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BVP Medical BVP Medical Solved Paper-2006

  • question_answer
    A cylindrical resonance tube open at both ends, has a fundamental frequency\[f\], in air. If half of the length is dipped vertically in water, the fundamental frequency of the air column will be:

    A) \[2f\]                                    

    B) \[3f/2\]

    C) \[f\]                                      

    D) \[f\]/2

    Correct Answer: C

    Solution :

                    Key Idea: When open end pipe is dipped in water vertically it becomes close end pipe. Fundamental frequency of open pipe,\[f=\frac{v}{2l}\] ???(i) When half length of tube is dipped vertically in water, then length of the air column becomes half \[\left( l=\frac{l}{2} \right)\] and the pipe becomes closed. So, new fundamental frequency of closed pipe                 \[f=\frac{v}{4l}=\frac{v}{4\left( \frac{l}{2} \right)}=\frac{v}{2l}\]         ??(ii) From Eqs. (i) and (ii), we get,                 \[f=f\] Hence, there will be no change in frequency.


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