Punjab Medical Punjab - MET Solved Paper-2003

  • question_answer
    The spring extends by \[x\] on loading, then energy stored by the spring is: (if \[T\] is the tension in spring and \[k\] is spring constant)

    A) \[\frac{{{T}^{2}}}{2k}\]                                 

    B) \[\frac{{{T}^{2}}}{2{{k}^{2}}}\]

    C) \[\frac{2k}{{{T}^{2}}}\]                                 

    D) \[\frac{2{{T}^{2}}}{k}\]

    Correct Answer: A

    Solution :

    The energy stored in the spring is                 \[E=\frac{1}{2}k{{x}^{2}}\]                                           ... (1) and expansion in the spring is                 \[x=\frac{T}{k}\]                                                               ... (2) From equations (1) and (2), we get                 \[E=\frac{1}{2}k\frac{{{T}^{2}}}{{{k}^{2}}}\]                    \[=\frac{1}{2}\frac{{{T}^{2}}}{k}=\frac{{{T}^{2}}}{2k}\]


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