Question

If L and R denote inductance and resistance respectively, then the dimension of $\dfrac{L}{R}$ is
$\text{A}\text{. }\left[ {{M}^{0}}{{L}^{0}}{{T}^{0}} \right]$
$\text{B}\text{. }\left[ {{M}^{0}}{{L}^{0}}T \right]$
$\text{C}\text{. }\left[ {{M}^{2}}{{L}^{0}}{{T}^{2}} \right]$
$\text{D}\text{. }\left[ ML{{T}^{2}} \right]$

Answer 1

Hint: First, calculate the dimensional formula of L and R by using the formulas V=iR and $E=L\dfrac{di}{dt}$ respectively. Once you find the dimensional formula of L and R, then divide the two dimensional formulas to find the dimensional formula of $\dfrac{L}{R}$.
Formula used:
V=iR
$\left[ V \right]=\dfrac{\left[ W \right]}{\left[ C \right]}$
$E=L\dfrac{di}{dt}$

Complete step by step solution:
To calculate the dimensional formula of $\dfrac{L}{R}$, we first need some expression for L and R in the terms of known quantities. With this we can then calculate the dimensional formula of L and R and divide them to find the dimensional formula of $\dfrac{L}{R}$.
To find the dimensional formula of R, let us use Ohm’s law. According to Ohm’s law, the potential difference (V) across a resistor and the current (i) flowing the resistor are related as V = iR, where R is the resistance of the resistor.
Hence, we get that $R=\dfrac{V}{i}$ …. (i).
Potential difference of V across two points is equal to the work done on a unit charge to move it between these two points. Hence, the dimensional formula of potential difference (V) is equal to the ratio of the dimensional formula of work to the dimensional formula charge.
i.e. $\left[ V \right]=\dfrac{\left[ W \right]}{\left[ C \right]}$.
Dimensional formula of work is $\left[ W \right]=\left[ M{{L}^{2}}{{T}^{-2}} \right]$ and the dimensional formula of charge is [C]=[IT], where [I] is the dimensional formula of current.
This means that, $\left[ V \right]=\dfrac{\left[ W \right]}{\left[ C \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}} \right]}{\left[ IT \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right]$.
From equation (i), we get that the dimensional formula is R is $\left[ R \right]=\dfrac{\left[ V \right]}{\left[ i \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right]}{\left[ I \right]}=\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-2}} \right]$.
$\Rightarrow \left[ R \right]=\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-2}} \right]$ ….. (ii).
Let us now calculate the dimensional formula of inductance L.
For this we will use Faraday’s law, which use that the emf induced in a coil is equal to $E=L\dfrac{di}{dt}$ ….. (iii), where l is the inductance of the coil and i is the current in the coil.
We can write equation (iii) as
$L=\dfrac{E}{\dfrac{di}{dt}}$
This implies that the dimensional formula of L is $\left[ L \right]=\dfrac{\left[ E \right]}{\left[ \dfrac{di}{dt} \right]}$.
Emf induced is a potential difference and we have already calculated the dimensional formula of potential difference.
Hence, $\left[ E \right]=\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right]$
$\left[ \dfrac{di}{dt} \right]=\dfrac{\left[ I \right]}{\left[ T \right]}=\left[ I{{T}^{-1}} \right]$
This gives us that $\left[ L \right]=\dfrac{\left[ E \right]}{\left[ \dfrac{di}{dt} \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-1}} \right]}{\left[ I{{T}^{-1}} \right]}=\left[ M{{L}^{2}}{{T}^{-2}}{{I}^{-2}} \right]$
$\Rightarrow \left[ L \right]=\left[ M{{L}^{2}}{{T}^{-2}}{{I}^{-2}} \right]$ …. (iv).
Divide equation (iv) by equation (iii).
$\Rightarrow \dfrac{\left[ L \right]}{\left[ R \right]}=\dfrac{\left[ M{{L}^{2}}{{T}^{-2}}{{I}^{-2}} \right]}{\left[ M{{L}^{2}}{{T}^{-3}}{{I}^{-2}} \right]}=\left[ {{M}^{0}}{{L}^{0}}T \right]$
Hence, the correct option is B.

Note: We can also find the dimension of $\dfrac{L}{R}$ by using the formula for current (i) flowing in a L-R circuit. The current (i) is given as $i={{i}_{\max }}{{e}^{-\dfrac{t}{\tau }}}$ …. (v).
Here, $\tau $ is called the time constant and its value for the L-R circuit is $\dfrac{L}{R}$.
If we look at equation (v), we see that e is raised to a power of $-\dfrac{t}{\tau }$ and power is just a number with no dimension. Hence, $\dfrac{t}{\tau }$ must have no dimension. For that $\tau $ must have the dimension of time, which means that $\dfrac{L}{R}$ has the dimension of time.