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A whistle producing sound waves of frequencies 9500 Hz and 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 up to a maximum of 10,000 Hz , the maximum value of v up to which he can hear the whistle is

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15 ms^-1

30 ms^-1

5 $\sqrt{2}$ ms^-1

15 $\sqrt{2}$ ms^-1

answer is D.

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Detailed Solution

$n^{1} = (\frac{V}{V - V_{s}}) n \Rightarrow 10, 000 = (\frac{V}{V - V_{s}}) 95, 00$

$\Rightarrow \frac{V}{V - V_{s}} = \frac{100}{95} = \frac{20}{19}$

$\Rightarrow 20 V - 20 V_{s} = 19 V$

$\Rightarrow V_{s} = \frac{V}{20} = \frac{300}{20} = 15 m / s$

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