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JEE Main 2013 Tes Paper

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Question : 4 of 8

Marks: +1, -0

Let

\left[\varepsilon_0\right]

denote the dimensional formula of the permittivity of vacuum. If

\mathrm{M}=

\mathrm{L}=

\mathrm{T}=

\mathrm{A}=

electric current, then :

$\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^2 \mathrm{~A}\right]$
$\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
$\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^2 \mathrm{~T}^{-1} \mathrm{~A}^{-2}\right]$
$\left[\varepsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^2 \mathrm{~T}^{-1} \mathrm{~A}\right]$

Solution:

\begin{array}{l}{\left[\varepsilon_0\right]=\left[\frac{\mathrm{C}^2}{\mathrm{~N}-\mathrm{m}_2}\right]=\frac{\mathrm{A}^2 \mathrm{~T}^2}{\mathrm{MLT}^{-2} \mathrm{~L}^2}} \\ {\left[\varepsilon_0\right]=\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2}\end{array}

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