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J & K CET Medical J & K - CET Medical Solved Paper-2006

  • question_answer
    The second overtone of an open pipe is in resonance with the first overtone of a dosed pipe of length 2 m. Length of the open pipe is:

    A)  4 m                                       

    B)  2 m

    C)  8 m                                       

    D)  1 m

    Correct Answer: A

    Solution :

                     Second overtone of open pipe of length I is \[{{n}_{o}}=\frac{v}{2l}\]                                              ...(i) First overtone of a closed pipe is \[{{n}_{c}}=\frac{v}{4l}=\frac{v}{4\times 2}\]                                 ?..(ii) Equating Eqs. (i) and (ii), we get                 \[\frac{v}{2l}=\frac{v}{8}\] \[\Rightarrow \]               \[l=4\,m\]


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