Q124 / 150 · MHT CET 2023 (11 May Shift 2) · Physics

A sound of frequency $480 \mathrm{~Hz}$ is emitted from the stringed instrument. The velocity of sound in air is $320 \mathrm{~m} / \mathrm{s}$. After completing 180 vibrations, the distance covered by a wave is

  • A
    $60 \mathrm{~m}$
  • B
    $90 \mathrm{~m}$
  • Correct answer
    C
    $120 \mathrm{~m}$
  • D
    $180 \mathrm{~m}$

Solution

Given: $\mathrm{v}=320 \mathrm{~m} / \mathrm{s}, \mathrm{f}=480 \mathrm{~Hz}, \mathrm{~N}=180$ $\mathrm{v}=\mathrm{f} \lambda$ $\therefore \quad \lambda=\frac{\mathrm{v}}{\mathrm{f}}$ Substituting the values, we get $\lambda=\frac{320}{480}=\frac{2}{3}$ $\therefore \quad$ The total distance covered after 180 vibrations is $\begin{aligned} & \mathrm{D}=\mathrm{N} \times \lambda \\ & \mathrm{D}=180 \times \frac{2}{3} \\ & \mathrm{D}=120 \mathrm{~m} \end{aligned}$

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

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