RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    A person is observing of two trains each of velocity is 4 m/s. A train is coming towards an observer frequency of each whistle is 240 Hz then the beats heard by an observer will be : (\[\upsilon \] = 320 m/s)  

    A)  zero                     

    B)         3    

    C)  6                            

    D)         5

    Correct Answer: C

    Solution :

    Number of beats\[\begin{align}   & =n\left\{ \frac{\upsilon }{\upsilon -{{\upsilon }_{s}}}-\frac{\upsilon }{\upsilon +{{\upsilon }_{s}}} \right\} \\  & =240\left\{ \frac{320}{320-4}-\frac{320}{320-4} \right\} \\  & =240\times 320\left\{ \frac{1}{316}-\frac{1}{324} \right\} \\  & =\frac{240\times 320\times 8}{316\times 324}=6 \\ \end{align}\]


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