Question:
A boy stands \(x\ \text{m}\) away from a high wall and claps his hands at a steady rate so that each clap coincides with the echo of the one before. If he makes \(60\) claps in \(2\) minutes and the velocity of sound is \(340\ \text{m/s}\), then the distance \(x\) from the wall is
A boy stands \(x\ \text{m}\) away from a high wall and claps his hands at a steady rate so that each clap coincides with the echo of the one before. If he makes \(60\) claps in \(2\) minutes and the velocity of sound is \(340\ \text{m/s}\), then the distance \(x\) from the wall is
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Find the time between claps. That interval equals the time for sound to go to the wall and back.
Updated On: Nov 12, 2026
- \(170\ \text{m}\)
- \(340\ \text{m}\)
- \(255\ \text{m}\)
- \(510\ \text{m}\)
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The Correct Option is B
Solution and Explanation
Step 1: Understanding the Concept
Each clap coincides with the echo of the previous one. So the time between claps equals the round-trip time of sound to the wall and back.
Step 2: Time between claps
\(60\) claps in \(2\) minutes means \(60\) claps in \(120\ \text{s}\), so one clap every \(2\ \text{s}\).
Step 3: Calculation
Round trip: \(\dfrac{2x}{v}=2\ \text{s}\), so \(x=\dfrac{340\times2}{2}=340\ \text{m}\).
Final Answer:
The wall is 340 m away. \[ \boxed{\text{(B) }\text{340 m}} \]
Each clap coincides with the echo of the previous one. So the time between claps equals the round-trip time of sound to the wall and back.
Step 2: Time between claps
\(60\) claps in \(2\) minutes means \(60\) claps in \(120\ \text{s}\), so one clap every \(2\ \text{s}\).
Step 3: Calculation
Round trip: \(\dfrac{2x}{v}=2\ \text{s}\), so \(x=\dfrac{340\times2}{2}=340\ \text{m}\).
Final Answer:
The wall is 340 m away. \[ \boxed{\text{(B) }\text{340 m}} \]
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