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JEE Main & Advanced AIEEE Solved Paper-2006

  • question_answer
    The potential energy of a 1 kg particle free to move along the x-axis is given by \[V(x)=\left( \frac{{{x}^{4}}}{4}-\frac{{{x}^{2}}}{2} \right)J\] The total mechanical energy of the particle is 2 J. Then, the maximum speed (in m/s) is     AIEEE  Solved  Paper-2006

    A) \[3/\sqrt{2}\]    

    B)        \[\sqrt{2}\]        

    C)        \[1/\sqrt{2}\]   

    D)        2

    Correct Answer: A

    Solution :

    The variation of potential energy is given as \[V(x)=\left( \frac{{{x}^{4}}}{4}-\frac{{{x}^{2}}}{2} \right)\]         ??..(i) For minimum value of V, \[\frac{dv}{dx}=0\] On differentiating Eq. (i), we get \[\frac{4{{x}^{3}}}{4}-\frac{2x}{2}=0\] \[\Rightarrow \] \[x=0,\,x=\pm 1\] So,     \[{{V}_{\min }}(x=\pm 1)=\frac{1}{4}-\frac{1}{2}=\frac{-1}{4}J\] \[\therefore \]\[{{K}_{\max }}+{{V}_{\min }}=\]Total mechanical energy (from energy conservation) \[{{K}_{\max }}=(1/4)+2\Rightarrow {{K}_{\max }}=9/4\] \[\Rightarrow \]\[\frac{m{{v}^{2}}}{2}=\frac{9}{4}\] \[\Rightarrow \]\[v=\frac{3}{\sqrt{2}}m/s\]                         \[(\because m=1\,kg)\]


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