NEET Physics Mathematical Tools, Units & Dimensions NEET PYQ-Mathematical Tools, Units and Dimensions

  • question_answer
    The dimension of \[\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}},\] where \[{{\varepsilon }_{0}}\] is permittivity of free space and E is electric field, is [AIPMT (S) 2010]

    A) \[[M{{L}^{2}}{{T}^{-2}}]\]

    B) \[[M{{L}^{-1}}{{T}^{-2}}]\] 

    C) \[[M{{L}^{-2}}{{T}^{-1}}]\]

    D) \[[ML{{T}^{-1}}]\]

    Correct Answer: A

    Solution :

    Dimensions of \[{{\varepsilon }_{0}}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]
    Dimensions of \[E=[ML{{T}^{-3}}{{A}^{-1}}]\]
    \[\therefore \] Dimensions of \[\frac{1}{2}{{\varepsilon }_{0}}{{E}^{2}}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]\[\times [{{M}^{2}}{{L}^{2}}{{T}^{-6}}{{A}^{-2}}]\]
                \[=[M{{L}^{-1}}{{T}^{-3}}]\]


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