Manipal Medical Manipal Medical Solved Paper-2014

  • question_answer
    A particle of mass m moves in a circular path radius r under the action of a force\[\frac{m\,{{\upsilon }^{2}}}{r}\]. The work done during its motion over half of the circumference of the circular path will be

    A) \[\left( \frac{m{{v}^{2}}}{r} \right)\times 2\pi r\]

    B) \[\left( \frac{m{{v}^{2}}}{r} \right)\times \pi r\]

    C) \[\left( \frac{2\pi r}{\frac{m{{v}^{2}}}{r}} \right)\]                               

    D) zero

    Correct Answer: D

    Solution :

    Work done by centripetal force is always zero.


You need to login to perform this action.
You will be redirected in 3 sec spinner