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Manipal Engineering Manipal Engineering Solved Paper-2009

  • question_answer
     A cylindrical tube open at both the ends has a fundamental frequency of 390 Hz in air. If \[\frac{1}{4}\]th of the tube is immersed vertically in water the fundamental frequency of air column is

    A)  260 Hz                                 

    B)  130 Hz         

    C)  390 Hz                 

    D)         520 Hz        

    Correct Answer: A

    Solution :

    Fundamental frequency of cylindrical open tube                 \[n=\frac{v}{2L}=390\,\,Hz\] When it is immersed in water it becomes a closed tube of length\[\frac{3}{4}th\]of the initial length. Therefore, its fundamental frequency is                 \[n=\frac{v}{4\left( \frac{3}{4}L \right)}=\frac{v}{3L}=\frac{2}{3}\left( \frac{v}{2L} \right)\]                 \[=\frac{2}{3}\,\times 390\,Hz=260\,Hz\]


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