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. `15 sqrt(2) ms^(-1)`
B. `15// sqrt(2) ms^(-1)`
C. `15 ms^(-1)`
D. `30 ms^(-1)`

Answer

C

Explanation

`because n'=((v)/(v-v_(s)))n`
`:. 10000=((300)/(300-v_(s)))(9500)`
`rArr 300-v_(s)=(300xx9500)/(10000)=285rArrv_(s)=15 ms^(-)`