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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

  • question_answer
    If one end of a wire is fixed with a rigid support and the other end is stretched by a force of 10 N, then the increase in length is 0.5 mm. The ratio of the energy of the wire and the work done in displacing it through 1.5 mm by the weight is

    A)  \[\frac{1}{3}\]                                  

    B)  \[\frac{1}{4}\]

    C)  \[\frac{1}{2}\]                  

    D)         1

    Correct Answer: C

    Solution :

    Work done in stretching a wire                 \[W=\frac{1}{2}Fl=\frac{1}{2}\times 10\times 0.5\times {{10}^{-3}}\]                 \[=2.5\times {{10}^{-3}}J\] Work done to displace it through 1.5 mm                 \[W=F\times l=5\times {{10}^{-3}}J\] The ratio of above two works\[=1:2\]


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