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**Hint:**Although known that the time constant of an $L - R$ circuit is given by $\tau = \dfrac{L}{R}$, but before we use that approach, it is important to find the dimensions of the individual elements and combining them to find the dimension of $\dfrac{L}{R}.$

**Formulae used:**

$E = \dfrac{1}{2}L{i^2}$

Where $E$ is the energy stored in an $L - R$ circuit, $L$ is the inductance and $i$ is the current in the circuit and is dimensionally represented by $A$.

$E = {i^2}RT$

Where $E$ is the energy stored in an $L - R$ circuit, $R$ is the resistance, $T$ is the time and $i$ is the current in the circuit and is dimensionally represented by $A$.

$\tau = \dfrac{L}{R}$

Where $\tau $ is the time constant of an $L - R$ circuit, $R$ is the resistance of the circuit and $L$ is the inductance of the circuit.

**Complete step by step solution:**

To find the dimensions of inductance, we will first find an equation that equates inductance with a quantity whose dimensions are well known and easily calculated, that is,

$E = \dfrac{1}{2}L{i^2}$

Where $E$ is the energy stored in an $L - R$ circuit, $L$ is the inductance and $i$ is the current in the circuit and is dimensionally represented by $A$ . Therefore,

$ \Rightarrow L = \dfrac{{2E}}{{{i^2}}}$

Since dimensions of $E = ML{T^{ - 2}}$ and dimension of $i = A$ , therefore the dimensional formula of inductance is ,

\[L = \dfrac{{M{\text{ }}L\;{T^ - }^2\;}}{{{A^2}}} = M{\text{ }}L\;{T^ - }^2\;{A^ - }^2\] $...\left( 1 \right)$

Similarly in the case of the resistance of the circuit

$E = {i^2}RT$

Where $E$ is the energy stored in an $L - R$ circuit, $R$ is the resistance, $T$ is the time and $i$ is the current in the circuit and is dimensionally represented by $A$ . Therefore,

$ \Rightarrow R = \dfrac{E}{{{i^2}t}}$

Since dimensions of $E = ML{T^{ - 2}}$ and dimension of $i = A$ , therefore the dimensional formula of inductance is ,

$R = \dfrac{{ML{T^{ - 2}}}}{{{A^2}T}} = ML{T^{ - 3}}{A^{ - 2}}$

To find the dimensional formula of \[\dfrac{L}{R}\] we simply multiply the individual dimensions, that is,

$ \Rightarrow \dim \left( {\dfrac{L}{R}} \right) = \dfrac{{ML{T^ - }^2{A^ - }^2}}{{ML{T^{ - 3}}{A^{ - 2}}}}$

$ \Rightarrow \dim \left( {\dfrac{L}{R}} \right) = {M^0}{L^0}{T^1}{A^0}$

$ \Rightarrow \dim \left( {\dfrac{L}{R}} \right) = T$

**Therefore, the dimension of \[\dfrac{L}{R}\] is dependent solely on time.**

Alternatively:

Since the time constant, $\tau = \dfrac{L}{R}$, therefore you can tell that the dimensions of \[\dfrac{L}{R}\] will be similar to that of $\tau $ as dimensional equality is only possible if the dimensions of both quantities are equal.

**Therefore the dimensions of \[\dfrac{L}{R}\] is $T$.**

**Note:**Dimensional analysis questions usually have multiple approaches possible, depending entirely on the ease of your application and knowledge. Dimensions of a particular quantity can be solved in multiple ways by using the right formula to relate that quantity to those quantities whose dimensions you’re sure of. In this question, you could’ve further expanded the formula or Resistance and Inductance to get to the four basic units: mass $\left( M \right)$ , time $\left( T \right)$ , length $(L)$ and current $\left( A \right)$ .

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