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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 34

  • question_answer
    A particle of mass 'm' moves on the x-axis under the influence of a force of attraction towards the origin given by \[\vec{F}=\frac{-k}{{{x}^{2}}}\hat{i}.\] If the particle starts from rest at \[x=a,\]the speed it will attain on reaching the point \[x=\frac{a}{2}\]will be

    A) \[\sqrt{\frac{2k}{ma}}\]        

    B)        \[\sqrt{\frac{k}{ma}}\]          

    C) \[\sqrt{\frac{4k}{ma}}\]              

    D)        \[\sqrt{\frac{k}{2ma}}\]

    Correct Answer: A

    Solution :

    [a] \[F=-\frac{k}{{{x}^{2}}}\] \[v.\frac{dv}{dx}=\frac{-k}{m{{x}^{2}}}\] \[\int\limits_{0}^{v}{v.dv}=\int\limits_{a}^{-a/2}{\frac{k}{m{{x}^{2}}}}\,\,dx\] \[v=\sqrt{\frac{2k}{ma}}\]     


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