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WB JEE Medical WB JEE Medical Solved Paper-2012

  • question_answer
    The frequency of the first overtone of a closed pipe of length \[{{l}_{1}}\]is equal to that of the first overtone of an open pipe of length \[{{l}_{2}}.\]The  ratio of their lengths \[({{l}_{1}}:{{l}_{2}})\] is

    A)  2: 3             

    B)  4:5

    C)  3:5              

    D)  3:4

    Correct Answer: D

    Solution :

     Here, \[{{v}_{1}}=\frac{(2n-1)}{4{{l}_{1}}}v\] \[{{v}_{2}}=\frac{n}{2{{l}_{2}}}v\] Hence, \[{{v}_{1}}={{v}_{2}}\] \[\frac{{{l}_{1}}}{{{l}_{2}}}=\frac{3}{4}\]


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