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JEE Main 2013

JEE Main 2013 Physics question on Units and Measurements

Let [ε0][\varepsilon_0] denote the dimensional formula of the permittivity of vacuum. If MM = mass, LL = length, TT = Time and AA = electric current, then
  1. A.
    [ε0]=[M−1L−3T2A][\varepsilon_0] = [M^{-1} L^{-3} T^2 A]
  2. B.
    [ε0]=[M−1L−3T4A2][\varepsilon_0] = [M^{-1} L^{-3} T^4 A^2]
  3. C.
    [ε0]=[M−2L2T−1A−2][\varepsilon_0] = [M^{-2} L^2 T^{-1} A^{-2}]
  4. D.
    [ε0]=[M−1L2T−1A2][\varepsilon_0] = [M^{-1} L^2 T^{-1} A^2]
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Correct option: A

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