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Let $ \varepsilon_0$ denote the dimensional formula of the permittivity of vacuum. If M = mass, L = length and T = time and A = electric current then,

$\begin {array} {1 1} (A)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^4A^2] & \quad (B)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A^{-2}] \\ (C)\;[ \varepsilon_0 ]=[ M^{-1}L^{2}T^{-1}A] & \quad (D)\;[ \varepsilon_0 ]=[ M^{-1}L^{-3}T^2A ] \end {array}$

 

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$ \large\frac{1}{4 \pi \varepsilon_0} $ $\large\frac{q^2}{r^2}$ $ = F$
$ \varepsilon_0 = \large\frac{[A^2T^2]}{[MLT^{-2}L^2]}$$ = M^{-1}L^{-3}T^4A^2$
Ans : (A)
answered Feb 15, 2014 by thanvigandhi_1
edited Mar 14, 2014 by thanvigandhi_1
 

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