[PermeabilityPermittivity]will have the dimensions of :
A
M0L0T0A0
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B
M2L2T4A2
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C
M2L4T−6A−4
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D
M−2L−4T6A4
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Solution
The correct option is CM2L4T−6A−4 Permeability =μ0=MLA−2T−2 .........(1) Permittivity =M−1L−3A2T4 .................(2) So, eqn(1)eqn(2)=MLA−2T−2M−1L−3A2T4=M2L+4A−4T−6 Hence, option C is correct.