Question

A sounding body emitting a frequency of `150 H_(Z)` is dropped from a height. During its fall under gravity it crosses a balloon moving upwards with a constant velocity of `2 m//s` one second after it started to fall . The difference in the frequency observer by the man in balloon just before and just afer crossing the body will be (velocity of sound `= 300 m//s`, `g = 10 m//s^(2))`
A. 12
B. 6
C. 8
D. 4

Answer 1

Correct Answer - A
`f=f_(0)((vpmv_(0))/(vpmv_(s)))`
When approaching `f_(a)=150[(300+2)/(300-10)]`
when receding `f_(r)=150[(300-2)/(300+10)]rArr f_(a)-f_(r)~=12` hence (A)