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JEE Main & Advanced AIEEE Solved Paper-2003

  • question_answer
    A metal wire of linear mass density of \[9.8\] g/m is stretched with a tension of 10 kg-wt between two rigid supports 1 m apart. The wire passes at its middle point between the poles of a permanent magnet and it vibrates in resonance when carrying an alternating current of frequency n. The frequency n of the alternating source is     AIEEE  Solved  Paper-2003

    A) 50 Hz    

    B)       100 Hz                  

    C) 200 Hz                  

    D) 25 Hz

    Correct Answer: A

    Solution :

    The wire will vibrate with the same frequency as that of source. This can be considered as an example of forced vibration.                     \[T=10\times 9.8\,N=98\,\,N\],                     \[m=9.8\times {{10}^{-3}}kg/m\] Frequency of wire, \[f=\frac{1}{2L}\sqrt{\left( \frac{T}{m} \right)}\]                                     (where, m = mass/unit length)                                     \[=\frac{1}{2\times 1}\sqrt{\left( \frac{98}{9.8\times {{10}^{-3}}} \right)}\]                                     = 50 Hz


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