Question

Which of the following combinations has the dimension of electrical resistance (ϵ₀ is the permittivity of vacuum and μ₀ is the permeability of vacuum)?
1. \(\sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} \)
2. \(\frac{{{\mu _0}}}{{{\epsilon_0}}}\)
3. \(\sqrt {\frac{{{\epsilon_0}}}{{{\mu _0}}}} \)
4. \(\frac{{{\epsilon_0}}}{{{\mu _0}}}\)

Answer 1

Correct Answer - Option 1 : \(\sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} \)

Concept:

Dimensional formula of resistance is M¹L²T^-3I^-2

Where m is the mass, I is the current, L is the length, and T is the time.                 

Calculation:

The dimensional formula of permittivity of vacuum, ϵ₀ = (M^-1L^-3T⁴A²)

The dimensional formula of permeability of vacuum, μ₀ = (MLT^-2A^-2)

The dimensional formula of electrical resistance, R = (ML²T^-3A^-2)

From question,

R ∝ ϵ₀ and μ₀

We need to select the suitable combination,

\( \Rightarrow R = k\sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} \) 

On substituting, dimensional formula on the above formula:

\( \Rightarrow M{L^2}{T^{ - 3}}{A^{ - 2}} = k\sqrt {\frac{{ML{T^{ - 2}}{A^{ - 2}}}}{{{M^{ - 1}}{L^{ - 3}}{T^4}{A^2}}}} \) 

\( \Rightarrow M{L^2}{T^{ - 3}}{A^{ - 2}} = k\sqrt {{M^2}{L^4}{T^{ - 6}}{A^{ - 4}}} \) 

⇒ ML²T^-3A^-2 = kML² T^-3 A^-2

Now, both sides have become equal.

Thus,

\(\therefore R \propto \sqrt {\frac{{{\mu _0}}}{{{\epsilon_0}}}} \)