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The dimensional formula of mobility is _______.
\[\begin{align}
  & \text{A}\text{.}{{M}^{-1}}{{L}^{1}}{{T}^{2}}{{A}^{1}} \\
 & \text{B}\text{.}{{M}^{1}}{{L}^{0}}{{T}^{-2}}{{A}^{-1}} \\
 & \text{C}\text{.}{{M}^{1}}{{L}^{-1}}{{T}^{-2}}{{A}^{-1}} \\
 & \text{D}\text{.}{{M}^{-1}}{{L}^{0}}{{T}^{2}}{{A}^{1}} \\
\end{align}\]

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Last updated date: 07th Sep 2024
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Answer
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Hint: Mobility is defined as the drift velocity of the particle per unit electric field present. So, the dimensional formula of mobility will be the dimensional formula of drift velocity divided by the dimensional formula of electric field. We know that the dimensional formula of a physical quantity is the expression of a physical quantity in terms of fundamental physical quantities.

Formula used:
\[\mu =\dfrac{{{v}_{d}}}{E}\]

Complete step by step answer:
Dimensional formula is the expression of a physical quantity in terms of fundamental physical quantities. Mass (M), Length (L), Time (T), Current (A) and temperature (K) are the fundamental quantities.
Mobility ($\mu$) is defined as the drift velocity ($v_d$) of the particle per unit electric field (E) present.
\[\mu =\dfrac{{{v}_{d}}}{E}\]
Clearly, we can see that the drift velocity is a type of velocity and it can have the same dimensional formula as that of velocity. So, the dimensional formula of drift velocity will be,
\[\left[ {{v}_{d}} \right]={{M}^{0}}{{L}^{1}}{{T}^{-1}}\]
Electric field at a point can be stated as the force per unit charge experienced by an infinitesimal positive test charge present at that point. That means, the dimensional formula of electric field will be,
\[\left[ E \right]=\dfrac{\left[ F \right]}{\left[ q \right]}=\dfrac{ML{{T}^{-2}}}{AT}={{M}^{1}}{{L}^{1}}{{T}^{-3}}{{A}^{-1}}\]
The dimensional formula of mobility is the dimensional formula of drift velocity divided by the dimensional formula of electric field. So,
\[\left[ \mu \right]=\dfrac{\left[ {{v}_{d}} \right]}{\left[ E \right]}=\dfrac{{{M}^{0}}{{L}^{1}}{{T}^{-1}}}{{{M}^{1}}{{L}^{1}}{{T}^{-3}}{{A}^{-1}}}={{M}^{-1}}{{L}^{0}}{{T}^{2}}{{A}^{1}}\]
Thus, we can see that option D is the correct answer.

Note: Students usually memorize the dimensional formulas. But, in my opinion, it is not a right practice. It is good to understand the physical quantities and relate them. Thus, these big formulas will become easier to us. It is the best and shortest way in practice. Dimensional analysis will sometimes help us to eliminate some wrong options. So, it is a good thing to understand this.