Q.

E2μ0 has the dimensions (E= electric field, μ0 = permeability of free space)

Moderate

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By Expert Faculty of Sri Chaitanya

M2L3T−2A2

MLT−4

ML3T−2

M−1L2TA−2

answer is 2.

(Detailed Solution Below)

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Detailed Solution

E2μ0=ε0E2ε0μ0= energy / volume (1/ speed of light )2⇒E2μ0= energy ( speed )2 volume ⇒E2μ0=ML2T−2L2T−2L3=MLT−4

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