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Question: The dimension of \(\sqrt {\dfrac{\mu }{\varepsilon }} \) where \(\mu \) is permeability and \(\varep...

The dimension of με\sqrt {\dfrac{\mu }{\varepsilon }} where μ\mu is permeability and ε\varepsilon is permittivity is same as:
A) Resistance
B) Inductance
C) Capacitance
D) None of these

Explanation

Solution

We are here considering the permeability and permittivity of that of a free space.
These have dimensional values that are already derived in terms of SI units.
Also,When the free space is considered, the permeability is the ability to make the flux line to pass though the space
.
Complete step by step answer:
The permeability is the ability of a body to pass magnetic lines through it. Its value depends on how much a particular body is permeable to the flux lines. The value of it varies accordingly with the materials. When the free space is considered, the permeability is the ability to make the flux line to pass though the space. The permittivity is the ability to make the electric field of lines to pass through the body. Like the permeability, the permittivity also changes with the materials. The permittivity of free space is the ability of the space to allow electric field lines to pass through.
The dimensional formula for permittivity is,
[ε]=[M−1L−3T4A2][\varepsilon ] = [{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]
The dimensional formula of permeability is,
[μ]=[MLT−2A−2][\mu ] = [ML{T^{ - 2}}{A^{ - 2}}]
Dividing the dimensional form of permeability by permittivity we get, [μ][ε]=[MLT−2A−2][M−1L−3T4A2]=[M2L4T−6A−4]\dfrac{{[\mu ]}}{{[\varepsilon ]}} = \dfrac{{[ML{T^{ - 2}}{A^{ - 2}}]}}{{[{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}]}} = [{M^2}{L^4}{T^{ - 6}}{A^{ - 4}}]
Now taking the root of the above equation, we have
[μ][ε]=[M2L4T−6A−4]=[ML2T−3A−2]\sqrt {\dfrac{{[\mu ]}}{{[\varepsilon ]}}} = \sqrt {[{M^2}{L^4}{T^{ - 6}}{A^{ - 4}}]} = [M{L^2}{T^{ - 3}}{A^{ - 2}}]
Now, let us take the dimensional formula of resistance. We have,
[R]=[ML2T−3A−2][R] = [M{L^2}{T^{ - 3}}{A^{ - 2}}]
Thus, we can observe that the dimension of the square root of permeability divided permittivity is the same as that of resistance.

So, the correct answer is “Option A”.

Note:
As we can see from the definition that both the permeability and permittivity is the absence of the resistance. So, when one gets divided by the other the dimensional formula of resistance is observed. Also, we know that relative permittivity and permeability are unit-less entities.