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JEE Main & Advanced Physics Elasticity Question Bank Work Done in Stretching a Wire

  • question_answer
    The length of a rod is 20 cm and area of cross-section \[2\,c{{m}^{2}}\]. The Young's modulus of the material of wire is \[1.4\times {{10}^{11}}\,N/{{m}^{2}}\]. If the rod is compressed by 5 kg-wt along its length, then increase in the energy of the rod in joules will be

    A)                 \[8.57\times {{10}^{-6}}\]

    B)                             \[22.5\times {{10}^{-4}}\]

    C)                 \[9.8\times {{10}^{-5}}\]

    D)                             \[45.0\times {{10}^{-5}}\]

    Correct Answer: A

    Solution :

    Energy = \[\frac{1}{2}Fl=\frac{1}{2}\times F\times \left( \frac{FL}{AY} \right)=\frac{1}{2}\times \frac{{{F}^{2}}L}{AY}\]             \[=\frac{1}{2}\times \frac{{{(50)}^{2}}\times 20\times {{10}^{-2}}}{2\times {{10}^{-4}}\times 1.4\times {{10}^{11}}}\]\[=8.57\times {{10}^{-6}}J\]


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