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WB JEE Medical WB JEE Medical Solved Paper-2015

  • question_answer
    The length of an open organ pipe is twice the length of another closed organ pipe. The fundamental frequency of the open pipe is 100 Hz. The frequency of the third harmonic of the closed pipe is

    A)  100 Hz

    B)  200 Hz

    C)  300 Hz

    D)  150 Hz

    Correct Answer: C

    Solution :

     Let the length of closed organ pipe be 1. Then, \[{{v}_{n}}=(2n-1)\frac{v}{4l}\] Its third harmonics (put n = 2) In above question, \[{{v}_{3}}=\frac{3}{4}\frac{v}{l}\] ?(i) Now, the length of open organ pipe is \[2l,\] then \[{{v}_{n}}=n\frac{v}{2(2l)}=\frac{nv}{4l}\] Its fundamental frequency. \[{{v}_{0}}=\frac{v}{2l}=100\] ?(ii) From Eq. (i) and (ii), we get \[{{v}_{3}}=\frac{3v}{4l}=3\times 100=300\,Hz\]


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