JEE Main & Advanced Sample Paper JEE Main - Mock Test - 21

  • question_answer
    A ring of mass M and radius a lies on a smooth horizontal surface. An insect of mass m sitting on it starts crawling on the ring with a constant speed. The trajectory of the centre of the ring is

    A) a circle of radius \[\left( \frac{Ma}{m} \right)\]                                

    B) a circle of radius \[\left( \frac{ma}{M+m} \right)\]

    C) a circle of radius \[\left( \frac{ma}{M} \right)\]

    D) a straight line

    Correct Answer: B

    Solution :

    [b] The initial position of the centre of mass of the system (ring + insect) is supposed at a distance x from the centre of the ring. Then \[x=\frac{ma}{M+m}.\] Now, net external force on the system is zero. Hence, the centre of mass will remain stationary. Hence, the centre of the ring will move in a circle of radius \[-\frac{ma}{M+m}\] and the insect will move in a circle of radius \[\frac{Ma}{M+m}\]


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