Submarine `A` is going with speed of `18km//hr`. Submarine `B` is chasing `A` with speed of `27 km//hr`. It sends frequency of `500Hz` and hears after reflection from `A`. The perceived frequency is :
`(V_("sound in water")=1500m//s)`

Options

A. `500Hz`
B. `502Hz`
C. `498Hz`
D. `504Hz`

Answer

To solve the problem, we need to determine the frequency perceived by submarine B after it sends a sound wave towards submarine A, which is moving away from it. We will use the Doppler effect formula for sound waves in water.

### Step-by-Step Solution:

1. **Convert Speeds from km/hr to m/s:**
- Speed of submarine A (V_A) = 18 km/hr = \( \frac{18 \times 1000}{3600} = 5 \, \text{m/s} \)
- Speed of submarine B (V_B) = 27 km/hr = \( \frac{27 \times 1000}{3600} = 7.5 \, \text{m/s} \)

2. **Identify the speed of sound in water:**
- Speed of sound in water (V_sound) = 1500 m/s

3. **Calculate the frequency perceived by submarine A (f_A):**
- The formula for the apparent frequency (f') when the source is moving away from the observer is given by:
\[ f' = f \left( \frac{V + V_{observer}}{V - V_{source}} \right) \]
- Here, the source is submarine A, and the observer is submarine B.
- Substituting the values:
- f = 500 Hz
- V = V_sound = 1500 m/s
- V_observer = V_B = 7.5 m/s (since B is approaching A)
- V_source = V_A = 5 m/s (since A is moving away from B)
- Thus,
\[ f_A = 500 \left( \frac{1500 + 7.5}{1500 - 5} \right) \]
- Calculate the denominator and numerator:
- Denominator: \( 1500 - 5 = 1495 \)
- Numerator: \( 1500 + 7.5 = 1507.5 \)
- Therefore,
\[ f_A = 500 \left( \frac{1507.5}{1495} \right) \]

4. **Calculate the frequency f_A:**
- Using a calculator:
\[ f_A \approx 500 \times 1.008 = 504 \, \text{Hz} \]

5. **Calculate the frequency perceived by submarine B after reflection (f_B):**
- Now, submarine A acts as a source emitting frequency f_A back towards submarine B.
- The formula for the frequency perceived by B (f_B) when the source is moving away from the observer is:
\[ f_B = f_A \left( \frac{V + V_{observer}}{V - V_{source}} \right) \]
- Here, the observer is submarine B and the source is submarine A.
- Substituting the values:
- f_A = 504 Hz (calculated above)
- V = V_sound = 1500 m/s
- V_observer = V_B = 7.5 m/s
- V_source = V_A = 5 m/s
- Thus,
\[ f_B = 504 \left( \frac{1500 + 7.5}{1500 - 5} \right) \]
- The denominator and numerator remain the same as before:
- Denominator: \( 1495 \)
- Numerator: \( 1507.5 \)
- Therefore,
\[ f_B = 504 \left( \frac{1507.5}{1495} \right) \]

6. **Calculate the frequency f_B:**
- Using a calculator:
\[ f_B \approx 504 \times 1.008 = 508 \, \text{Hz} \]

### Final Answer:
The perceived frequency by submarine B after reflection from submarine A is approximately **508 Hz**.