Question

A man standing between two parallel hills, claps his hand and hears successive echoes at regular intervals of 1 s. If velocity of sound is 340 m s–1, then the distance between the hills is

• A. 100 m
• B. 170 m
• C. 510 m
• D. 340 m

Answer 1

Answer: (C)

510 m

Answer 2

Let the man M be at a distance x from hill $H_{1}$and y from hill $H_{2}$as shown in figure, Let $y > x.$

The time interval between the original sound and echoes from $H_{1}$and $H_{2}$will be respectively

$t_{1} = \frac{2x}{v}andt_{2} = \frac{2y}{v}$

Where v is the velocity of sound. The distance between the hills is $x + y = \frac{v}{2}\lbrack t_{1} + t_{2}\rbrack = \frac{340}{3}\lbrack 1 + 2\rbrack = 510m$