Uttarakhand PMT Uttarakhand PMT Solved Paper-2009

  • question_answer
    A particle of mass m moves in a circular path radius r under the action of a force \[\frac{m{{v}_{2}}}{r}.\] The work done during its morion 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)  \[\frac{(2\pi r)}{\left( \frac{m{{v}^{2}}}{r} \right)}\]

    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