When the observer moves towards a stationary source with velocity V₁, the apparent frequency of emitted note is F₁. When observer moves away from the source with velocity V₁, the apparent frequency is F₂. If V is the velocity of sound in air and F₁/F₂ = 2 then V/V₁ is equal to (a) 3 (b) 2 (c) 1 (d) 4

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

When the observer moves towards a stationary source with velocity V₁… | MHT CET 2025 (26 Apr, Shift 2)

Q15MHT CET 2025 · 26 APR · SHIFT 2Medium

Doppler Effect

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

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Answer: (a)

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

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Doppler Effect question from MHT CET 2025 (26 Apr, Shift 2)

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