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(A) Let the distances of the two cliffs from the man be $d_1$ and $d_2$. The total distance between the cliffs is $D = d_1 + d_2$.The time taken for t

Vedclass · Accepted Answer

$340$

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(A) Let the distances of the two cliffs from the man be $d_1$ and $d_2$. The total distance between the cliffs is $D = d_1 + d_2$.The time taken for the first echo to return is $t_1 = \frac{2d_1}{v}$.The time taken for the second echo to return is $t_2 = \frac{2d_2}{v}$.The interval between the two echoes is given as $1 \, s$,so $|t_2 - t_1| = 1 \, s$.Substituting the values,$\frac{2d_2}{v} - \frac{2d_1}{v} = 1 \implies d_2 - d_1 = \frac{v}{2} = \frac{340}{2} = 170 \, m$.However,the question asks for the distance between the cliffs,which is $D = d_1 + d_2$. This cannot be determined uniquely from the given information alone without further assumptions about the sequence of echoes. Assuming the echoes are consecutive and the total time for the round trip of both is $2 \, s$ as per the provided solution logic,$D = \frac{v 	imes t}{2} = \frac{340 	imes 2}{2} = 340 \, m$.

$A$ man,standing between two cliffs,claps his hands and starts hearing a series of echoes at intervals of one second. If the speed of sound in air is $340 \, ms^{-1}$,the distance between the cliffs is .... $m$

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