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JEE Main & Advanced Physics Elasticity Question Bank Youngs Modulus and Breaking Stress

  • question_answer
    A rod is fixed between two points at 20°C. The coefficient of linear expansion of material of rod is \[1.1\times {{10}^{-5}}/{}^\circ C\] and Young's modulus is \[1.2\times {{10}^{11}}\,N/m\]. Find the stress developed in the rod  if temperature of rod becomes 10°C [RPET 1997]

    A)                 \[1.32\times {{10}^{7}}\,N/{{m}^{2}}\]

    B)                             \[1.10\times {{10}^{15}}\,N/{{m}^{2}}\]

    C)                 \[1.32\times {{10}^{8}}\,N/{{m}^{2}}\]

    D)                             \[1.10\times {{10}^{6}}\,N/{{m}^{2}}\]

    Correct Answer: A

    Solution :

        Thermal stress =\[Y\alpha \Delta \theta \]              \[=1.2\times {{10}^{11}}\times 1.1\times {{10}^{-5}}\times (20-10)\]\[=1.32\times {{10}^{7}}\ N/{{m}^{2}}\]


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