Question

A person standing between two parallel cliffs fires a gun and hears two echoes, first echo after 1 second and second echo after 3 second. The distance between the two cliffs is (The velocity of sound \(= 330\) m/s)

• A. \(330\) m
• B. \(660\) m
• C. \(990\) m
• D. \(1320\) m

Hint:

Each echo travels to a cliff and back, so distance = v t / 2 for each cliff. Add the two distances.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept:
An echo is sound that goes to a cliff and comes back. So the distance to each cliff is half of the distance travelled by the sound before the echo is heard.

Step 2: Key Formula or Approach:
\[ d = \frac{vt}{2} \]

Step 3: Detailed Explanation:
The person stands between two parallel cliffs. The first echo comes from the nearer cliff, and the second echo from the farther cliff.
Nearer cliff:
\[ d_1 = \frac{330\times1}{2} = 165 \text{ m} \]
Farther cliff:
\[ d_2 = \frac{330\times3}{2} = 495 \text{ m} \]
The distance between the two cliffs is the sum:
\[ d = d_1 + d_2 = 165 + 495 = 660 \text{ m} \]
Option (A) 330 m is only the distance travelled in 1 s. Option (C) 990 m would be \(330\times3\), the path length of the second echo, and (D) 1320 m is twice the correct value.

Final Answer:
The distance between the cliffs is 660 m, option (B). \[ \boxed{660 \text{ m (B)}} \]