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Punjab Medical Punjab - MET Solved Paper-2008

  • question_answer
    A uniform wire of linear density \[0.004kg\text{-}{{m}^{-1}}\] when stretched between two rigid supports with a tension \[3.6\times {{10}^{2}}N\] resonates with a frequency of\[420Hz\]. The next harmonic frequency with which the wire resonates is\[490Hz\]. The length of the wire in metres is

    A) \[1.41\]                               

    B) \[2.14\]

    C) \[2.41\]                               

    D) \[3.14\]

    Correct Answer: B

    Solution :

    \[n=\frac{1}{2l}\sqrt{\frac{T}{M}}\] or            \[490-420=\frac{1}{2\times l}\sqrt{\frac{3.6\times {{10}^{2}}}{0.004}}\] or                        \[70=\frac{1}{2l}\times 300\] or                            \[l=\frac{300}{2\times 70}=2.14m\]


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