The apparent frequency of a note is 200 Hz when a listener is moving with a velocity of 40 `ms​^(-1)` towards a stationary source. When he moves away from the same source with the same speed, the apparent frequency of the same note is 160 Hz. The velocity of sound in air (in m/s) is :-

Options

A. 340
B. 330
C. 360
D. 320

Answer

D

Explanation

When listener is moving towards the source,
`upsilon.=(v+v_(o))/(v)upsilonor200=(v+40)/(v)upsilon" "...(i)`
where `upsilon`. is apparent frequency and `upsilon` is original frequency of source and v is velocity of sound in air.
When listener is moving away from the source,
`upsilon..=(v-v_(o))/(v)upsilonor160=(v-40)/(v)upsilon" "...(ii)`
Divide (i) by (ii), we get `(200)/(160)=(v+40)/(v-40)`
Solving, we get `v=360m//s`