A whistle producing sound waves of frequencies $$9500\,Hz$$ and above is approaching a stationary person with speed $$v\,ms^{-1}$$. 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 the whistle is

According to the Doppler effect, when a sound source approaches a stationary observer, the apparent frequency $$f'$$ heard by the observer is given by the formula:

$$f' = f \left( \frac{v_{\text{sound}}}{v_{\text{sound}} - v_{\text{source}}} \right)$$
Substitute the given values into the Doppler effect formula:

$$10000 = 9500 \left( \frac{300}{300 - v} \right)$$

$$20 = 19 \left( \frac{300}{300 - v} \right)$$

$$20(300 - v) = 19 \times 300$$

$$6000 - 20v = 5700$$

Rearrange the terms to solve for $$20v$$:

$$20v = 6000 - 5700$$

$$20v = 300$$

$$v = 15\text{ m/s}$$

The maximum value of $$v$$ up to which the person can hear the whistle is 15 m/s(option D)