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(A) \[\left[ {{L}^{-1}}T \right]\]

(B) \[\left[ L{{T}^{-1}} \right]\]

(C) \[\left[ {{L}^{{1}/{2}\;}}{{T}^{{1}/{2}\;}} \right]\]

(D) \[\left[ L{{T}^{{1}/{2}\;}} \right]\]

Answer
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The oscillation of electric field and magnetic field in vacuum creates electromagnetic waves. The light is electromagnetic waves in free space. The velocity of the light is denoted by \[c\].The velocity of light \[c\] is constant therefore \[c\] depends on electric and magnetic constant. The electric constant is known by the permeability of free space (\[{{\mu }_{0}}\]) and magnetic constant is known by permittivity of free space (\[{{\varepsilon }_{0}}\]).

\[c=\dfrac{1}{\sqrt{\left( {{\mu }_{0}}{{\varepsilon }_{0}} \right)}}\]

\[\therefore {{({{\mu }_{0}}{{\varepsilon }_{0}})}^{{-1}/{2}\;}}=c\]

The unit of the \[c\] is \[{m}/{s}\;\]. So the unit of given quantity is \[{m}/{s}\;\].

The dimensional equivalent of unit \[{m}/{s}\;\] is \[\left[ L{{T}^{-1}} \right]\]

Some important values if these constants:

Permeability of free space (\[{{\mu }_{0}}\]) = \[4\pi \times {{10}^{-7}}N{{A}^{-2}}\]

Permittivity of free space (\[{{\varepsilon }_{0}}\]) = \[8.85419\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}\]

Velocity of light (\[c\]) = \[2.99792\times {{10}^{8}}m{{s}^{-1}}\]

The dimensional relationship does not mean that the relation between both side quantities are linearly the same, it only compares the dimensions. Here the relation between the speed of light and \[({{\mu }_{0}}{{\varepsilon }_{0}})\] is not only inversely proportional but the square root of \[({{\mu }_{0}}{{\varepsilon }_{0}})\] inversely proportional to speed of light but the dimensional does not predict it.

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