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

  • question_answer
    An area of cross-section of rubber string is \[2\,c{{m}^{2}}\]. Its length is doubled when stretched with a linear force of \[2\times {{10}^{5}}\]dynes. The Young's modulus of the rubber in \[dyne/c{{m}^{2}}\] will be                                      [MP PET 1985]

    A)                 \[4\times {{10}^{5}}\]

    B)                             \[1\times {{10}^{5}}\]

    C)                 \[2\times {{10}^{5}}\]      

    D)                 \[1\times {{10}^{4}}\]

    Correct Answer: B

    Solution :

        If length of the wire is doubled then strain = 1 Y = \[\text{Stress}=\frac{\text{Force}}{\text{Area}}\]= \[\frac{2\times {{10}^{5}}}{2}={{10}^{5}}\frac{dyne}{c{{m}^{2}}}\]


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