Question

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.

Answer 1

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
$=\frac{3}{10}\mathrm{second}$.

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

$$\begin{gathered} \mathrm{We}\mathrm{know}\mathrm{that}: \\ \mathrm{Velocity}v=\frac{S}{t} \end{gathered}$$

$\Rightarrow v=\frac{50}{\left({\displaystyle \frac{3}{20}}\right)}=333\mathrm{m}/\mathrm{s}$

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