Question

A man standing unsymmetrically between two parallel cliffs, claps his hand and starts hearing a series of echoes at intervals of 1 s. If the speed of sound in air is 340 m$s^{-1}$, then the distance between the two prallel cliffs, is

• A. 170 m
• B. 340 m
• C. 510 m
• D. 680 m

Answer 1

Answer: (C)

510 m

Let the man $M$ be at distance $x$ from hill $H_1$ and $y$ from hill $H_2$ as shown in figure. Let $y>x$.
The time interval between the original sound and echos from $H_1$ and $H_2$ will be respectively
$t_1=\frac{2x}{v}$ and $t_2=\frac{2y}{v}$, where $v$ is the velocity of sound.
The distance between the hills is
$x+y=\frac{v}{2}[t_1+t_2]=\frac{340}{2}[1+2]=510$