JEE Main & Advanced JEE Main Paper (Held on 08-4-2019 Morning)

  • question_answer
    In SI units, the dimesions of\[\sqrt{\frac{{{\in }_{0}}}{{{\mu }_{0}}}}\]is: [JEE Main 8-4-2019 Morning]

    A) \[{{A}^{-1}}TM{{L}^{3}}\]                    

    B) \[{{A}^{2}}{{T}^{3}}{{M}^{-1}}{{L}^{-2}}\]

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

    D) \[A{{T}^{-3}}M{{L}^{3/2}}\]

    Correct Answer: B

    Solution :

    dimension of\[\sqrt{\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}\] \[[{{\varepsilon }_{0}}]=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\] \[[{{\mu }_{0}}]=[ML{{T}^{-2}}{{A}^{-2}}]\] dimensions of\[\sqrt{\frac{{{\varepsilon }_{0}}}{{{\mu }_{0}}}}={{\left[ \frac{{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}}{ML{{T}^{-2}}{{A}^{-2}}} \right]}^{\frac{1}{2}}}\] \[={{[{{M}^{-2}}{{L}^{-4}}{{T}^{6}}{{A}^{4}}]}^{1/2}}\] \[=[{{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{A}^{2}}]\]


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