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A man standing in front of a mountain be...

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.

A

i.`330 m`
ii. `330 m//s`

B

i. `300 m`
ii. `720 m//s`

C

i. `240 m`
ii. `300 m//s`

D

i. `270 m`
ii. 270 m//s`

Text Solution

Verified by Experts

The correct Answer is:
A, B, D

Drumming frequency ` = 40 "cycle//min" = 40 "cycle//60s"`
Drumming time period
`T = (1)/(f) = ( 60 s)/( 40 "cycle") = (3)/(4) s//"cycle"`
( time duration between consecutive drumming)
During this time interval , if sound goes to mountain and comes back then echo will not be heard distinctly.
`(3)/(4) = ( 2 l)/( v)` (i)
Now if he moves `90 m`. This situation arises at ` t = 60 "cycle//min"`,
`T = (1)/(f) = 1 s//"cycle"`
for this case sound goes to mountain and comes back after time `T//2`:
`(1)/(2) = ( 2 ( l - 90))/( v)` (ii)
Solving Eqs. (i) and (ii)
so , `l = 270 m`
`v = 720 m//s`
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