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Punjab Medical Punjab - MET Solved Paper-2010

  • question_answer
    A thin ring of mass \[2.7kg\] and radius \[8cm\] rotates about an axis through its centre and perpendicular to the plane of the ring at\[1.5rev/s\]. Calculate the kinetic energy of the ring.

    A) \[0.763\,\,J\]                     

    B) \[0.345\,\,J\]

    C) \[1.5\,\,J\]                         

    D)  zero

    Correct Answer: A

    Solution :

    For a thin ring                 \[KE=\frac{1}{2}I{{\omega }^{2}}\] Here,\[I=m{{r}^{2}}=2.7{{(0.08)}^{2}}=0.0172kg-{{m}^{2}}\]          \[\omega =1.5rev/s=3\pi \,\,rad/s\] After substituting the values, we get        \[KE=0.763\,\,J\]


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