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Question
An observer standing in front of a vertical wall claps 10 times one second. He adjusts his distance from the wall in such a way that the second of his clapping coincides with the echo. This happens when bis distance from the wall is 17.5m. What is the velocity of sound heard by him?
Solution
Given: d = 17.5m
An observer claps 10 times in one second, i.e., he claps at an interval of 1/10th of a second or 0.1 s. As the sound of the echo coincides with the clap this means that the echo is heard after 0.1s.
Therefore, t = 0.1s
v = `(2"d")/"t"=(2xx17"s")/0.1` = 350 m/s.
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