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 with the next clap. If he has to clap 10 times during every 3 seconds, find the velocity of sound in air.

Answer

A, C

Explanation

He has to clap 10 times in 3 seconds.
so, time interval between two clap
`=(3/(10 second))`
so, the time taken to go the wall
`=(v=(3/(20second))`
We know `s=vt`
`rarrV=s/t=50/(3/20)=333m/s`