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NEET 2009
Exam
NEET
Year
2009
Subject
Physics
Chapter
Waves and Sound

NEET 2009 · Physics · Waves and Sound

The driver of a car travelling with speed 30 ~ms^ -1 towards a hill...

The driver of a car travelling with speed 30 ms−130 \mathrm{~ms}^{-1} towards a hill sounds a horn of frequency 600 Hz600 \mathrm{~Hz}. If the velocity of sound in air is 330 ms−1330 \mathrm{~ms}^{-1}, the frequency of reflected sound as heard by driver isThe driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If theThe driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If theThe driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If theThe driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If the

Options

  1. A550 Hz550 \mathrm{~Hz}
  2. B555.5 Hz555.5 \mathrm{~Hz}
  3. C720 Hz720 \mathrm{~Hz}
  4. D500 Hz500 \mathrm{~Hz}

Correct answer

C. 720 Hz720 \mathrm{~Hz}

Step-by-step solution

Key Idea Apply Doppler's effect. Whenever there is a relative motion between a source of sound and the observer (listner), the frequency of sound heard by the observer is different from the actual frequency of sound emitted by source. The driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If the
The driver of a car travelling with speed 30 ~ms^ -1 towards a hill sounds a horn of frequency 600 ~Hz . If the
[for case I] n′=vv−30n\quad \mathbf{n}^{\prime}=\frac{\mathrm{v}}{\mathrm{v}-30} \mathbf{n} [for case II] n′′=v+30vn′n^{\prime \prime}=\frac{v+30}{v} n^{\prime} From Eqs. (i) and (ii), we get n′′=v+30v−30n=360300×600=720Hz\begin{aligned} \mathrm{n}^{\prime \prime}=\frac{\mathrm{v}+30}{\mathrm{v}-30} \mathrm{n}& =\frac{360}{300} \times600 \\ & =720 \mathrm{ Hz} \end{aligned}
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