Correct Answer - 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`
