Question

# Let [ε0] denote the dimensional formula of the permittivity of the vaccum and [μ0] denote the permeability of the vacuum. If M= mass, L= length, T= time and I= electric current, then(ε0]= M−1L−3T2I(ε0]=M−1L−3T4I2(μ0]=MLT−2I−2(μ0]=ML2T−1I

Solution

## The correct options are B (ε0]=M−1L−3T4I2 C (μ0]=MLT−2I−2 Unit of ε0=C2N−m2(ε0]=I2T4ML3=(M−1L−3T4I2] We know that μ0ϵ0= 1c2 (where c is the velocity of light) Using this we can say [μ0]= [MLT−2I−2]

Suggest corrections