Question:
A man standing between two parallel cliffs fires a gun. He hear two echos one after \(3\) second and the other after \(5\) second. The separation between the two cliffs is \(1360\) m, the speed of sound is
A man standing between two parallel cliffs fires a gun. He hear two echos one after \(3\) second and the other after \(5\) second. The separation between the two cliffs is \(1360\) m, the speed of sound is
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Each echo travels twice the distance to its cliff, and the two distances add up to the separation of the cliffs.
Updated On: Nov 12, 2026
- \(320\) m/s
- \(330\) m/s
- \(340\) m/s
- \(350\) m/s
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The Correct Option is C
Solution and Explanation
Step 1: Understanding the Concept
An echo returns after the sound travels to the cliff and back, so the distance to a cliff is \(\frac{vt}{2}\).
Step 2: The two cliffs
The first echo gives \(d_{1}=\frac{3v}{2}\) and the second gives \(d_{2}=\frac{5v}{2}\).
Step 3: Calculation
The cliffs are \(1360\) m apart, so \(d_{1}+d_{2}=\frac{3v+5v}{2}=4v=1360\).
Step 4: Result
\(v=340\) m/s, which is option (C).
Final Answer:
The speed of sound is 340 m/s. \[ \boxed{\text{(C) }340\text{ m/s}} \]
An echo returns after the sound travels to the cliff and back, so the distance to a cliff is \(\frac{vt}{2}\).
Step 2: The two cliffs
The first echo gives \(d_{1}=\frac{3v}{2}\) and the second gives \(d_{2}=\frac{5v}{2}\).
Step 3: Calculation
The cliffs are \(1360\) m apart, so \(d_{1}+d_{2}=\frac{3v+5v}{2}=4v=1360\).
Step 4: Result
\(v=340\) m/s, which is option (C).
Final Answer:
The speed of sound is 340 m/s. \[ \boxed{\text{(C) }340\text{ m/s}} \]
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