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BITSAT 2021
Exam
BITSAT
Year
2021
Subject
Physics
Chapter
Waves and Sound

BITSAT 2021 · Physics · Waves and Sound

The frequency of whistle of an engine appears to be (4 / 5)^ th of...

The frequency of whistle of an engine appears to be (4/5)th (4 / 5)^{\text{th }} of initial frequency when it crosses a stationary observer. If the velocity of sound is 330m/s,thenthespeedofenginewillbe330 \mathrm{ m} / \mathrm{s} , then the speed of engine will be

Options

  1. A30m/s30 \mathrm{ m} / \mathrm{s}
  2. B36.6m/s36.6 \mathrm{ m} / \mathrm{s}
  3. C40m/s40 \mathrm{ m} / \mathrm{s}
  4. D330m/s330 \mathrm{ m} / \mathrm{s}

Correct answer

B. 36.6m/s36.6 \mathrm{ m} / \mathrm{s}

Step-by-step solution

n′=nvv−vs………(1)n′′=nvv+vs……..(2)n^{\prime}=\frac{n v}{v-v_{s}} \ldots \ldots \ldots(1)n^{\prime \prime}=\frac{n v}{v+v_{s}} \ldots \ldots..(2) From (1) and (2), n′n′′=v+vsv−vs……..(3)\frac{n^{\prime}}{n^{\prime \prime}}=\frac{v+v_{s}}{v-v_{s}} \ldots \ldots..(3) According to question, n′n′=54vs=?v=330m/s……….(4)\frac{n^{\prime}}{n^{\prime}}=\frac{5}{4}v_{s}=? v=330 \mathrm{ m} / \mathrm{s} \ldots \ldots \ldots.(4) From eq. (3) and (4) 54=[330+vs330−vs]⇒9vs=330∴vs=36.6m/s\frac{5}{4}=\left[\frac{330+v_{s}}{330-v_{s}}\right] \Rightarrow 9 v_{s}=330\therefore v_{s}=36.6 \mathrm{ m} / \mathrm{s}
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