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MGIMS WARDHA MGIMS WARDHA Solved Paper-2008

  • question_answer
    A bob of mass 10 kg is attached to wire 0.3 m long. Its breaking stress is \[4.8\times {{10}^{7}}N/{{m}^{2}}.\]The area of cross-section of the wire is 10-6 m2. The maximum angular velocity with which it can be rotated in a horizontal circle is

    A) 8rad/s     

    B) 4rad/s

    C) 2 rad/s                                 

    D) 1 rad/s

    Correct Answer: B

    Solution :

    Centripetal force = breaking force \[\Rightarrow \]\[m{{\omega }^{2}}r=\]breaking stress x cross-sectional area \[\Rightarrow \]               \[m{{\omega }^{2}}r=P\times A\] \[\Rightarrow \]               \[\omega =\sqrt{\frac{P\times A}{mr}}\]                 \[=\sqrt{\frac{4.8\times {{10}^{7}}\times {{10}^{-6}}}{10\times 0.3}}\]                 \[\omega =4\,rad/s\]


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