A whistle producing sound waves of frequencies $9500 H z$ and above is approaching a stationary person with speed $V m / s$ . The velocity of sound in air is $300 m / s$ . If the person can hear frequencies upto a maximum of $10, 000 H z$ , the maximum value of $V$ upto which he can hear the whistle is

30

m / s

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5 \sqrt{2} m / s

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15 / \sqrt{2} m / s

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15

m / s

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Solution

The correct option is D $15$ $m / s$
According to Doppler effect:

$n^{'} = n [\frac{v - v_{0}}{v - v_{s}}]$

Where, $v_{0}$ is speed of observer
$v_{s}$ is speed of source

$n^{'} = 10000$ and $n = 9500$

$\Rightarrow$ $10000 = 9500 [\frac{300}{300 - V}]$

$\Rightarrow V = 15 m / s$

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A whistle producing sound waves of frequencies 9500 Hz and above is approaching a stationary person with speed Vm/s. The velocity of sound in air is 300 m/s. 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

The correct option is D 15 m/s According to Doppler effect:n′=n[v−v0v−vs]Where, v0 is speed of observer vs is speed of source n′=10000 and n=9500⇒ 10000=9500[300300−V]⇒V=15 m/s

The correct option is D $15$ $m / s$
According to Doppler effect:

$n^{'} = n [\frac{v - v_{0}}{v - v_{s}}]$

Where, $v_{0}$ is speed of observer
$v_{s}$ is speed of source

$n^{'} = 10000$ and $n = 9500$

$\Rightarrow$ $10000 = 9500 [\frac{300}{300 - V}]$

$\Rightarrow V = 15 m / s$