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BVP Medical BVP Medical Solved Paper-2003

  • 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)  \[36\pi \times {{10}^{5}}N\]                      

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

    Correct Answer: A

    Solution :

                    Using the realtion \[F=\frac{YAl}{L}\]                           ??(i) Here:  \[Y=9\times {{10}^{10}}N/{{m}^{2}},\,\,\,d=4\times {{10}^{-3}}m,\] \[l=0.1%L\] 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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