Haryana PMT Haryana PMT Solved Paper-2003

  • question_answer
    A circular ring of mass \[m\] and radius \[r\] is rolling on a smooth horizontal surface with speed \[u\]. Its kinetic energy is :

    A) \[\frac{1}{8}m{{u}^{2}}\]                               

    B) \[\frac{1}{4}m{{u}^{2}}\]

    C)  \[\frac{1}{4}m{{u}^{2}}\]                            

    D)  \[m{{u}^{2}}\]

    Correct Answer: B

    Solution :

                    K.E. of rotation \[=\frac{1}{2}I{{\omega }^{2}}=\frac{1}{2}\frac{m{{r}^{2}}}{2}\times {{\omega }^{2}}\] \[=\frac{1}{4}m{{r}^{2}}\times \frac{{{\upsilon }^{2}}}{{{r}^{2}}}=\frac{1}{4}m{{\upsilon }^{2}}\]


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