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: Total path of the two echoes:
In 1 s the first echo covers $330$ m and in 3 s the second covers $990$ m. Each is a round trip.

Step 2: Add the round trips:
The combined round trips equal $330 + 990 = 1320$ m. Each cliff's distance is counted twice (there and back), so the gap between the cliffs is half of that.

Step 3: Compute:
$\frac{1320}{2} = 660$ m.

Step 4: Check against the echo times:
The first echo from 165 m away returns after $2\times165/330 = 1$ s. The second echo from 495 m away returns after $2\times495/330 = 3$ s. Both match the given times, so 165 + 495 = 660 m is consistent.

Final Answer:
Option (B). \[ \boxed{660 \text{ m (B)}} \]