A car moving towards an approaching bike. The bike is moving with a velocity `30ms^(-1)`. Frequency of horn sounded by the car is 100 Hz, while it is observed by bike rider as 120 Hz. The actual duration of horn is 6 s. Find the time interval for which the bike rider hears it. (Take velocity of sound in air as `330ms^(-1)`)

Options

A. 5s
B. 6s
C. 7s
D. Cannot be calcualted with given information

Answer

To solve the problem, we need to determine the time interval for which the bike rider hears the horn of the car. We are given the following information:

- Frequency of the horn sounded by the car, \( f_s = 100 \, \text{Hz} \)
- Frequency observed by the bike rider, \( f' = 120 \, \text{Hz} \)
- Actual duration of the horn, \( t_s = 6 \, \text{s} \)
- Velocity of sound in air, \( v = 330 \, \text{m/s} \)
- Velocity of the bike, \( v_b = 30 \, \text{m/s} \)

### Step-by-step Solution:

1. **Calculate the number of cycles emitted by the car in 6 seconds**:
\[ N = f_s \times t_s = 100 \, \text{Hz} \times 6 \, \text{s} = 600 \, \text{cycles} \]

2. **Determine the time interval for the bike rider to hear these 600 cycles**:
The bike rider hears the frequency \( f' = 120 \, \text{Hz} \). We can find the time \( t \) it takes for the bike rider to hear 600 cycles using the formula:
\[ N = f' \times t \]
Rearranging this gives:
\[ t = \frac{N}{f'} = \frac{600 \, \text{cycles}}{120 \, \text{Hz}} = 5 \, \text{s} \]

### Conclusion:
The time interval for which the bike rider hears the horn is \( 5 \, \text{s} \).