A boat is travelling in a river with a speed `10m//s` along with stream flowing 2m/s. From this boat, a sound transmitter is lowered into the river through a rigid support. The wavelength of the sound emitted from the transmitter inside the water is `14.45mm`. Assume that attenuation of sound in water and air is negligible.
(a) What will be the frequency detected by a receiver kept inside the river downstream?
(b) The transmitter and the receiver are now pulled up into air. the air is blowing with a speed `5m//s` in the direction opposite the river stream. Determine the frequency of the sound detected by the receiver.
(Temperature of the air and water = `20^(@)C`, Density of river water `= 10^(3) kg//m^(3)`,
Bulk modulus of the water `= 2.088 xx 10^(9)`Pa, gas constant `R = 8.31 J//mol-K`,
Mean molecular mass of air `= 28.8 xx 10^(-3) kg//mol`, `C_(P)//C_(V)` for air `= 1.4`)

Explanation

To solve the problem step by step, we will break it down into two parts as per the question.

### Part (a): Frequency Detected by the Receiver in the River

1. **Calculate the Velocity of Sound in Water**:
The formula for the velocity of sound in a medium is given by:
\[ c = \sqrt{\frac{K}{\rho}} \]
where \( K \) is the bulk modulus and \( \rho \) is the density of the medium.
Given:
- Bulk modulus \( K = 2.088 \times 10^9 \, \text{Pa} \)
- Density \( \rho = 10^3 \, \text{kg/m}^3 \)

Substituting the values:
\[ c = \sqrt{\frac{2.088 \times 10^9}{10^3}} = \sqrt{2.088 \times 10^6} \approx 1445 \, \text{m/s} \]

2. **Calculate the Frequency of the Sound Emitted**:
The frequency \( f_0 \) can be calculated using the wavelength \( \lambda \):
\[ f_0 = \frac{c}{\lambda} \]
Given:
- Wavelength \( \lambda = 14.45 \, \text{mm} = 14.45 \times 10^{-3} \, \text{m} \)

Substituting the values:
\[ f_0 = \frac{1445}{14.45 \times 10^{-3}} \approx 100000 \, \text{Hz} = 10^5 \, \text{Hz} \]

3. **Apply the Doppler Effect Formula**:
The formula for the frequency detected by the observer is:
\[ f' = f_0 \left( \frac{c + v_m}{c + v_s} \right) \]
where:
- \( v_m = 2 \, \text{m/s} \) (velocity of the medium)
- \( v_s = 10 \, \text{m/s} \) (velocity of the source)

Substituting the values:
\[ f' = 10^5 \left( \frac{1445 + 2}{1445 - 10} \right) = 10^5 \left( \frac{1447}{1435} \right) \]

Calculating the fraction:
\[ f' \approx 10^5 \times 1.0084 \approx 100840 \, \text{Hz} \]

### Part (b): Frequency Detected by the Receiver in Air

1. **Calculate the Velocity of Sound in Air**:
The formula for the velocity of sound in air is:
\[ c = \sqrt{\frac{\gamma R T}{M}} \]
where:
- \( \gamma = 1.4 \)
- \( R = 8.31 \, \text{J/(mol K)} \)
- \( T = 20 \, \text{°C} = 293 \, \text{K} \)
- \( M = 28.8 \times 10^{-3} \, \text{kg/mol} \)

Substituting the values:
\[ c = \sqrt{\frac{1.4 \times 8.31 \times 293}{28.8 \times 10^{-3}}} \approx 344 \, \text{m/s} \]

2. **Apply the Doppler Effect Formula for Air**:
In this case, the air is blowing opposite to the direction of the sound, so:
- \( v_m = -5 \, \text{m/s} \) (velocity of the medium)
- \( v_s = 10 \, \text{m/s} \) (velocity of the source)

The formula remains the same:
\[ f' = f_0 \left( \frac{c + v_m}{c + v_s} \right) \]

Substituting the values:
\[ f' = 10^5 \left( \frac{344 - 5}{344 - 10} \right) = 10^5 \left( \frac{339}{334} \right) \]

Calculating the fraction:
\[ f' \approx 10^5 \times 1.0149 \approx 101490 \, \text{Hz} \]

### Final Answers:
- (a) The frequency detected by the receiver inside the river downstream is approximately **100840 Hz**.
- (b) The frequency detected by the receiver in air is approximately **101490 Hz**.