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CET Karnataka Medical CET - Karnataka Medical 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\[n=\frac{v}{2L}=390Hz\] 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 \[n=\frac{v}{4\left( \frac{3}{4}L \right)}=\frac{v}{3L}=\frac{2}{3}\left( \frac{v}{2L} \right)\] \[=\frac{2}{3}\times 390Hz=260Hz\]


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