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

  • question_answer
    The force required to stretch a steel wire of \[1\,c{{m}^{2}}\] cross-section to 1.1 times its length would be         \[(Y=2\times {{10}^{11}}\,N{{m}^{-2}})\]                                       [MP PET 1992]

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

    B)                 \[2\times {{10}^{3}}\,N\]

    C)                 \[2\times {{10}^{-6}}N\]  

    D)                 \[2\times {{10}^{-7}}\,N\]

    Correct Answer: A

    Solution :

        \[F=A\times Y\times \text{strain}\]=\[1\times {{10}^{-4}}\times 2\times {{10}^{11}}\times 0.1=2\times {{10}^{6}}N\]


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