bullets are fixed at regualr intervals of 10 second from a car moving with a speed of ` 30 ms^(-1)` towards another car approaching with a speed of 60 m/s . The interval at which the firing can be reported is (Speed of sound = `330 ms^(-1)`

Options

A. 7.7 s
B. 8.32 s
C. 6.7 s
D. 12 s

Answer

To solve the problem step by step, we will use the principles of wave motion and the Doppler effect.

### Step 1: Identify the Given Information
- Speed of the first car (source) = \( v_s = 30 \, \text{m/s} \)
- Speed of the second car (observer) = \( v_o = 60 \, \text{m/s} \)
- Speed of sound = \( v = 330 \, \text{m/s} \)
- Time interval between bullets fired from the first car = \( T = 10 \, \text{s} \)

### Step 2: Calculate the Frequency of Bullet Firing
The frequency \( f \) of the bullets being fired can be calculated using the formula:
\[ f = \frac{1}{T} \]
Substituting the value of \( T \):
\[ f = \frac{1}{10} = 0.1 \, \text{Hz} \]

### Step 3: Apply the Doppler Effect Formula
Since both the source and observer are moving towards each other, we can use the Doppler effect formula for the apparent frequency \( f' \):
\[ f' = f \left( \frac{v + v_o}{v - v_s} \right) \]
Substituting the known values:
\[ f' = 0.1 \left( \frac{330 + 60}{330 - 30} \right) \]

### Step 4: Simplify the Expression
Calculating the numerator and denominator:
\[ f' = 0.1 \left( \frac{390}{300} \right) \]
\[ f' = 0.1 \times 1.3 = 0.13 \, \text{Hz} \]

### Step 5: Calculate the New Time Interval
The new time interval \( T' \) can be found using the relationship between frequency and time:
\[ T' = \frac{1}{f'} \]
Substituting the value of \( f' \):
\[ T' = \frac{1}{0.13} \approx 7.69 \, \text{s} \]

### Step 6: Round the Result
Rounding \( T' \) to two decimal places gives:
\[ T' \approx 7.7 \, \text{s} \]

### Conclusion
The interval at which the firing can be reported is approximately \( 7.7 \, \text{s} \).

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