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VMMC VMMC Medical Solved Paper-2005

  • question_answer
    The diameter of brass road is 4 mm. Youngs modulus of brass is \[9\times {{10}^{9}}\,N/{{m}^{2}}\]. The force required to stretch 0.1% of its length is:

    A) 360 \[\pi \]N

    B)                        36 N                      

    C) 36\[\pi \]x 105 N              

    D)        \[144\,\pi \times {{10}^{3}}N\]

    Correct Answer: A

    Solution :

    Using the reaction \[F=\frac{\Upsilon Al}{L}\]                                                           ?(i) Here: \[\Upsilon =9\times {{10}^{10}}N/{{m}^{2}},\,d=4\times {{10}^{-3}}m,\] \[l=0.1%\,L\] From equation (i), we get \[F=\frac{9\times {{10}^{10}}\times \pi {{(2\times {{10}^{-3}})}^{2}}\times 0.1}{100}\] \[=360\,\pi \,N\]


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