When the observer moves towards a stationary source with velocity ' V₁ ', the apparent frequency of the emitted note is ' F₁ '. When the observer moves away from the source with velocity ' V₁ ', the apparent frequency is ' F₂ '. If 'V' is the speed of sound in air and F₁/F₂ = 2, then V/V₁ = (a) 2 (b) 3 (c) 6 (d) 8

Answer: (b). Observer moving: F₁ = F (V + V₁)/(V), F₂ = F (V - V₁)/(V). (F₁)/(F₂) = (V + V₁)/(V - V₁) = 2 ⇒ V = 3V₁ So the answer is (b) 3.

When the observer moves towards a stationary source with velocity '… | MHT CET 2026 (16 Apr, Shift 1)

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Doppler Effect

When the observer moves towards a stationary source with velocity ' $V_1$ ', the apparent frequency of the emitted note is ' $F_1$ '. When the observer moves away from the source with velocity ' $V_1$ ', the apparent frequency is ' $F_2$ '. If 'V' is the speed of sound in air and $F_1/F_2 = 2$ , then $V/V_1 =$

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Answer: (b)

Observer moving: $F_1 = F\,\dfrac{V + V_1}{V}$ , $F_2 = F\,\dfrac{V - V_1}{V}$ .
$\dfrac{F_1}{F_2} = \dfrac{V + V_1}{V - V_1} = 2 \Rightarrow V = 3V_1$
So the answer is (b) 3.

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Doppler Effect question from MHT CET 2026 (16 Apr, Shift 1)

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