Question

Dimension of $\sqrt{\raisebox{1ex}{${\epsilon }_0$}\!\left/ \!\raisebox{-1ex}{${\mu }_0$}\right.}$ are.

• A. $\left[M{L}^2{T}^{-3}{A}^{-2}\right]$
• B. $\left[M^{-1}{L}^{-2}{T}^3{A}^2\right]$
• C. $\left[M^2{L}^2{T}^{-3}{A}^{-2}\right]$
• D. $\left[M^{-1}{L}^2{T}^3{A}^2\right]$

Answer 1

The correct option is B

$\left[M^{-1}{L}^{-2}{T}^3{A}^2\right]$

Step1: Given data.

Equation of given constant$=$$\sqrt{\raisebox{1ex}{${\epsilon }_0$}\!\left/ \!\raisebox{-1ex}{${\mu }_0$}\right.}$

Step2: Find the dimension of the given equation.

We have,

Equation of given constant$=$$\sqrt{\raisebox{1ex}{${\epsilon }_0$}\!\left/ \!\raisebox{-1ex}{${\mu }_0$}\right.}$

Where ${\epsilon }_0$ is the permittivity of free space and ${\mu }_0$ is the permeability of free space.

Then, Divide the numerator and denominator of the given constant equation by ${\epsilon }_0$ we get.

$\sqrt{\frac{{\epsilon }_0}{{\mu }_0}}=\sqrt{\frac{{\epsilon }_0{\epsilon }_0}{{\mu }_0{\epsilon }_0}}$

$\sqrt{\frac{{\epsilon }_0}{{\mu }_0}}=\frac{{\epsilon }_0}{\sqrt{{\mu }_0{\epsilon }_0}}$ …..$\left(i\right)$

We know that,

$C=\frac{1}{\sqrt{{\mu }_0{\epsilon }_0}}$ ……$\left(ii\right)$

Where $C$ is the speed of light, ${\epsilon }_0$ is the permittivity of free space and ${\mu }_0$ is the permeability of free space.

Substitute equation $\left(ii\right)$ in $\left(i\right)$ we get.

$\sqrt{\frac{{\epsilon }_0}{{\mu }_0}}={\epsilon }_0\times C$ …….$\left(iii\right)$

Since the S.I unit of ${\epsilon }_0\times C$ is,

${\epsilon }_0\times C=\frac{Co{l}^2}{Nm}\times \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.$

Where $Col$ is the unit of Charge, $N$ is the unit of force, $m$ is the meter, and $s$ is second.

Now, according to the dimension formula.

${\epsilon }_0\times C=\frac{Co{l}^2}{N{m}^2}\times \raisebox{1ex}{$m$}\!\left/ \!\raisebox{-1ex}{$s$}\right.=\left[\frac{I^2{T}^{2}L{T}^{-1}}{ML{T}^{-2}{L}^2}\right]$

${\epsilon }_0\times C=\left[M^{-1}{L}^{-2}{T}^3{A}^2\right]$ ……$\left(iv\right)$

From equation $\left(iii\right)$ and $\left(iv\right)$ dimension of the given equation as.

$\sqrt{\frac{{\epsilon }_0}{{\mu }_0}}=\left[M^{-1}{L}^{-2}{T}^3{A}^2\right]$

Hence, option B is correct. Dimension of $\sqrt{\raisebox{1ex}{${\epsilon }_0$}\!\left/ \!\raisebox{-1ex}{${\mu }_0$}\right.}$ are $\left[M^{-1}{L}^{-2}{T}^3{A}^2\right]$.