RAJASTHAN PMT Rajasthan - PMT Solved Paper-2011

  • question_answer
    A uniform thin ring of mass 0.4 kg rolls without slipping on a horizontal surface with a linear velocity of 10 cm/s. The kinetic energy of the ring is

    A)  \[4\times {{10}^{-3}}\]                                 

    B)  \[4\times {{10}^{-2}}J\]

    C)  \[2\times {{10}^{-3}}J\]               

    D)         \[2\times {{10}^{-2}}J\]

    Correct Answer: A

    Solution :

    Given,\[m=0.4\,\,kg\]and\[v=10\,\,cm/s\] Kinetic energy\[=\frac{1}{2}m{{v}^{2}}(1+1)\] Kinetic energy\[=m{{v}^{2}}\]                            \[=0.4\times {{(0.1)}^{2}}\]                            \[=4\times {{10}^{-3}}J\]


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