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BVP Medical BVP Medical Solved Paper-2000

  • question_answer
    Two whistles A and B produces notes of frequencies 660 Hz and 594 Hz respectively. There is listener at the mid point of the line joining them. Now the whistle B and the listener start moving with speed 30 m/s towards the whistle A. If speed of sound be 330 m/s, how many beats will be heard by the listener :

    A)  4

    B)  5            

    C)  6                            

    D)  7

    Correct Answer: C

    Solution :

    As B and listener are moving with the same velocity, frequency of the note from B remains unchanged. Frequency of the note from A, will be \[n=\left( \frac{\upsilon -{{\upsilon }_{1}}}{\upsilon -{{\upsilon }_{s}}} \right)n=\left( \frac{330-30}{330-0} \right)660=600Hz\] Hence, number of beats heard by the listener is \[600-594=6\]


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