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Question

A siren emitting a sound of frequency 800 Hz800\text{ Hz} moves away from an observer towards a cliff at a speed of 15 ms115\text{ ms}^{-1}. The frequency of sound that the observer hears in the echo reflected from the cliff will be: (Take, velocity of sound in air = 330 ms1330\text{ ms}^{-1})

A

800 Hz800\text{ Hz}

B

838 Hz838\text{ Hz}

C

885 Hz885\text{ Hz}

D

765 Hz765\text{ Hz}

Step-by-Step Solution

The siren is moving towards the cliff, so the cliff acts as a stationary observer receiving sound from a moving source. The apparent frequency ff' reaching the cliff is given by the Doppler effect formula: f=f0(vvvs)f' = f_0 \left(\frac{v}{v - v_s}\right) where: f0=800 Hzf_0 = 800\text{ Hz} (actual frequency of the siren) v=330 ms1v = 330\text{ ms}^{-1} (speed of sound) vs=15 ms1v_s = 15\text{ ms}^{-1} (speed of the source)

f=800×(33033015)=800×330315=800×2221838.09 Hzf' = 800 \times \left(\frac{330}{330 - 15}\right) = 800 \times \frac{330}{315} = 800 \times \frac{22}{21} \approx 838.09\text{ Hz}

The cliff then acts as a stationary source of sound, reflecting it back to the stationary observer. The frequency of the echo heard by the observer will be the same as the frequency received by the cliff. Therefore, the frequency of the echo heard by the observer is approximately 838 Hz838\text{ Hz}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

PHYSICSWAVEemittingfrequencyobservertowardsfrequency

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