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

  • question_answer
    A block of mass m is placed on a surface with a vertical cross section given by \[y=\frac{{{x}^{3}}}{6}.\]If the coefficient of friction is 0.5, the maximum height above the ground at which the block can be placed without slipping is: [JEE MAIN 2014]

    A) \[\frac{1}{3}m\]

    B) \[\frac{1}{2}m\]

    C) \[\frac{1}{6}m\]

    D) \[\frac{2}{3}m\]

    Correct Answer: C

    Solution :

    \[mg\sin \theta ={{\mu }_{s}}mg\cos \theta \]
    [when partied is just balanced]
    \[\Rightarrow \]\[\tan \theta ={{\mu }_{s}}\]
    \[\tan \theta =\frac{dy}{dx}=\left( \frac{{{x}^{2}}}{2} \right)\]
    \[\therefore \]\[\frac{{{x}^{2}}}{2}=0.5\Rightarrow x=1\]\[\Rightarrow \]\[y=\frac{1}{6}\]

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