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Let [ ε 0 ] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then -

1. [ ε 0 ] = [M –1 L 2 T –1 A –2 ]
2. [ ε 0 ] = [M –1 L 2 T –1 A]
3. [ ε 0 ] = [M –1 L –3 T 2 A]
4. [ ε 0 ] = [M –1 L –3 T 4 A 2 ]

Correct answer: [ ε 0 ] = [M –1 L –3 T 4 A 2 ]

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