Question

Dimension of $\sqrt{\dfrac{{{\varepsilon }_{o}}}{{{\mu }_{o}}}}$
$A.M{{L}^{2}}{{T}^{-3}}{{A}^{-2}}$
$B.{{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{A}^{2}}$
$C.{{M}^{2}}{{L}^{2}}{{T}^{-3}}{{A}^{-2}}$
$D.{{M}^{-1}}{{L}^{2}}{{T}^{3}}{{A}^{2}}$

Answer 1

Hint: We have to find the dimension of $\sqrt{\dfrac{{{\varepsilon }_{o}}}{{{\mu }_{o}}}}$. It means we have to find the dimension of the square root of electric permittivity and magnetic permittivity. We will relate these to quantities with the relation with speed of light and then solve for the easy solution.

Complete answer:
We have to find the dimension of the required quantity with the help of following relation:-
$c=\dfrac{1}{\sqrt{{{\mu }_{o}}{{\varepsilon }_{o}}}}$……………… $(i)$
Where, $c$is speed of light, ${{\varepsilon }_{o}}$is permittivity of free space or electric permittivity and ${{\mu }_{o}}$is magnetic permittivity.
In dimensional analysis $M$ is mass, $L$ is length, $T$ is time and $A$ is electric current.
We have to find dimension of $\sqrt{\dfrac{{{\varepsilon }_{o}}}{{{\mu }_{o}}}}$…………… $(ii)$
Simplifying equation $(ii)$ by multiplying numerator and denominator by ${{\varepsilon }_{o}}$we get:
$\implies \sqrt{\dfrac{{{\varepsilon }_{o}}\times {{\varepsilon }_{o}}}{{{\mu }_{o}}\times {{\varepsilon }_{o}}}}$
$\implies \dfrac{{{\varepsilon }_{o}}}{\sqrt{{{\mu }_{o}}\times {{\varepsilon }_{o}}}}$
$\implies {{\varepsilon }_{o}}\times \dfrac{1}{\sqrt{{{\mu }_{o}}\times {{\varepsilon }_{o}}}}$……………… $(iii)$
Using $(i)$ we get write $(iii)$ as follows
${{\varepsilon }_{o}}\times c$………………. $(iv)$
We know that the dimension of ${{\varepsilon }_{o}}$ is ${{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}$and dimension of $c$ is ${{L}^{1}}{{T}^{-1}}$. Putting these values in equation $(iii)$ we get,
${{M}^{-1}}{{L}^{-3}}{{T}^{4}}{{A}^{2}}\times {{L}^{1}}{{T}^{-1}}$
$\implies {{M}^{-1}}{{L}^{-2}}{{T}^{3}}{{A}^{2}}$

So, the correct answer is “Option B”.

Additional Information:
We should also have basic knowledge of dimensional analysis. The relationship between different physical quantities by identifying their base quantities is called dimensional analysis. The concept of dimensional analysis is based on the following base quantities:-
Mass, length, time and electric current.
We should also know the concept of base(Fundamental)quantities and derived quantities also.
Base Quantities- The quantities that are independent of other quantities and can represent themselves are called base quantities. Examples are Mass. Length, Time, Electric Current, Amount of substance, Temperature and Luminous Intensity.

Derived Quantities- The quantities that are dependent on other quantities and can’t represent themselves are called derived quantities. Examples are Speed, Acceleration, Force, etc.

We should also have basic knowledge of dimensional analysis. The relationship between different physical quantities by identifying their base quantities is called dimensional analysis. The concept of dimensional analysis is based on the following base quantities:-
Mass, length, time and electric current.

We should also know the concept of base(Fundamental)quantities and derived quantities also.
Base Quantities- The quantities that are independent of other quantities and can represent themselves are called base quantities. Examples are Mass. Length, Time, Electric Current, Amount of substance, Temperature and Luminous Intensity.
Derived Quantities- The quantities that are dependent on other quantities and can’t represent themselves are called derived quantities. Examples are Speed, Acceleration, Force, etc.

Note:
We have solved this problem with the help of the relation with speed of light. We can directly solve the problem by putting the respective dimensions of both the quantities also but the calculation is a bit difficult. Dimensions of quantities should be applied correctly.