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A man standing in front of a mountain beats a drum at regular intervals. The drumming rate is gradually increased and he finds that echo is not heard distinctly when the rate becomes `40` per minute. He then moves near to the mountain by `90` metres and finds that echo is again not heard distinctly when the drumming rate becomes `60` per minute. Calculate (a) the distance between the mountain and the initial position of the man and (b) the velocity of sound.

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Correct Answer - `270 m`
Time interval between beats (initially) `= (60)/(40) = 1.5` sec
Since `t = (2S)/(v)`
`1.5 = (2S)/(2v)`
where `S` is the initial distance and `v` is the speed of sound . As the man approches `90m` towards the mountain
`(2(S - 90))/(v) = 1` since drumming rate is `60//min`
Solving for `S` from the two relations we get
`(2S)/(2(S - 90)) = 1.5 , (S - 90) /(S) = (2)/(3)`
`rArr 3S - 3 xx 90 = 2S , S = 270 m`

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