Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

# 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 - Physics

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.

#### 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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#### APPEARS IN

HC Verma Class 11, 12 Concepts of Physics 1
Chapter 16 Sound Waves
Q 3 | Page 353