Q129 / 150 · MHT CET 2024 (10 May Shift 1) · Physics

When the listener moves towards stationary source with velocity ' $\mathrm{V}_1$ ', the apparent frequency of emitted note is ' $F_1$ '. When observer moves away from the source with velocity ' $\mathrm{V}_1$ ', apparent frequency is ' $\mathrm{F}_2$ '. If' V is the velocity of sound in air and $\frac{F_1}{F_2}=2$ then $\frac{V}{V_1}$ is

  • A
    2
  • Correct answer
    B
    3
  • C
    4
  • D
    5

Solution

$\begin{aligned} & F_1=F\left(\frac{V+V_1}{V}\right) \\ F_2 & =F\left(\frac{V-V_1}{V}\right) \\ \therefore \quad & \frac{F_1}{F_2}=\left(\frac{V+V_1}{V-V_1}\right)=2 \\ \therefore \quad & \frac{V}{V_1}=3 \end{aligned}$ ...(given)

Hence, the correct answer is option: 'B'.

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