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Waves — Medium JEE physics MCQ

**[JEE Mains 2026]** A detector moves from point A towards a stationary source and then moves away to point B. The observed frequencies differ by 20%. If the source emits a frequency of 500 Hz, and the speed of the detector is v_d, what is the ratio v_d/v (where v is the speed of sound)?
  1. A. 0.05
  2. B. 0.1
  3. C. 0.15
  4. D. 0.2

Solution

When approaching: f₁ = f₀(v + v_d)/v. When receding: f₂ = f₀(v - v_d)/v. Difference: f₁ - f₂ = f₀ × 2v_d/v = 20% of average ≈ 0.2 × 500 = 100 Hz. So 500 × 2(v_d/v) = 100, giving v_d/v = 0.1.

PHYSICS

medium • PYQ Reworded
Question
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[JEE Mains 2026] A detector moves from point A towards a stationary source and then moves away to point B. The observed frequencies differ by 20%. If the source emits a frequency of 500 Hz, and the speed of the detector is vd_d, what is the ratio vd_d/v (where v is the speed of sound)?
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Waves — Medium JEE Physics MCQ | MyGoalPrep