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

  • question_answer
    On increasing the length by 0.5 mm in a steel wire of length 2 m and area of cross-section \[2\,m{{m}^{2}}\], the force required is [Y for steel\[=2.2\times {{10}^{11}}\,N/{{m}^{2}}]\]]                        [MP PET/PMT 1988]

    A)                 \[1.1\times {{10}^{5}}\,N\]

    B)                             \[1.1\times {{10}^{4}}\,N\]

    C)                 \[1.1\times {{10}^{3}}\,N\]

    D)                             \[1.1\times {{10}^{2}}\,N\]

    Correct Answer: D

    Solution :

         \[F=\frac{YAl}{L}=\frac{2.2\times {{10}^{11}}\times 2\times {{10}^{-6}}\times 5\,\times {{10}^{-4}}}{2}\]\[=1.1\times {{10}^{2}}N\]   


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