If man were standing unsymmetrically between parallel cliffs, claps his hands and starts hearing a series of echoes at intervals of 1 s. If speed of sound in air is 340 ms-1, the distance between two cliffs would be ......

Options

A. (A) 340 m
B. (B) 510 m
C. (C) 170 m
D. (D) 680 m

Answer

B

Explanation

Here distance between two cliffs is supppose d which is equal to x + y.
Suppose `x lt y`
Here when sound is produced from position S at time `t =0,` first echo is heard at time `t_(1) -0= t _(1)=1s` (as per statement) as secons echo is heard at time `t _(2) -0 =t _(2) =t _(1) +1 (because` successive echose are heard at an interval of 1 s and so ` t _(2) -t_(1) = 1s)`
Now, `2x =vt _(2)`
`2y = vt _(2)`
`therefore 2x + 2y =v (t _(1) + t _(2))`
`therefore 2 (c + y ) = v (t _(1) + y _(2))`
`therefore 2d = v (t _(1) + t _(2))`
`therefore d = (v (t _(1) + t _(2)))/( 2) = (340 (1+2))/( 2) = 510 m`