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JCECE Engineering JCECE Engineering Solved Paper-2004

  • question_answer
    A stretched string is \[1\,\,m\] long. It?s mass per unit length is \[0.5\,\,g/m\]. It is stretched with a force of \[20\,\,N\]. It is plucked at a distance of \[25\,\,cm\] from one end. The frequency of note emitted by it will be:

    A) \[400\,\,Hz\]                     

    B) \[300\,\,Hz\]

    C) \[200\,\,Hz\]                     

    D) \[100\,\,Hz\]

    Correct Answer: C

    Solution :

    As stretched string of 1m long is plucked at a distance of \[25\,\,cm\] from end, so there will be \[2\] loops. Frequency,         \[n=\frac{p}{2l}\sqrt{\frac{T}{m}}\] Given,   \[p=2,\,\,l=1\,\,m,\,\,T=20N,\]                                                 \[m=0.5\times {{10}^{-3}}kg/m\] \[\therefore \]  \[n=\frac{2}{2\times 1}\sqrt{\frac{20}{0.5\times {{10}^{-3}}}}\] \[\Rightarrow \]               \[n=200\,\,Hz\]


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