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JEE Main 2013 Question Paper

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Question : 73

Total: 90

Let [ε0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:

[ε0]=[M−1L−3T4A2]
[ε0]=[M−1L2T−1A−2]
[ε0]=[M−1L2T−1A]
[ε0]=[M−1L−3T2A]

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