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WB JEE Medical WB JEE Medical Solved Paper-2006

  • question_answer
    A cube of side 40 mm has its upper face displaced by 0.1 mm by a tangential force of 8  kN. The shearing modulus of cube is :

    A)  \[2\times {{10}^{9}}\,N{{m}^{-2}}\]                       

    B)  \[4\times {{10}^{9}}\,N{{m}^{-2}}\]

    C)  \[8\times {{10}^{9}}\,N{{m}^{-2}}\]                       

    D)  \[16\times {{10}^{9}}\,N{{m}^{-2}}\]

    Correct Answer: A

    Solution :

                     Shearing modulus of cube \[\eta =\frac{Fl}{A\Delta l}=\frac{8\times {{10}^{3}}\times 40\times {{10}^{-3}}}{(40\times {{10}^{-3}})\times (0.1\times {{10}^{-3}})}\] \[=2\times {{10}^{9}}\,N{{m}^{-2}}\]


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