \text{m} - 0.33 \, \text{m} = 1.17 \, \text{m} \]

### Final Answer:
The difference in wavelength transmitted to wavelength reflected is \( 1.17 \, \text{m} \).

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