Q99 / 150 · JEE Main 2006 · Physics
A whistle producing sound waves of frequencies $9500 \mathrm{~Hz}$ and above is approaching a stationary person with speed $v \mathrm{~ms}^{-1}$. The velocity of sound in air is $300 \mathrm{~ms}^{-1}$. If the person can hear frequencies upto a maximum of $10,000 \mathrm{~Hz}$, the maximum value of $\mathrm{v}$ upto which he can hear the whistle is
- A$30 \mathrm{~ms}^{-1}$
- B$15 \sqrt{2} \mathrm{~ms}^{-1}$
- C$15 / \sqrt{2} \mathrm{~ms}^{-1}$
- Correct answerD$15 \mathrm{~ms}^{-1}$
Solution
$f_{a p p}=\frac{f(300)}{300-v} \Rightarrow v=15 \mathrm{~m} / \mathrm{s}$
Hence, the correct answer is option: 'D'.