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JEE Main 2009 (AIEEE) · Doppler Effect

Doppler Effect question from JEE Main 2009 (AIEEE)

Q75JEE Main 2009 · AIEEEDifficult

Doppler Effect

A motor cycle starts from rest and accelerates along a straight path at 2 m/s2.2\text{ m/s}^2. At the starting point of the motor cycle there is a stationary electric siren. 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= 330\text{ ms}^{-1})

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Answer: (a)

  1. Observer moving away from a stationary source: f′=fv−uvf' = f\frac{v-u}{v}
  2. 0.94=330−u330⇒u=19.80.94 = \frac{330 - u}{330} \Rightarrow u = 19.8 m/s
  3. From rest: s=u22a=(19.8)24≈98s = \frac{u^2}{2a} = \frac{(19.8)^2}{4} \approx 98 m
  4. So the answer is (a) 9898 m.
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