Two cars are approaching each other with same speed of ` 20 m//s `. A man in car A fires bullets at regular intervals of 10 seconds. What will be the time interval noted by a man in car B between 2 bullets?
(velocity of sound =340 m/s )

Options

A. 11. 1s
B. 10s
C. 8.9 s
D. 12 s

Answer

To solve the problem, we need to find the time interval noted by a man in car B between two bullets fired from car A. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Given Information
- Speed of both cars (Car A and Car B) = 20 m/s
- Time interval between bullets fired from Car A = 10 seconds
- Velocity of sound = 340 m/s

### Step 2: Calculate the Frequency of Bullets Fired from Car A
The frequency \( f \) of bullets fired can be calculated using the formula:
\[ f = \frac{1}{T} \]
where \( T \) is the time interval between bullets. Given \( T = 10 \) seconds, we have:
\[ f = \frac{1}{10} = 0.1 \text{ Hz} \]

### Step 3: Apply the Doppler Effect Formula
Since both cars are approaching each other, we will use the Doppler effect formula for sound:
\[ f' = f \cdot \frac{v + v_o}{v - v_s} \]
where:
- \( f' \) = apparent frequency observed by the man in car B
- \( f \) = actual frequency (0.1 Hz)
- \( v \) = speed of sound (340 m/s)
- \( v_o \) = speed of observer (Car B, 20 m/s)
- \( v_s \) = speed of source (Car A, 20 m/s)

### Step 4: Substitute the Values into the Doppler Effect Formula
Substituting the values into the formula:
\[ f' = 0.1 \cdot \frac{340 + 20}{340 - 20} \]
\[ f' = 0.1 \cdot \frac{360}{320} \]
\[ f' = 0.1 \cdot 1.125 = 0.1125 \text{ Hz} \]

### Step 5: Calculate the Time Period for the Apparent Frequency
The time period \( T' \) corresponding to the apparent frequency \( f' \) can be calculated using:
\[ T' = \frac{1}{f'} \]
Substituting the value of \( f' \):
\[ T' = \frac{1}{0.1125} \approx 8.89 \text{ seconds} \]

### Conclusion
The time interval noted by a man in car B between two bullets is approximately **8.89 seconds**.

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