A person feels 2.5$ difference of frequency of a motor-car horn. If the motor-car is moving to the person and the velocity of sound is 320 m/sec, then the velocity of car will be

8 m/s (approx)

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800 m/s

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7 m/s

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6 m/s (approx)

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Solution

The correct option is B 8 m/s (approx)
By Doppler's formula $n^{'} = \frac{n v}{(v - v_{s})}$
Since, source is moving towards the listener so $n^{'} > n$ .
If n = 100 then n' = 102.5
$\Rightarrow 102.5 = \frac{100 \times 320}{(320 - v_{s})} \Rightarrow v_{s} = 8$ m/sec

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A person feels 2.5$ difference of frequency of a motor-car horn. If the motor-car is moving to the person and the velocity of sound is 320 m/sec, then the velocity of car will be

The correct option is B 8 m/s (approx)By Doppler's formula n′=nv(v−vs)Since, source is moving towards the listener so n′>n.If n = 100 then n' = 102.5⇒102.5=100×320(320−vs)⇒vs=8 m/sec

The correct option is B 8 m/s (approx)
By Doppler's formula $n^{'} = \frac{n v}{(v - v_{s})}$
Since, source is moving towards the listener so $n^{'} > n$ .
If n = 100 then n' = 102.5
$\Rightarrow 102.5 = \frac{100 \times 320}{(320 - v_{s})} \Rightarrow v_{s} = 8$ m/sec