Sum

A man stands before a large wall at a distance of 50.0 m and claps his hands at regular intervals. Initially, the interval is large. He gradually reduces the interval and fixes it at a value when the echo of a clap merges every 3 seconds, find the velocity of sound in air.

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#### Solution

Given:

Distance of the large wall from the man *S* = 50 m

He has to clap 10 times in 3 seconds.

So, time interval between two claps will be \[= \frac{3}{10}\text { second }\]

Therefore, the time taken \[\left( t \right)\] by sound to go the wall is \[t = \frac{3}{20}\text { second }\]

\[\text { We know that: } \]

\[\text { Velocity } v = \frac{S}{t}\]

\[\Rightarrow v = \frac{50}{\left( \frac{3}{20} \right)} = 333 m/s\]

Hence, the velocity of sound in air is 333 m/s.

Concept: Wave Motion

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