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

  • question_answer
    The coefficient of linear expansion of brass and steel are \[{{\alpha }_{1}}\] and \[{{\alpha }_{2}}\]. If we take a brass rod of length \[{{l}_{1}}\] and steel rod of length \[{{l}_{2}}\] at 0°C, their difference in length \[({{l}_{2}}-{{l}_{1}})\] will remain the same at a temperature if        [EAMCET (Med.) 1995]

    A)                 \[{{\alpha }_{1}}{{l}_{2}}={{\alpha }_{2}}{{l}_{1}}\]

    B)                             \[{{\alpha }_{1}}l_{2}^{2}={{\alpha }_{2}}l_{1}^{2}\]

    C)                 \[\alpha _{1}^{2}{{l}_{1}}=\alpha _{2}^{2}{{l}_{2}}\]

    D)                             \[{{\alpha }_{1}}{{l}_{1}}={{\alpha }_{2}}{{l}_{2}}\]

    Correct Answer: D

    Solution :

        \[{{L}_{2}}={{l}_{2}}(1+{{\alpha }_{2}}\Delta \theta )\] and \[{{L}_{1}}={{l}_{1}}(1+{{\alpha }_{1}}\Delta \theta )\]             \[\Rightarrow ({{L}_{2}}-{{L}_{1}})=({{l}_{2}}-{{l}_{1}})+\Delta \theta ({{l}_{2}}{{\alpha }_{2}}-{{l}_{1}}{{\alpha }_{1}})\] Now \[({{L}_{2}}-{{L}_{1}})=({{l}_{2}}-{{l}_{1}})\] so, \[{{l}_{2}}{{\alpha }_{2}}-{{l}_{1}}{{\alpha }_{1}}=0\]


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