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

  • question_answer
    A brass rod of cross-sectional area \[1\,c{{m}^{2}}\] and length 0.2 m is compressed lengthwise by a weight of 5 kg. If Young's modulus of elasticity of brass is \[1\times {{10}^{11}}\,N/{{m}^{2}}\] and \[g=10\,m/{{\sec }^{2}}\], then increase in the energy of the rod will be                                                            [MP PMT 1991]

    A)                 \[{{10}^{-5}}\]J

    B)                 \[2.5\times {{10}^{-5}}\]J

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

    D)                             \[2.5\times {{10}^{-4}}\]J

    Correct Answer: B

    Solution :

    \[U=\frac{1}{2}\times \frac{{{\text{(stress)}}^{\text{2}}}}{Y}\times \text{volume}\]= \[\frac{1}{2}\times \frac{{{F}^{2}}\times A\times L}{{{A}^{2}}\times Y}\] =\[\frac{1}{2}\times \frac{{{F}^{2}}L}{AY}=\frac{1}{2}\times \frac{{{(50)}^{2}}\times 0.2}{1\times {{10}^{-4}}\times 1\times {{10}^{11}}}\]= \[2.5\times {{10}^{-5}}J\]


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