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JEE Main 2009
Exam
JEE Main
Year
2009
Subject
Physics
Chapter
Waves and Sound

JEE Main 2009 · Physics · Waves and Sound

A motor cycle starts from rest and accelerates along a straight...

A motor cycle starts from rest and accelerates along a straight path at 2 m/s22 \mathrm{~m} / \mathrm{s}^2. At the starting point of the motor cycle there is a stationary electric sire. How far has the motor cycle gone when the driver hears the frequency of the siren at 94%94 \% of its value when the motor cycle was at rest? (speed of sound =330 ms−1)\left.=330 \mathrm{~ms}^{-1}\right)

Options

  1. A49 m49 \mathrm{~m}
  2. B98 m98 \mathrm{~m}
  3. C147 m147 \mathrm{~m}
  4. D196 m196 \mathrm{~m}

Correct answer

B. 98 m98 \mathrm{~m}

Step-by-step solution

Motor cycle, u=0,a=2 m/s2u=0, a=2 \mathrm{~m} / \mathrm{s}^2 Observer is in motion and source is at rest. ⇒n′=nv−vOv+vS⇒94100n=n330−vO330⇒330−v0=330×94100⇒vO=330−94×3310=33×610 m/ss=v2−u22a=9×33×33100=9×1089100≈98 m.\begin{aligned}& \Rightarrow n^{\prime}=n \frac{v-v_O}{v+v_S} \Rightarrow \frac{94}{100}n=n \frac{330-v_O}{330} \Rightarrow330-v_0=\frac{330 \times 94}{100}\\ & \Rightarrow v_O=330-\frac{94 \times 33}{10}=\frac{33 \times 6}{10} \mathrm{~m}/ \mathrm{s}\\ & s=\frac{v^2-u^2}{2 a}=\frac{9 \times 33 \times 33}{100}=\frac{9 \times 1089}{100} \approx98 \mathrm{~m} . \end{aligned}
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