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NEET Sample Paper NEET Sample Test Paper-67

  • 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

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

    B) \[\left[ M{{L}^{-1}}{{T}^{-}}^{2} \right]\]

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

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

    Correct Answer: B

    Solution :

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


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