To solve the problem step by step, we will break it down into two parts as specified in the question.
### Part (a): Frequency detected by a receiver kept inside the river downstream
1. **Calculate the speed of sound in water (C)**:
The speed of sound in a medium can be calculated using the formula:
\[
C = \sqrt{\frac{B}{\rho}}
\]
where \(B\) is the bulk modulus and \(\rho\) is the density of the medium. Given:
- Bulk modulus of water, \(B = 2.088 \times 10^9 \, \text{Pa}\)
- Density of water, \(\rho = 10^3 \, \text{kg/m}^3\)
Substituting the values:
\[
C = \sqrt{\frac{2.088 \times 10^9}{10^3}} = \sqrt{2.088 \times 10^6} \approx 1444.5 \, \text{m/s}
\]
2. **Identify velocities**:
- Velocity of the boat (source), \(V_s = 10 \, \text{m/s}\) (moving downstream)
- Velocity of the river (medium), \(V_m = 2 \, \text{m/s}\) (also downstream)
- Velocity of the receiver (observer), \(V_o = 0 \, \text{m/s}\) (stationary in the water)
3. **Use the Doppler effect formula**:
The frequency detected by the receiver can be calculated using the Doppler effect formula for sound in a moving medium:
\[
f' = f \frac{C + V_m}{C + V_m - V_s}
\]
where \(f\) is the original frequency emitted by the transmitter.
4. **Calculate the original frequency (f)**:
The original frequency can be calculated using the wavelength:
\[
f = \frac{C}{\lambda}
\]
Given that the wavelength \(\lambda = 14.45 \, \text{mm} = 14.45 \times 10^{-3} \, \text{m}\):
\[
f = \frac{1444.5}{14.45 \times 10^{-3}} \approx 100,000 \, \text{Hz} \, (10^5 \, \text{Hz})
\]
5. **Substitute values into the Doppler effect formula**:
\[
f' = 10^5 \frac{1444.5 + 2}{1444.5 + 2 - 10}
\]
\[
f' = 10^5 \frac{1446.5}{1436.5} \approx 10^5 \times 1.007
\]
\[
f' \approx 100,700 \, \text{Hz}
\]
### Part (b): Frequency detected by the receiver when both are in the air
1. **Calculate the speed of sound in air (C)**:
The speed of sound in air can be calculated using the formula:
\[
C = \sqrt{\frac{\gamma R T}{M}}
\]
where:
- \(\gamma = 1.4\)
- \(R = 8.31 \, \text{J/(mol K)}\)
- \(T = 20 + 273.15 = 293.15 \, \text{K}\)
- Mean molecular mass of air, \(M = 28.8 \times 10^{-3} \, \text{kg/mol}\)
Substituting the values:
\[
C = \sqrt{\frac{1.4 \times 8.31 \times 293.15}{28.8 \times 10^{-3}}} \approx 344 \, \text{m/s}
\]
2. **Identify velocities in air**:
- Velocity of the boat (source), \(V_s = 10 \, \text{m/s}\) (moving downstream)
- Velocity of the air (medium), \(V_m = -5 \, \text{m/s}\) (upstream)
- Velocity of the receiver (observer), \(V_o = 0 \, \text{m/s}\) (stationary in the air)
3. **Use the Doppler effect formula again**:
\[
f'' = f' \frac{C + V_m}{C + V_m - V_s}
\]
Substituting the values:
\[
f'' = 100,700 \frac{344 - 5}{344 - 5 - 10}
\]
\[
f'' = 100,700 \frac{339}{329} \approx 100,700 \times 1.0305
\]
\[
f'' \approx 103,000 \, \text{Hz}
\]
### Final Answers:
- (a) The frequency detected by the receiver kept inside the river downstream is approximately **100,700 Hz**.
- (b) The frequency detected by the receiver when both are in the air is approximately **103,000 Hz**.
