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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 10

  • question_answer
    The position of a particle moving along a straight line is given by \[x(t)=\frac{A}{B}(1-{{e}^{At}}),\] where B is constant and\[A>0\]. The dimensions of \[\frac{{{A}^{3}}}{B}\] is same as

    A) Linear momentum         

    B) Moment of inertia     

    C) Relative velocity            

    D) Acceleration

    Correct Answer: D

    Solution :

    [d] At is dimensionless, so \[[At]=1\Rightarrow [A]=[{{T}^{-1}}]\] Also \[[x]=\left[ \frac{A}{B} \right],\] so \[[B]=[{{L}^{-1}}{{T}^{-1}}]\] Therefore, \[\left[ \frac{{{A}^{3}}}{B} \right]=\left[ \frac{{{T}^{-3}}}{{{L}^{-1}}{{T}^{-1}}} \right]=\left[ L{{T}^{-2}} \right]\]


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