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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    A mass of 6.5kg in hanging from the end of a 60 cm long steel wire \[(Y=2\times {{10}^{11}}Pa)\] with area of corss-section 0.05 cm2. When it is revolving in a vertical circle it has an angular velocity of 2 revolutions per second, at the bottom of the circle. Approximate elongation of the wire (in metres) when the mass is at its lowest point of the trajectory is:

    A) \[8\times {{10}^{-4}}\]                                  

    B)  \[4\times {{10}^{-4}}\]

    C) \[8\times {{10}^{-5}}\]                                  

    D)  \[4\times {{10}^{-5}}\]

    Correct Answer: D

    Solution :

                     Forces acting on the mass are  the tension T and  the weight W. At the lowest point, \[T-W=\frac{m{{v}^{2}}}{r}\]                 or            \[T=W+mr{{\omega }^{2}}\]                       \[(\because v=r\omega )\] \[=6.5\times 9.8+6.5\times 0.60\times {{(2)}^{2}}\] \[=63.7+15.6\] \[=79.3\,N\]                 We have, \[Y=\frac{TL}{Al}\] \[\Rightarrow \]               \[l=\frac{TL}{Ay}\] \[=\frac{79.3\times 0.60}{0.05\times {{10}^{-4}}\times 2\times {{10}^{11}}}\] \[=475.8\times {{10}^{-7}}m\] \[4.75\times {{10}^{-5}}\,m\] \[\approx 4\times {{10}^{-5}}\,m\]


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