JEE Main & Advanced Physics NLM, Friction, Circular Motion Question Bank Second Law of Motion

  • question_answer
    If rope of lift breaks suddenly, the tension exerted by the surface of lift                                     [AFMC 2002] (a = acceleration of lift)

    A)             mg      

    B)             \[m\,(g+a)\]

    C)             \[m(g-a)\]

    D)               \[0\]

    Correct Answer: D

    Solution :

                    If rope of lift breaks suddenly, acceleration becomes equal to g so that tension, \[T=m(g-g)=0\]


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