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

Apparent frequency `v. = ((V)/(V - U_(s)) ) .v`
10,000 = `((300)/(300 - U_(s)) ) . 9500`
`rArr U_(s) = 15 ms^(-1)`.
So correct choice is (C) .