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

  • question_answer
    When a force is applied on a wire of uniform cross-sectional area \[3\times {{10}^{-6}}\,{{m}^{2}}\]and length 4m, the increase in length is 1 mm. Energy stored in it will be \[(Y=2\times {{10}^{11}}\,N/{{m}^{2}})\] [MP PET 1995; Pb. PET 2002]

    A) 6250 J                   

    B)                 0.177 J

    C)                 0.075 J

    D)                             0.150 J

    Correct Answer: C

    Solution :

    \[U=\frac{1}{2}\times \frac{YA{{l}^{2}}}{L}\]\[=\frac{1}{2}\times \frac{2\times {{10}^{11}}\times 3\times {{10}^{-6}}\times {{(1\times {{10}^{-3}})}^{2}}}{4}\]             \[=0.075\ J\]


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