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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2010

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

    A)  \[360\,\pi N\]                                  

    B)  36 N

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

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

    Correct Answer: A

    Solution :

    The force required to stretch                 \[F=\frac{\gamma Al}{L}\]                 \[=\frac{9\times {{10}^{10}}\times \pi \times 4\times {{10}^{-6}}\times 0.1}{100}\]                 \[=360\pi N\]


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