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Question

A man standing on a platform observes that the frequency of the sound of a whistle emitted by a moving train is less by 140 Hz that the frequency emitted when the train is stationary. If the velocity of sound in air is 330 m/s and the speed of the train is 70 m/s, the frequency of the whistle is

A
571 Hz
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B
800 Hz
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C
400 Hz
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D
260 Hz
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Solution

The correct option is B 800 Hz
Let f0 be the frequency of sound heard by the observer, when both the source of sound and observer are at rest.
Let v be the velocity of sound in the stationary medium and vs be the velocity of the source of sound.

Since the frequency of whistle heard by the observer is less, we can say that the train is moving away from observer.
When a source is moving away from a stationary observer, the frequency heard by the observer is given by
fapp=f0(vv+vs) ......(1)

From the data given in the question,
Decrease in frequency f0fapp=140 Hz

Using (1) in the above equation,

f0(vsv+vs)=140

Given,
Velocity of sound in stationary medium v=330 m/s

Velocity of source of sound vs=70 m/s

Substituting these in the above equation, we get

f0(70330+70)=140

f0=140×40070=800 Hz

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