Q. A man standing between two parallel hills, claps his hand and hears successive echoes at regular intervals of $11s.$ If velocity of sound is $340ms^{- 1},$ then the distance between the hills is

A) $100m$
B) $170\ m$
C) $510\ m$
D) $340\ m$
Explanation

(c) Let the man $M$ be at a distance $x$ from hill $H_{1}$ and $y$ from hill $H_{2}$ as shown in figure. Let $y > x.$
[image:15_image90.png]
The time interval between the original sound and echoes 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
$x + y = \frac{v}{2}\left( t_{1} + t_{2} \right) = \frac{340}{2}\lbrack 1 + 2\rbrack = 510\ m$

Dr. Abubacker
Reviewed by Dr. Abubacker

Physics Educator & NEET/JEE Specialist

Ph.D. in Physics with 19+ years of teaching experience.

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