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JEE Main & Advanced JEE Main Paper (Held On 8 April 2017)

  • question_answer
    A uniform disc of radius R and mass M is free to rotate only about its axis. A string is wrapped over its rim and a body of mass m is tied to the free end of the string as shown in the figure. The body is released from rest. Then the acceleration of the body is -                                 [JEE Online 08-04-2017]

    A) \[\frac{2Mg}{2m+M}\]                 

    B) \[\frac{2Mg}{2M+m}\]

    C) \[\frac{2mg}{2M+m}\]                 

    D) \[\frac{2mg}{2m+M}\]

    Correct Answer: D

    Solution :

    mg - T = ma \[RT=I\propto \] \[RT=\frac{M{{R}^{2}}}{2}.\frac{a}{R}\] \[T=\frac{Ma}{2}\] \[mg-\frac{Ma}{2}=ma\] \[mg=a\left( \frac{M}{2}+m \right)\] \[mg=a\left( \frac{M+2m}{2} \right)\] \[a=\frac{2mg}{M+2m}\]


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