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Punjab Medical Punjab - MET Solved Paper-2001

  • question_answer
    A bob of mass \[10\,\,kg\] is attached to wire \[0.3\] long. Its breaking stress is\[4.8\times {{10}^{7}}N{{m}^{2}}\]. The area of cross section of the wire is\[{{10}^{-6}}{{m}^{2}}\]. The maximum angular velocity with which it can be rotated in a horizontal circle

    A) \[8\,\,rad/\sec \]                            

    B) \[4\,\,rad/\sec \]

    C) \[2\,\,rad/\sec \]                            

    D) \[1\,\,rad/\sec \]

    Correct Answer: B

    Solution :

    Using the relation Force = breaking stress\[\times \]area                 \[=(4.8\times {{10}^{7}})({{10}^{-6}})=48\,\,N\] ... (1) Using the relation                 \[F=Mr{{\omega }^{2}}\]                                              ... (2)                 \[Mr{{\omega }^{2}}=48\]                 \[10\times 0.3\times {{\omega }^{2}}=48\]                 \[{{\omega }^{2}}=\frac{48}{10\times 0.3}=16\]      or           \[{{\omega }^{2}}=16\]                 \[\omega =4\,\,rad/\sec \]


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