A policeman on duty detects a drop of `10%` in the pitch of the horn of a moving car as it crosses him. If the velocity of sound is `330 m//s`, the speed of the car will be

Options

A. `17.4 m//s `
B. `20.4 m//s`
C. `18.7 m//s `
D. `16.4 m//s `

Answer

To solve the problem, we need to determine the speed of the car based on the drop in pitch detected by the policeman. The drop in pitch of the sound can be analyzed using the Doppler effect.

### Step-by-Step Solution:

1. **Understanding the Problem**:
The problem states that there is a drop of 10% in the pitch of the horn of a moving car as it crosses the policeman. The velocity of sound is given as 330 m/s.

2. **Determine the Change in Frequency**:
If the pitch drops by 10%, it means the frequency of the sound detected by the policeman is 90% of the original frequency (f).
\[ f' = f - 0.1f = 0.9f \]

3. **Using the Doppler Effect Formula**:
The Doppler effect formula for a source moving away from a stationary observer is given by:
\[ f' = f \left(\frac{v}{v + v_s}\right) \]
where:
- \( f' \) = observed frequency
- \( f \) = source frequency
- \( v \) = speed of sound (330 m/s)
- \( v_s \) = speed of the source (car)

4. **Substituting the Known Values**:
Since we know \( f' = 0.9f \), we can substitute this into the Doppler effect equation:
\[ 0.9f = f \left(\frac{330}{330 + v_s}\right) \]

5. **Canceling Out the Frequency**:
We can cancel \( f \) from both sides (assuming \( f \neq 0 \)):
\[ 0.9 = \frac{330}{330 + v_s} \]

6. **Cross-Multiplying**:
Cross-multiplying gives us:
\[ 0.9(330 + v_s) = 330 \]

7. **Expanding the Equation**:
Expanding the left side:
\[ 297 + 0.9v_s = 330 \]

8. **Isolating \( v_s \)**:
Now, we isolate \( v_s \):
\[ 0.9v_s = 330 - 297 \]
\[ 0.9v_s = 33 \]
\[ v_s = \frac{33}{0.9} = 36.67 \, \text{m/s} \]

9. **Final Answer**:
The speed of the car is approximately \( 36.67 \, \text{m/s} \).