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JCECE Medical JCECE Medical Solved Paper-2011

  • question_answer
    Two sources A and B are sounding notes of frequency 680 Hz. A listener moves from A to B with a constant velocity \[\upsilon .\]If the speed of sound 340 m/s, what must be the value of \[\upsilon .\]so that he hears 10 beats per second?

    A)  2.0 m/s          

    B)  2.5 m/s

    C)  3.0 m/s          

    D)  3.5 m/s

    Correct Answer: B

    Solution :

     Apparent frequency due to source A \[n=\frac{v-{{v}_{0}}}{v}\] \[=\frac{v-4}{v}\times n\] Apparent frequency due to source B \[n=\frac{v+u}{v}\] \[=\frac{v-u}{v}\times n\] \[\therefore \] \[n-n=\frac{2u}{v}\times n={{I}_{0}}\] \[u=\frac{10u}{2n}\] \[=\frac{10\times 340}{2\times 680}\] \[=2.5\,m/s\]


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