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JEE Main & Advanced Sample Paper JEE Main Sample Paper-23

  • question_answer
    The apparent frequency of the whistle of an engine changes in the ratio \[9:8\] as the engine passes a stationary observer. If the velocity of the sound is \[340\,m{{s}^{-1}}\], then the velocity of the  engine is :

    A)  \[40\,m{{s}^{-1}}\]                            

    B)  \[20\,m{{s}^{-1}}\]

    C)  \[34\,m{{s}^{-1}}\]                            

    D)  \[180\,m{{s}^{-1}}\]

    Correct Answer: B

    Solution :

    From Doppler's effect received frequency \[\frac{9}{8}=\frac{340+{{V}_{s}}}{340-{{V}_{s}}}\,\left[ \because \,\,\frac{{{F}_{approach}}}{{{F}_{seperation}}}\,=\frac{\left( \frac{V}{V-{{V}_{S}}} \right)}{\left( \frac{V}{V+{{V}_{S}}} \right)}\,=\frac{V+{{V}_{S}}}{V-{{V}_{S}}}=\frac{9}{8} \right]\]\[\Rightarrow \,9(340-{{v}_{s}})\,=8\times (340\,+{{V}_{S}})\]           \[{{V}_{S}}=\frac{340}{17}=20m{{s}^{-1}}\]


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