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JEE Main & Advanced Physics Mathematical Tools, Units & Dimensions Question Bank Mock Test - Units Dimenstions and Measurment

  • question_answer
    The position of a particle at time \[t\]is given by the relation \[x(t)=(\frac{v_{0}^{{}}}{\alpha })(1-c_{{}}^{-\alpha t})\]where \[v_{0}^{{}}\] ls a constant and a > 0. The dimensions of \[v_{0}^{{}}\] and \[\alpha \] are respectively

    A) \[M_{{}}^{0}L_{{}}^{1}T_{{}}^{-1}and\,T_{{}}^{-1}\]    

    B) \[M_{{}}^{0}L_{{}}^{1}T_{{}}^{0}and\,T_{{}}^{-1}\]

    C) \[M_{{}}^{0}L_{{}}^{1}T_{{}}^{-1}and\,LT_{{}}^{-2}\]  

    D) \[M_{{}}^{0}L_{{}}^{1}T_{{}}^{-1}and\,LT_{{}}^{{}}\]

    Correct Answer: A

    Solution :

    [a] Dimension of \[\alpha t=[{{M}^{0}}{{L}^{0}}{{T}^{0}}]\therefore \,\,\,\,[\alpha ]=[{{T}^{-1}}]\] Again\[\left[ \frac{{{v}_{0}}}{a} \right]=[L]\]so \[[{{v}_{0}}]=[L{{T}^{-1}}]\]


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