Question

The dimensional formula of ${{\mu }_{\circ }}{{\in }_{\circ }}$ is
$\begin{align}
  & a){{M}^{0}}{{L}^{-2}}{{T}^{2}} \\
 & b){{M}^{0}}{{L}^{2}}{{T}^{-2}} \\
 & c){{M}^{0}}{{L}^{1}}{{T}^{-1}} \\
 & d){{M}^{0}}{{L}^{-1}}{{T}^{1}} \\
\end{align}$

Answer 1

Hint: We use${{\mu }_{\circ }}$to denote the permeability of free space in vacuum and ${{\in }_{\circ }}$ is used to denote the permittivity of free space in vacuum. The product of both of these is used related to the speed of light. We know the dimensions of speed i.e. m/s. Hence we will determine the dimensional formula of the above product accordingly.

Formula used:
$c=\dfrac{1}{\sqrt{{{\mu }_{\circ }}{{\in }_{\circ }}}}$

Complete step-by-step answer:
To begin with let us first write the expression, relating the permeability of free space and permittivity of free space in terms of the speed of light. Let us say the speed of light in vacuum is ‘c’. Then product of ${{\mu }_{\circ }}{{\in }_{\circ }}$is related by the expression,
$c=\dfrac{1}{\sqrt{{{\mu }_{\circ }}{{\in }_{\circ }}}}$
Now let us express the above product i.e. ${{\mu }_{\circ }}{{\in }_{\circ }}$ in terms of speed of light.
$\begin{align}
  & c=\dfrac{1}{\sqrt{{{\mu }_{\circ }}{{\in }_{\circ }}}} \\
 & \text{Squaring on both sides,} \\
 & \Rightarrow {{c}^{2}}=\dfrac{1}{{{\left( \sqrt{{{\mu }_{\circ }}{{\in }_{\circ }}} \right)}^{2}}} \\
 & \Rightarrow {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{c}^{2}}} \\
\end{align}$
The speed has a dimensional formula of ${{M}^{0}}{{L}^{1}}{{T}^{-1}}$. Substituting this in the above equation we get,
$\begin{align}
  & {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{c}^{2}}} \\
 & {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{\left( {{M}^{0}}{{L}^{1}}{{T}^{-1}} \right)}^{2}}} \\
 & \Rightarrow {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{M}^{0}}{{L}^{2}}{{T}^{-2}}} \\
 & \Rightarrow {{\mu }_{\circ }}{{\in }_{\circ }}={{M}^{0}}{{L}^{-2}}{{T}^{2}} \\
\end{align}$

So, the correct answer is “Option a”.

Note: To solve the above question we need to know the laws of exponents. Here are some of the laws of exponents which are used in the above question. If ${{\left( ab \right)}^{n}}={{a}^{n}}{{b}^{n}}$ .One more very useful law of exponent is ${{\left( \dfrac{a}{b} \right)}^{n}}=\dfrac{{{a}^{n}}}{{{b}^{n}}}={{a}^{n}}{{b}^{-n}}$ . We know that the speed of light in air is $3\times {{10}^{8}}m/s$. Hence the product of the permeability if free space in vacuum and the permittivity of free space in vacuum we get numerically equal to,
$\begin{align}
  & {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{c}^{2}}} \\
 & \Rightarrow {{\mu }_{\circ }}{{\in }_{\circ }}=\dfrac{1}{{{(3\times {{10}^{8}})}^{2}}}=0.11\times {{10}^{-16}} \\
\end{align}$
It is to be noted that the numerical value of permeability of free space in vacuum is equal to ${{\mu }_{\circ }}=4\pi \times {{10}^{-7}}Tm/A$ and the value of the permittivity of free space in vacuum is ${{\in }_{\circ }}=8.854\times {{10}^{-12}}{{C}^{2}}{{N}^{-1}}{{m}^{-2}}$.