A whistle producing sound waves of frequencies `9500 Hz` and above is approaching a stationary person with speed `vms^(-1)`. The velocity of sound in air is `300 ms^(-1)`. If the person can hear frequencies upto a maximum of `10,000 Hz`.The maximum value of `v` upto which he can hear whistle is

Options

A. `15sqrt(2)m//s`
B. `(15)/(sqrt(2))m//s`
C. 15m/s
D. 30m/s

Answer

C

Explanation

When observer moves towards stationary observer, then apparent frequency
`f' = f[(v)/(v-v_(s)) ] rArr (f')/(f) = [(v)/(v - v_(s))]`
Where v is the velocity of sound in air.
` (10000)/(9500)= (300)/(33-v_(s)) rArr (300 - v ) = 285 rArr v=15m//s`