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CET Karnataka Engineering CET - Karnataka Engineering Solved Paper-2005

  • question_answer
    The apparent frequency of a note is 200 Hz, when a listener is moving with a velocity of \[40\text{ }m{{s}^{-1}}\] 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 \[n=\frac{\upsilon +{{\upsilon }_{o}}}{\upsilon }\times n\Rightarrow \,\,\,200=\frac{\upsilon +40}{\upsilon }\times n\]                ... (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                 \[=\frac{200}{160}=\frac{\upsilon +40}{\upsilon -40}\]                 \[=5\,\upsilon -200=4\upsilon +160\] \[\therefore \]  \[\upsilon =360\,m/s\]


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