EAMCET Medical EAMCET Medical Solved Paper-1996

  • question_answer
    Dimensional formula of \[\frac{{{e}^{4}}}{{{\varepsilon }_{0}}^{2}{{m}_{p}}{{m}_{e}}^{2}{{c}^{2}}G}\] is: (where e = charge, \[{{m}_{p}}\]and \[{{m}_{e}}\], are masses \[{{\varepsilon }_{0}}=\] permittivity of free space, c = velocity of light and G = gravitational constant):

    A)  [M]                                      

    B)  [T]

    C)  [\[\pi \]]                             

    D)  [I4 T2 M]

    Correct Answer: B

    Solution :

                     Not Available


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