A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with velocity v meters per sound. The velocity of sound in air is 300 m/s. If person can hear frequencies up to 10000 Hz, the maximum value of v up to which he can hear the 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^(-)`