Question

The dimension of resistance $R$ in terms of ${\mu }_o$ and ${\epsilon }_o$ are:
A.) $\sqrt{{\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}}$
B.) $\sqrt{{\displaystyle \frac{{\epsilon }_o}{{\mu }_o}}}$
C.) ${\displaystyle \frac{{\epsilon }_o}{{\mu }_o}}$
D.) ${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}$

Answer 1

Hint: In this question, we first write the dimensional formula for resistance as $\left[M^1{L}^2{T}^{-3}{A}^{-2}\right]$, permittivity as $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$, and permeability as $\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]$.

Complete step-by-step answer:

Now we divide the dimension of permeability by the dimension of permittivity and get $${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}=\left[M^2{L}^4{T}^{-6}{A}^{-4}\right]$$. As we compare this to the dimensional formula of resistance we get ${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}$ gives us the square the dimensional formula of resistance. So we conclude $${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}=R^2$$ and $\sqrt{{\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}}=R$.
Therefore option A is correct.

In this question, we were asked to find the resistance $R$ in terms of permeability and permittivity of free space and hence, therefore we will be using the values of permeability $\left({\mu }_o\right)$ and permittivity $\left({\epsilon }_o\right)$ for free space.
The permittivity $\left(\epsilon \right)$ is the measure of how easily a material can get polarized in an electric field. Whereas the permeability $\left(\mu \right)$ is the measure of how easily a material can get magnetized in a magnetic field.

We will now see the values and dimensional formulae of that are permeability $\left({\mu }_o\right)$ and permittivity $\left({\epsilon }_o\right)$ for free space that is
The dimensional formula for resistance is $\left[M^1{L}^2{T}^{-3}{A}^{-2}\right]$
The permittivity of free space equals ${\epsilon }_o=8.85{\displaystyle \frac{Farad}{m}}$ and its dimensional formula is equal to $\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]$
Similarly, the permeability of free space equals ${\mu }_o=1.26{\displaystyle \frac{Henry}{m}}$ and its dimensional formula is equal to $\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]$

Now, we will divide the dimension of permeability by the dimension of permittivity and get
$${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}={\displaystyle \frac{\left[M^1{L}^1{T}^{-2}{A}^{-2}\right]}{\left[M^{-1}{L}^{-3}{T}^4{A}^2\right]}}$$
$$\Rightarrow {\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}=\left[M^2{L}^4{T}^{-6}{A}^{-4}\right]$$
Now when we compare this to the dimensional formula of resistance that is $\left[M^1{L}^2{T}^{-3}{A}^{-2}\right]$,
We will notice that what we have got is the nothing but the square of the dimensional formula of resistance.

Therefore, we can conclude that the ratio ${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}$ gives us the square of the resistance that is
$${\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}=R^2$$
So the square root will give us the resistance that is
$\sqrt{{\displaystyle \frac{{\mu }_o}{{\epsilon }_o}}}=R$

Hence, option A is correct.

Note: For these types of questions we need to be well versed with the concepts of permittivity and permeability. We need to know their dimensional formulas and the dimensional formula of values that can be derived from permittivity and permeability that is resistance and velocity of free space.