Question

A boy stands between the two walls and claps his hands as shown below:

If ‘$x$’ is less than $60m$ and the time difference between the first and the second echo is $0.25s$, then calculate the value of ‘$x$’. (Velocity of sound in air is $344m{s}^{-1}$)

• A. $36$
• B. $34$
• C. $40$
• D. $17$

Answer 1

The correct option is D

$17$

Given,

Speed of sound in air = $344m{s}^{-1}$
Time difference between the first and second echo = $0.25s$

Since it is given that $x$ is less than $60m$, echo is first heard from wall 1.

Let the time taken for the first echo be $t_1$

Total distance covered by the sound wave before reaching the boy from the first wall = $2x$

$\therefore$$t_1=\frac{2x}{344}$

Let the time taken for the second echo be $t_2$

Total distance covered by the sound wave before reaching the boy from the second wall = $120m$

$\therefore$$t_2=\frac{120m}{344}$

$t_2-t_1=Timedifferencebetweenfirstandsecondecho$

$\frac{120-2x}{344}=0.25$

$120-2x=86$

$\Rightarrow 2x=34$

$x=17$