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JEE Main & Advanced Sample Paper JEE Main Sample Paper-18

  • question_answer
    A small ball rests at the bottom of a watch glass of radius\[R\]. It is displaced through a small distance \[x\] from this position and released, then the total distance covered before it comes to the bottom and rests there is [Coefficient of friction between watch glass surface and the ball is\[\mu ,\]\[x<<R\]] is

    A)  \[\frac{{{x}^{2}}}{2\mu R}\]                      

    B)  \[\frac{{{x}^{2}}}{3\mu R}\]

    C)  \[\frac{{{x}^{2}}}{\mu R}\]                         

    D)  None of these

    Correct Answer: A

    Solution :

     Let it travel distance s. \[mgh=\int{\mu g\cos \theta .\,ds\approx \mu mgs}\] \[mgh=mg\left[ R-\sqrt{{{R}^{2}}-{{x}^{2}}} \right]=mg\frac{{{x}^{2}}}{2R}\] \[\therefore \]  \[s=\frac{{{x}^{2}}}{2\mu R}\]


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