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Chhattisgarh PMT Chhattisgarh PMT Solved Paper-2004

  • question_answer
    A wire of length L and cross-sectional area A is made of a material of Young's modulus Y. It is stretched by an amount x. The work done is

    A)  \[YxA/2L\]                         

    B)  \[Y{{x}^{2}}A/L\]

    C) \[Y{{x}^{2}}A/2L\]                           

    D)  \[2Y{{x}^{2}}A/L\]

    Correct Answer: C

    Solution :

    Work done in stretching a wire \[w=\frac{1}{2}Fx\] So, Young's modulus                 \[Y=\frac{F/A}{x/L}=\frac{FL}{Ax}\] or            \[F=\frac{YAx}{L}\] So, work done \[W=\frac{1}{2}\frac{YAx}{L}\times x\]                                 \[=\frac{YA{{x}^{2}}}{2L}\]


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