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JCECE Engineering JCECE Engineering Solved Paper-2015

  • question_answer
    Two tuning forks \[P\] and \[Q\] sounded together and 6 beats per second are heard. \[P\] is in unison with a \[30\,\,cm\] air column open at both ends and \[Q\] is in resonance when length of air column is increased by\[2\,\,cm\]. The frequencies of forks \[P\] and \[Q\] are

    A)  \[90\,\,Hz\]and\[84\,\,Hz\]

    B) \[100\,\,Hz\]and\[106\,\,Hz\]

    C) \[96\,\,Hz\] and\[90\,\,Hz\]

    D) \[206\,\,Hz\] and\[200\,\,Hz\]

    Correct Answer: C

    Solution :

    We know that\[f=\frac{v}{2l}\] (Fundamental frequency of an open organ pipe). Number of beats heard\[=6\] \[\Rightarrow \]               \[{{f}_{P}}-{{f}_{Q}}=6\] or            \[\frac{v}{2(0.3)}-\frac{v}{2(0.32)}=6\] or            \[v=\frac{2\times 0.3\times 0.32\times 6}{0.32-0.3}\] Now,     \[{{f}_{p}}=\frac{v}{2(0.3)}=\frac{0.32\times 6}{0.02}=96\,\,Hz\] and        \[{{f}_{Q}}=\frac{v}{2(0.32)}=\frac{0.3\times 6}{0.02}=90\,\,Hz\]


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