A person observes a change of 2.5% in frequency of sound of horn of a car . If the car is apporaching forward the person sound velocity is 320 m/s then velocity of car in m/ s wil be appromately

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Solution

The correct option is D 8
Doppler formula n' $= \frac{n v}{v - v_{s}}$ $n^{'} > n$
if $n 2100 n^{'} = 102.5$
Since source is moving towards distance so
$102.5 = \frac{100 \times 320}{320 - v_{s}}$
$∴ v_{s} = 8$ m/sec.

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A person observes a change of 2.5% in frequency of sound of horn of a car . If the car is apporaching forward the person sound velocity is 320 m/s then velocity of car in m/ s wil be appromately

The correct option is D 8Doppler formula n'=nvv−vs n′>nif n2100n′=102.5Since source is moving towards distance so102.5=100×320320−vs∴vs=8m/sec.

The correct option is D 8
Doppler formula n' $= \frac{n v}{v - v_{s}}$ $n^{'} > n$
if $n 2100 n^{'} = 102.5$
Since source is moving towards distance so
$102.5 = \frac{100 \times 320}{320 - v_{s}}$
$∴ v_{s} = 8$ m/sec.