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MGIMS WARDHA MGIMS WARDHA Solved Paper-2010

  • question_answer
    A 4 m long copper wire of cross-sectional area 1.2 cm2 is stretched by a force of \[4.8\times {{10}^{3}}N,\]Young's    modulus    for    copper   is \[y=1.2\times {{10}^{11}}N/{{m}^{2}}\]the increase in length of the wire is

    A)  1.32mm                              

    B)  0.8mm

    C)  0.48 mm                             

    D)  5.36 mm

    Correct Answer: A

    Solution :

                     \[Stress=\frac{Force}{Area}\] \[=\frac{4.8\times {{10}^{3}}}{1.2\times {{10}^{-4}}}=4\times {{10}^{7}}N/{{m}^{2}}\] Young's modulus, \[Y=\frac{Stress}{Strain}\] \[\Rightarrow \] \[Strain=\frac{4\times {{10}^{7}}}{1.2\times {{10}^{11}}}=3.3\times {{10}^{-4}}\] \[\therefore \]Increase in length = longitudinal strain x initial length \[=(3.3\times {{10}^{-4}})\times 4\]                 \[=13.2\times {{10}^{-4}}m=1.32\,mm\]


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