JEE Main & Advanced Physics NLM, Friction, Circular Motion JEE PYQ - NLM Friction Circular Motion

  • question_answer
    Two masses \[{{m}_{1}}=5kg\] and \[{{m}_{2}}=10kg\], connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of \[{{m}_{2}}\] to stop the motion is : [JEE Main Online 08-04-2018]

    A) \[\text{43}\text{.3 kg}\]

    B) \[\text{10}\text{.3 kg}\]

    C) \[\text{18}\text{.3 kg}\]

    D) \[\text{27}\text{.3 kg}\]

    Correct Answer: D

    Solution :

    [d] \[{{m}_{1}}g=\mu ({{m}_{2}}+m)g\]
    \[\therefore \text{ }m\text{ }=\text{ }23.3\text{ }kg\]
    Value greater than this and minimum in the given options is 27.3 kg

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