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

  • question_answer
    A wire of length 50 cm and cross sectional area of 1 sq. mm is extended by 1 mm. The required work will be \[(Y=2\times {{10}^{10}}\,N{{m}^{-2}})\]                                       [RPET 1999]

    A)                 \[6\times {{10}^{-2}}\,J\]

    B)                             \[4\times {{10}^{-2}}\,J\]

    C)                 \[2\times {{10}^{-2}}\,J\]

    D)                             \[1\times {{10}^{-2}}\,J\]

    Correct Answer: C

    Solution :

    \[W=\frac{YA{{l}^{2}}}{2L}=\frac{2\times {{10}^{10}}\times {{10}^{-6}}\times {{({{10}^{-3}})}^{2}}}{2\times 50\times {{10}^{-2}}}\]\[=2\times {{10}^{-2}}\ J\]


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