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JEE Main 2019 (10 Apr, Shift 2) · Doppler Effect

Doppler Effect question from JEE Main 2019 (10 Apr, Shift 2)

Q53JEE Main 2019 · 10 APR · SHIFT 2Medium

Doppler Effect

A source of sound S is moving with a velocity of 50 m/s towards a stationary observer. The observer measures the frequency of the source as 1000 Hz. What will be the apparent frequency of the source when it is moving away from the observer after crossing him? (Take velocity of sound in air is 350 m/s)

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Answer: (a)

  1. Approaching: 1000=f 350350−50⇒f=1000×300350≈857 Hz1000 = f\,\dfrac{350}{350 - 50} \Rightarrow f = \dfrac{1000 \times 300}{350} \approx 857\text{ Hz}.
  2. Receding: f′=f 350350+50=1000×300400f' = f\,\dfrac{350}{350 + 50} = 1000 \times \dfrac{300}{400}
  3. So the answer is (a) 750 Hz.
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