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BHU PMT BHU PMT (Mains) Solved Paper-2008

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

    A)  1.41                                      

    B)  2.14

    C)   2.41                                     

    D)  3.14

    Correct Answer: B

    Solution :

                     \[\therefore \] \[n=\frac{1}{2l}\sqrt{\frac{T}{m}}\] 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           \[l=\frac{300}{2\times 70}=2.14\,m\]


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