A person hears the sound of the explosion of a bomb after 5 s and another person hears the sound after 6 s. The ratio of minimum and maximum possible distance between the person is _____. ( velocity of sound in air is `330 ms^(-1)`)

Options

A. `1 : 1`
B. `1 : 2`
C. `1 : 11`
D. `5 : 6`

Explanation

To solve the problem, we need to find the minimum and maximum possible distances between the two persons based on the time they heard the explosion of the bomb. We will use the speed of sound in air, which is given as 330 m/s.

### Step-by-Step Solution:

1. **Identify the times heard by each person**:
- First person hears the explosion after 5 seconds.
- Second person hears the explosion after 6 seconds.

2. **Calculate the time difference**:
- The time difference between when the first and second person hears the sound is:
\[ \Delta t = 6 \text{ seconds} - 5 \text{ seconds} = 1 \text{ second} \]

3. **Calculate the minimum distance**:
- The minimum distance (d1) can be calculated using the time it took for the first person to hear the sound:
\[ d_1 = \text{speed} \times \text{time} = 330 \, \text{m/s} \times 5 \, \text{s} = 1650 \, \text{m} \]

4. **Calculate the maximum distance**:
- The maximum distance (d2) is calculated using the total time taken for the sound to reach both persons:
\[ d_2 = \text{speed} \times \text{total time} = 330 \, \text{m/s} \times (5 \, \text{s} + 6 \, \text{s}) = 330 \, \text{m/s} \times 11 \, \text{s} = 3630 \, \text{m} \]

5. **Calculate the ratio of minimum to maximum distance**:
- The ratio of the minimum distance (d1) to the maximum distance (d2) is:
\[ \text{Ratio} = \frac{d_1}{d_2} = \frac{1650 \, \text{m}}{3630 \, \text{m}} = \frac{1}{11} \]

### Final Answer:
The ratio of the minimum and maximum possible distance between the two persons is \(\frac{1}{11}\).

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