Question

The dimensions of $\dfrac{{\text{R}}}{L}$is? Here, ${\text{R}}$ = electric resistance, $L$ = self inductance.
A. $[{{\text{T}}^{ - 2}}]$
B. $[{{\text{T}}^{ - 1}}]$
C. $[{\text{M}}{{\text{L}}^{ - 1}}]$
D. $[{\text{T}}]$

Answer 1

Hint
The concept of the given question is based on the dimensional formula. The dimensional formula of the ratio of the electric resistance to self inductance is being asked. Firstly we will find the dimensional formulae of electric resistance and the self inductance. Then, we will find the ratio of the same. This will give us the answer.
$\Rightarrow R = \dfrac{{\text{V}}}{{\text{I}}}$
where $R$ is resistance, $V$ is voltage and $I$ is current.
$\Rightarrow V = {\text{E}}{\text{.ds}}$
where ${\text{E}}$ is the electric field and $s$ is the distance.
$\Rightarrow {\text{E}} = \dfrac{{\text{F}}}{q}$
where ${\text{F}}$ is the force and $q$ is the charge.
$\Rightarrow q = {\text{I}} \times t$ where $t$ is the time.
and $\phi = {\text{BA}} = {\text{LI}}$
Where ${\text{B}}$ is magnetic field, $A$ is area, $L$ is the self inductance and $I$ is the current.

Complete step by step answer
As the dimensional formula of the ratio of electric resistance to self inductance is asked, we will find the dimensional formulae of electric resistance and self inductance first and then, we will divide them.
Firstly, let us compute the dimensional formula of electric resistance.
By Ohm’s law, the resistance is the ratio of voltage by current.
So, we have the formula of resistance as,
$\Rightarrow {\text{R}} = \dfrac{{\text{V}}}{{\text{I}}}$…… (1)
The voltage is the product of electric field and distance. So we can write,
$\Rightarrow V = {\text{E}}{\text{.ds}}$
And the electric field is the ratio of force by charge,
$\Rightarrow {\text{E}} = \dfrac{{\text{F}}}{q}$
The charge is the product of current and time.
$\Rightarrow q = {\text{I}} \times {\text{t}}$
So on combining all the formulae, so, we get the expression for voltage as,
$\Rightarrow V = \dfrac{{\text{F}}}{{{\text{I}} \times {\text{t}}}}{\text{.ds}}$
Now substituting the value of voltage in the equation (1), we get,
$\Rightarrow {\text{R}} = \dfrac{{{\text{F}}{\text{.ds}}}}{{{{\text{I}}^2}t}}$
Now we need to substitute the dimensional formulae of force, distance, current and time.
So, we get the dimensional formula of the electric resistance as,
$\Rightarrow {\text{R}} = \dfrac{{{\text{[ML}}{{\text{T}}^{ - 2}}{\text{]}} \times [{\text{L}}]}}{{{{{\text{[A]}}}^2} \times [{\text{T}}]}}$
On cancelling the common terms from the numerator and the denominator we get,
$\Rightarrow {\text{R}} = {\text{[M}}{{\text{L}}^2}{{\text{T}}^{ - 3}}{A^{ - 2}}{\text{]}}$
Secondly, now let us compute the dimensional formula of self inductance.
The electric flux is given by the formulae,
$\Rightarrow \phi = {\text{LI}}$
It is also given by,
$\Rightarrow \phi = {\text{BA}}$
Combining the above equations and expressing the formula in terms of self inductance, we get,
$\Rightarrow {\text{L}} = \dfrac{{{\text{BA}}}}{{\text{I}}}$
Now we can substitute the dimensional formulae of magnetic field, area and current.
So, we get the dimensional formula of the self inductance as follows,
$\Rightarrow {\text{L}} = \dfrac{{{\text{[M}}{{\text{T}}^{ - 2}}{{\text{A}}^{ - 1}}{\text{][}}{{\text{L}}^2}{\text{]}}}}{{{\text{[A]}}}}$
Again on cancelling the common terms we get,
$\Rightarrow {\text{L}} = {\text{[M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}{A^{ - 2}}{\text{]}}$
So now we can combine the dimensional formulae of the electric resistance and the self inductance by taking their ratio. Hence we have,
$\Rightarrow \dfrac{{\text{R}}}{{\text{L}}} = \dfrac{{\left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 3}}{{\text{A}}^{ - 2}}} \right]}}{{\left[ {{\text{M}}{{\text{L}}^2}{{\text{T}}^{ - 2}}{{\text{A}}^{ - 2}}} \right]}}$
So on cancelling the above given terms from the numerator and the denominator we get
$\Rightarrow \dfrac{{\text{R}}}{{\text{L}}} = \left[ {{{\text{T}}^{ - 1}}} \right]$
Hence the dimensional formula of the ratio of electric resistance to self inductance is given by $[{T^{ - 1}}]$
Thus, option (B) is correct.

Note
The equation which relates the fundamental units in terms of the derived units is called dimensional formula. In mechanics, the length, mass and the time are taken as the fundamental units and are represented by the letters $L$, $M$ and $T$.