A man stands between two parallel cliffs (not in middle). When he claps his hands, he hears two echoes one after `1` second and the other after `2` second. If the velocity of sound in air is `330 ms^(-1)` , the width of the valley is

Options

A. `330 m`
B. `494 m`
C. `660 m`
D. `990 m`

Explanation

To solve the problem, we need to determine the width of the valley based on the time it takes for the echoes to return to the man after he claps his hands.

### Step-by-Step Solution:

1. **Understanding the Echoes**:
- The man hears two echoes: one after 1 second and another after 2 seconds.
- The first echo (after 1 second) comes from the cliff that is closer to him, and the second echo (after 2 seconds) comes from the cliff that is farther away.

2. **Using the Speed of Sound**:
- The speed of sound in air is given as \( v = 330 \, \text{m/s} \).

3. **Calculating Distances**:
- For the first echo (1 second):
\[ \text{Distance to the first cliff and back} = v \times t = 330 \, \text{m/s} \times 1 \, \text{s} = 330 \, \text{m} \]
Since this distance is for the round trip (to the cliff and back), the distance to the first cliff is:
\[ d_1 = \frac{330}{2} = 165 \, \text{m} \]

- For the second echo (2 seconds):
\[ \text{Distance to the second cliff and back} = v \times t = 330 \, \text{m/s} \times 2 \, \text{s} = 660 \, \text{m} \]
Again, since this distance is for the round trip, the distance to the second cliff is:
\[ d_2 = \frac{660}{2} = 330 \, \text{m} \]

4. **Calculating the Width of the Valley**:
- The total width of the valley \( D \) is the sum of the distances to both cliffs:
\[ D = d_1 + d_2 = 165 \, \text{m} + 330 \, \text{m} = 495 \, \text{m} \]

5. **Final Answer**:
- The width of the valley is \( 495 \, \text{m} \).