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JEE Main & Advanced Sample Paper JEE Main Sample Paper-39

  • question_answer
    Using mass(M), length(L), time(T) and electric current  as fundamental quantities the dimensions of permittivity will be

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

    B)  \[ML{{T}^{-2}}{{A}^{-2}}\]

    C)  \[{{M}^{-1}}{{L}^{-3}}{{T}^{+4}}{{A}^{2}}\]               

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

    Correct Answer: C

    Solution :

     Force, \[F=\frac{1}{4\pi {{\varepsilon }_{0}}}\frac{{{q}_{1}}{{q}_{2}}}{{{r}^{2}}}\]\[\Rightarrow \]\[{{\varepsilon }_{0}}=\frac{{{q}_{1}}.{{q}_{2}}}{4\pi F{{r}^{2}}}\] So dimension of \[{{\varepsilon }_{0}}\] \[=\frac{{{[AT]}^{2}}}{[ML{{T}^{-2}}][{{L}^{2}}]}=[{{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}]\]


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