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NEET NEET SOLVED PAPER 2015 (C)

  • question_answer
    A particle of unit mass undergoes one-dimensional motion such that its velocity varies according to \[v(x)=\beta {{x}^{-2n}}\] where, \[\beta \] and n are constants and \[x\] is the position of the particle. The acceleration of the particle as a function of x, is given by                                                                                                           

    A)  \[-2n{{\beta }^{2}}\,{{x}^{-2n-1}}\]        

    B)  \[-2n{{\beta }^{2}}\,{{x}^{-4n-1}}\]

    C)  \[-2{{\beta }^{2}}\,\,{{x}^{-2n+1}}\]       

    D)         \[-2n{{\beta }^{2}}\,\,\,{{e}^{-4n+1}}\]

    Correct Answer: B

    Solution :

                    Given, \[v=\beta {{x}^{-2n}}\] \[a=\frac{dv}{dt}=\frac{dx}{dt}.\frac{dv}{dx}\] \[\Rightarrow \,\,a=v\frac{dv}{dx}=(\beta {{x}^{-2n}})(-2n\beta {{x}^{-2n-1}})\] \[\Rightarrow \,\,a=-2n{{\beta }^{2}}{{x}^{-4n}}^{-1}\]


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