# NEET Physics Units and Measurement Questions Solved

Let $\left[{\epsilon }_{0}\right]$ denotes the dimensional formula of the permittivity of the vacuum and $\left[{\mu }_{0}\right]$ that of the permeability of the vacuum. If M = mass, L = length, T = Time and I = electric current, then

(1) $\left[{\epsilon }_{0}\right]={M}^{-1}{L}^{-3}{T}^{2}I$

(2) $\left[{\epsilon }_{0}\right]={M}^{-1}{L}^{-3}{T}^{4}{I}^{2}$

(3) $\left[{\mu }_{0}\right]=ML{T}^{-2}{I}^{-3}$

(4) $\left[{\mu }_{0}\right]=M{L}^{2}{T}^{-1}I$

Concept Videos :-

#1 | Basic Concepts & Examples
#2 | Dimensional Analysis : Remaining

Concept Questions :-

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