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AMU Medical AMU Solved Paper-2013

  • question_answer
    A uniform wire of cross-sectional area A and Youngs modulus Y is stretched within the elastic limit. If S is the stress in the wire, the elastic energy density stored in the wire in terms of the given parameters is

    A)  \[\frac{S}{2Y}\]                               

    B)  \[\frac{2Y}{{{S}^{2}}}\]

    C)  \[\frac{{{S}^{2}}}{2Y}\]      

    D)  \[\frac{{{S}^{2}}}{Y}\]

    Correct Answer: C

    Solution :

                     The elastic energy density stored in wire                 = Energy stored in unit volume of wire                 \[=\frac{1}{2}\times \] stress \[\times \] strain                 \[=\frac{1}{2}\frac{{{\left( Stress \right)}^{2}}}{Y}=\frac{{{S}^{2}}}{2Y}\]               


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