
The dimensions of electromotive force in terms of current $A$ are:
A) $\left[ {M{L^{ - 2}}{A^{ - 2}}} \right]$
B) $\left[ {M{L^2}{T^{ - 2}}{A^2}} \right]$
C) $\left[ {M{L^2}{T^{ - 2}}{A^{ - 2}}} \right]$
D) $\left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
Answer
232.8k+ views
Hint: To solve this question we should know about the base quantities which are used to form the dimensional formulae of any quantity. Also we should know how electromotive force is calculated i.e., the quantities involved in its calculation and their dimensional formulae.
Formulae used:
$V = \dfrac{W}{q}$
Here $V$ is the potential difference across the cell or the electromotive force of the cell, $W$ is the work done by the charge and $q$ is the charge.
Complete answer:
To solve this question we should know what electromotive force is. Electromotive force or the EMF, for short, of a cell is defined as the electric potential produced either by an electrochemical cell or by changing the magnetic field.
We know that,
$V = \dfrac{W}{q}$
Here $V$ is the potential difference across the cell or the electromotive force of the cell, $W$ is the work done by the charge and $q$ is the charge.
Let this be equation 1.
The potential difference gives us the value of the electromotive force or EMF of a cell. So,
$ \Rightarrow V = \dfrac{W}{q}$
Let this be equation 1.
This will give the value of electromotive force or EMF of a cell.
We know that the dimensional formulae of
$\left[ q \right] = \left[ {AT} \right]$
$\left[ W \right] = \left[ {M{L^2}{T^{ - 2}}} \right]$
Substituting the values of the above quantities in the equation 1 we get,
$ \Rightarrow \left[ V \right] = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {AT} \right]}}$
$ \Rightarrow \left[ V \right] = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
So the answer will be option (D).
Note: To solve questions related to dimensional analysis of any quantity, break the quantity into its smaller known units. Use the dimensional formulae of the smaller known units to find the dimensional formulae of the given quantity. Electromotive force is the energy per unit electric charge. It is the force driving all electrons. Flow of electrons is due to this force.
Formulae used:
$V = \dfrac{W}{q}$
Here $V$ is the potential difference across the cell or the electromotive force of the cell, $W$ is the work done by the charge and $q$ is the charge.
Complete answer:
To solve this question we should know what electromotive force is. Electromotive force or the EMF, for short, of a cell is defined as the electric potential produced either by an electrochemical cell or by changing the magnetic field.
We know that,
$V = \dfrac{W}{q}$
Here $V$ is the potential difference across the cell or the electromotive force of the cell, $W$ is the work done by the charge and $q$ is the charge.
Let this be equation 1.
The potential difference gives us the value of the electromotive force or EMF of a cell. So,
$ \Rightarrow V = \dfrac{W}{q}$
Let this be equation 1.
This will give the value of electromotive force or EMF of a cell.
We know that the dimensional formulae of
$\left[ q \right] = \left[ {AT} \right]$
$\left[ W \right] = \left[ {M{L^2}{T^{ - 2}}} \right]$
Substituting the values of the above quantities in the equation 1 we get,
$ \Rightarrow \left[ V \right] = \dfrac{{\left[ {M{L^2}{T^{ - 2}}} \right]}}{{\left[ {AT} \right]}}$
$ \Rightarrow \left[ V \right] = \left[ {M{L^2}{T^{ - 3}}{A^{ - 1}}} \right]$
So the answer will be option (D).
Note: To solve questions related to dimensional analysis of any quantity, break the quantity into its smaller known units. Use the dimensional formulae of the smaller known units to find the dimensional formulae of the given quantity. Electromotive force is the energy per unit electric charge. It is the force driving all electrons. Flow of electrons is due to this force.
Recently Updated Pages
Circuit Switching vs Packet Switching: Key Differences Explained

JEE General Topics in Chemistry Important Concepts and Tips

JEE Extractive Metallurgy Important Concepts and Tips for Exam Preparation

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

Electricity and Magnetism Explained: Key Concepts & Applications

Trending doubts
JEE Main 2026: Session 2 Registration Open, City Intimation Slip, Exam Dates, Syllabus & Eligibility

JEE Main 2026 Application Login: Direct Link, Registration, Form Fill, and Steps

JEE Main Marking Scheme 2026- Paper-Wise Marks Distribution and Negative Marking Details

Understanding the Angle of Deviation in a Prism

Hybridisation in Chemistry – Concept, Types & Applications

How to Convert a Galvanometer into an Ammeter or Voltmeter

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

Dual Nature of Radiation and Matter Class 12 Physics Chapter 11 CBSE Notes - 2025-26

Understanding Uniform Acceleration in Physics

Understanding the Electric Field of a Uniformly Charged Ring

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

Derivation of Equation of Trajectory Explained for Students

