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CEE Kerala Engineering CEE Kerala Engineering Solved Paper-2004

  • question_answer
    A pipe open at both the ends produces a note of fundamental frequency\[{{f}_{1}}\]. When the pipe is kept with\[\frac{3}{4}\]th of its length in water, it produces a note of fundamental frequency\[{{f}_{2}}\]. The ratio of \[\frac{{{f}_{1}}}{{{f}_{2}}}\]is:

    A)  \[\frac{4}{3}\]                                  

    B)  \[\frac{3}{4}\]

    C)  \[2\]                    

    D)         \[\frac{1}{2}\]

    E)  \[\frac{3}{2}\]

    Correct Answer: D

    Solution :

    Accordingly, \[{{f}_{1}}=\frac{v}{2l}\]                                               ?. (1) and        \[{{f}_{2}}=\frac{v}{4l/4}=\frac{v}{l}\]                     ?.. (2) Hence,     \[\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{1}{2}\]


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