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WB JEE Medical WB JEE Medical Solved Paper-2014

  • question_answer
    A car is moving with a speed of \[72\,km-{{h}^{-1}}\]towards a roadside source that emits sound at a frequency of 850 Hz. The car driver listens to the sound while approaching the source and again while moving away from the source after crossing it. If the velocity of sound is \[340\,m{{s}^{-1}},\] the difference of the two frequencies, the  driver hears is

    A)  50 Hz

    B)  85 Hz 

    C)  100 Hz

    D)  150 Hz

    Correct Answer: C

    Solution :

     By Dopplers Effect When observer is moving with velocity \[{{v}_{0}},\]towards a source at rest then approach frequency \[{{N}_{Approach}}=N\left( \frac{v+{{v}_{0}}}{v} \right)\] where \[N=850\,Hz,\,v=340\,m{{s}^{-1}},\,{{v}_{0}}=72\,km\,{{h}^{-1}}\] \[=20\,m{{s}^{-1}}=850\,\left( \frac{340+20}{340} \right)\] Similarly, when observer is moving away from source, then \[{{N}_{Separation}}=N\left( \frac{v-{{v}_{0}}}{v} \right)=850\left( \frac{340-20}{340} \right)\] The different of the two frequency, \[{{N}_{Approach}}-{{N}_{Separation}}\] \[=850\left( \frac{360}{340} \right)-850\left( \frac{320}{340} \right)\] \[=\frac{850}{340}\times 40=\frac{850}{8.5}=100\,Hz\]


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