# Let $\left[\epsilon_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If $\mathrm{M}=$ mass, $\mathrm{L}=$ length, $\mathrm{T}=$ time and $\mathrm{A}=$ electric current, then:

Question:

Let $\left[\epsilon_{0}\right]$ denote the dimensional formula of the permittivity of vacuum. If $\mathrm{M}=$ mass, $\mathrm{L}=$ length, $\mathrm{T}=$ time and $\mathrm{A}=$ electric current, then:

1. $\epsilon_{0}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{2} \mathrm{~A}\right]$

2. $\epsilon_{0}=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^{4} \mathrm{~A}^{2}\right]$

3. $\epsilon_{0}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{1} \mathrm{~A}^{2}\right]$

4. $\epsilon_{0}=\left[\mathrm{M}^{1} \mathrm{~L}^{2} \mathrm{~T}^{1} \mathrm{~A}\right]$

JEE Main Previous Year Single Correct Question of JEE Main from Physics Physical World, Units and Measurements chapter.

JEE Main Previous Year 2013

Correct Option: 2

Solution:

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