The apparent frequency of the whistle of an engine changes in the ratio 6:5 as the engine passes a stationary observer. If the velocity of sound is `330(m)/(s)`, then the velocity of the engine is

Options

A. `3(m)/(s)`
B. `30(m)/(s)`
C. `0.33(m)/(s)`
D. `660(m)/(s)`

Explanation

To solve the problem, we can use the Doppler effect equations for sound. The apparent frequency changes as the source of sound (the engine) moves towards and then away from a stationary observer.

### Step-by-Step Solution:

1. **Understanding the Problem**:
- The apparent frequency changes in the ratio of 6:5 as the engine moves past a stationary observer.
- The speed of sound in air is given as \( V = 330 \, \text{m/s} \).

2. **Setting Up the Equations**:
- When the engine is moving towards the observer, the apparent frequency \( f' \) can be expressed as:
\[ f' = f \frac{V}{V - V_s} \]
- When the engine is moving away from the observer, the apparent frequency \( f'' \) can be expressed as:
\[ f'' = f \frac{V}{V + V_s} \]

3. **Using the Given Ratio**:
- We know that:
\[ \frac{f'}{f''} = \frac{6}{5} \]
- Substituting the expressions for \( f' \) and \( f'' \):
\[ \frac{f \frac{V}{V - V_s}}{f \frac{V}{V + V_s}} = \frac{6}{5} \]
- The \( f \) cancels out:
\[ \frac{V}{V - V_s} \cdot \frac{V + V_s}{V} = \frac{6}{5} \]
- Simplifying gives:
\[ \frac{V + V_s}{V - V_s} = \frac{6}{5} \]

4. **Cross-Multiplying**:
- Cross-multiplying to eliminate the fraction:
\[ 5(V + V_s) = 6(V - V_s) \]
- Expanding both sides:
\[ 5V + 5V_s = 6V - 6V_s \]

5. **Rearranging the Equation**:
- Bringing all terms involving \( V_s \) to one side and constant terms to the other:
\[ 5V_s + 6V_s = 6V - 5V \]
\[ 11V_s = V \]

6. **Substituting the Value of V**:
- Substitute \( V = 330 \, \text{m/s} \):
\[ 11V_s = 330 \]
- Solving for \( V_s \):
\[ V_s = \frac{330}{11} = 30 \, \text{m/s} \]

### Final Answer:
The velocity of the engine is \( 30 \, \text{m/s} \).