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CET Karnataka Medical CET - Karnataka Medical Solved Paper-2005

  • question_answer
    The apparent frequency of a note is 200 Hz, when a listener is moving with a velocity of 40 ms-1 towards a stationary source. When he moves away from the same source with the same speed, the apparent frequency of the same note is 160 Hz. The velocity of sound in air in m/s is:

    A)  340

    B)  330

    C)  360

    D)  320

    Correct Answer: C

    Solution :

    When listener is moving towards the source then apparent frequency \[n=\frac{\upsilon +{{\upsilon }_{0}}}{\upsilon }\times n\Rightarrow 200=\frac{\upsilon +40}{\upsilon }\times n\] ?(i) where v = velocity of sound in air n = actual frequency of sound source Similarly, when listener is moving away, then     \[160=\frac{\upsilon -40}{\upsilon }\times n\] ...(ii) From Eqs. (i) and (ii), we have \[\frac{200}{160}=\frac{\upsilon +40}{\upsilon -40}\] \[5\upsilon -200=4\upsilon +160\] \[\therefore \] \[\upsilon =360m/s\]


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