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MHT CET 2026 (12 May, Shift 1) · Doppler Effect

Doppler Effect question from MHT CET 2026 (12 May, Shift 1)

Q2MHT CET 2026 · 12 MAY · SHIFT 1Easy

Doppler Effect

A whistle giving out $400$ Hz approaches a stationary observer at a speed of $34$ m/s. The frequency heard by the observer (in Hz) is nearly (speed of sound $= 340$ m/s)

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Answer: (c)

Source approaching: $n' = n\frac{v}{v - v_s}$ .
$n' = 400 \times \frac{340}{340 - 34} = 400 \times \frac{340}{306}$
$n' \approx 444$ Hz. So the answer is (c) $444$ .

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