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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
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The correct answer is:
$15 \mathrm{~ms}^{-1}$
$15 \mathrm{~ms}^{-1}$
$f_{a p p}=\frac{f(300)}{300-v} \Rightarrow v=15 \mathrm{~m} / \mathrm{s}$
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