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Physics · 2013
JEE Main 2013 · 2013-04-07 Offline · Q38
Let [ε_0] denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length, T = time and A = electric current, then:
Let $\displaystyle \left[\epsilon_0\right]$ denote the dimensional formula of the permittivity of vacuum. If $\displaystyle \mathrm{M}=$ mass, $\displaystyle \mathrm{L}=$ length, $\displaystyle \mathrm{T}=$ time and $\displaystyle \mathrm{A}=$ electric current, then :
Official answer
(3)
$\displaystyle \left[\epsilon_0\right]=\left[\mathrm{M}^{-1} \mathrm{~L}^{-3} \mathrm{~T}^4 \mathrm{~A}^2\right]$
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