Question

What are the dimensions of voltage in terms of mass ($M$), length ($L$) and time ($T$) and ampere ($A$)?
$\text{A}\text{. }\\!\\![\\!\\!\text{ }{{\text{M}}^{1}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-2}}}{{\text{A}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }$
$\text{B}\text{. }\\!\\![\\!\\!\text{ }{{\text{M}}^{1}}{{\text{L}}^{\text{2}}}{{\text{T}}^{3}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }$
$\text{C}\text{. }\\!\\![\\!\\!\text{ }{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{A}}^{\text{1}}}\text{ }\\!\\!]\\!\\!\text{ }$
$\text{D}\text{. }\\!\\![\\!\\!\text{ }{{\text{M}}^{1}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }$

Answer 1

In this question write the formula of voltage in terms of force, displacement and charge to calculate dimensions of voltage and then use the dimensional formula of force, displacement and charge to calculate dimensions of voltage.
Voltage in an electric field is defined as the change in potential energy of a unit positive charge when moved from one point to another.

Formula used:
$\Delta V\text{=}\dfrac{W}{q}$, $W=\overrightarrow{F}.\text{ }\overrightarrow{d}$and $V=\dfrac{\overrightarrow{F.}\text{ }\overrightarrow{d}}{q}$

Complete step-by-step answer:
We already know that the potential difference or voltage in an electric field is the work to be done on a unit positive charge to move it from one point to another. Mathematically it can be written as:
$\Delta V\text{=}\dfrac{W}{q}$
Where
$\Delta V=Voltage$
$W=\text{ Work done}$
$q=\text{ charge}$
We also know that work is said to be done when a force applied on an object moves it. It is given by:
$W=\overrightarrow{F}.\text{ }\overrightarrow{d}$
Where
$\overrightarrow{F}=\text{ Force applied}$
$\overrightarrow{d}=\text{ Displacement of object}$
On combining both above equations, we get
$V=\dfrac{\overrightarrow{F.}\text{ }\overrightarrow{d}}{q}$
Now as we know, force acting on the moving object is the product of its mass ($m$) and the acceleration ($a$). Therefore $\vec{F}=m\vec{a}$.
We know that the dimension of mass ($m$) is [${{\text{M}}^{1}}$] and Dimension of acceleration ($a$) is [${{\text{L}}^{1}}{{\text{T}}^{-2}}$]. Therefore dimension of force is [${{\text{M}}^{1}}{{\text{L}}^{1}}{{\text{T}}^{-2}}$].
SI unit of displacement is meter, therefore the dimension of displacement is [${{\text{L}}^{\text{1}}}$].
Charge can be written as product of current (dimensional formula is [${{\text{A}}^{1}}$]) and time (dimensional formula is [${{\text{T}}^{1}}$]). Therefore dimensional formula for charge is [${{\text{A}}^{\text{1}}}{{\text{T}}^{\text{1}}}$].
On combining the above data, we get dimension of $V$as
$V=\dfrac{\text{ }\\!\\![\\!\\!\text{ }{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{1}}}{{\text{T}}^{\text{-2}}}\text{ }\\!\\!]\\!\\!\text{ }\\!\\![\\!\\!\text{ }{{\text{L}}^{\text{1}}}\text{ }\\!\\!]\\!\\!\text{ }}{\text{ }\\!\\![\\!\\!\text{ }{{\text{A}}^{\text{1}}}{{\text{T}}^{\text{1}}}\text{ }\\!\\!]\\!\\!\text{ }}$
Now on simplifying we have
$\text{V= }\\!\\![\\!\\!\text{ }{{\text{M}}^{\text{1}}}{{\text{L}}^{\text{2}}}{{\text{T}}^{\text{-3}}}{{\text{A}}^{\text{-1}}}\text{ }\\!\\!]\\!\\!\text{ }$

So, the correct answer is “Option D”.

Note: Dimensions of any physical quantity are the expressions showing the powers to which fundamental units are to be raised to obtain unit derived quantity. Dimensions are very useful in unit conversions. There are seven fundamental physical quantities. Students should try to remember the basic fundamental quantities and their dimensional formula.