If a source emits sound of frequency 500 Hz and moves at 20 m/s toward a stationary observer (speed of sound 340 m/s), the observed frequency is closest to
A{'text': '500 Hz', 'label': 'A'}
B{'text': '470 Hz', 'label': 'B'}
C{'text': '531 Hz', 'label': 'C'}
D{'text': '520 Hz', 'label': 'D'}
Answer & Solution
Correct answer: C. {'text': '531 Hz', 'label': 'C'}
1. Doppler effect for source moving toward stationary observer.
2. f_observed = f_source × v / (v − v_source).
3. = 500 × 340 / (340 − 20) = 500 × 340 / 320.
4. = 500 × 1.0625 = 531.25 Hz.
_Source: NCERT Class 11 Physics, Ch 14 "Waves", §14 doppler ext_
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