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EAMCET Medical EAMCET Medical Solved Paper-2006

  • question_answer
    The kinetic energy of a body rotating at 300 revolutions per minute is 62.8 J. Its angular momentum  \[(\text{in kg}\,{{m}^{2}}{{s}^{-1}})\]is approximately:

    A)  1                                            

    B)  2

    C)  4                                            

    D)  8

    Correct Answer: C

    Solution :

                     Kinetic energy o rotating body \[K=\frac{1}{2}I{{\omega }^{2}}=62.8\,J\]                 Angular momentum,                                 \[L=I\omega \]                                 \[=\left( \frac{1}{2}I{{\omega }^{2}} \right)\times \frac{2}{\omega }\]                                 \[=\frac{62.8\times 2}{(2\times \pi \times 300/60)}\] \[=4\,kg\,{{m}^{2}}{{s}^{-1}}\]


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