
Check the dimensional correctness of the relation:
$mgh = \dfrac{1}{2}m{v^2}$, where letters have their usual meanings.
Answer
217.8k+ views
Hint: The dimensions of physical quantities are the powers to which the fundamental quantities are raised to represent that quantity. All physical quantities can be expressed as a combination of seven fundamental or base quantities. Here we have to find the dimensional correctness of the given relation.
Complete step by step solution:
Here we are given the expressions of kinetic energy and potential energy.
The expression for P.E is given as, $P.E = mgh$.
Let us first find the dimensional formula for potential energy for $mgh\;$
The mass,
$m = \left[ M \right]$
Acceleration can be written as,
$a = \dfrac{d}{{{t^2}}}$
where $d$ stands for the distance and $t$ stands for the time.
The dimensional formula for acceleration can be written as,
$a = \dfrac{L}{{{T^2}}} = L{T^{ - 2}}$
The dimensional formula for the acceleration due to gravity is, therefore,
$g = \left[ {L{T^{ - 2}}} \right]$
The dimensional formula for height
$h = \left[ L \right]$
The dimensional formula for $mgh\;$ can be written as,
$mgh = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$m = \left[ M \right]$
The formula for velocity
$v = \dfrac{d}{t}$
where $d$ stands for the displacement and $t$ stands for time.
The dimensional formula for velocity will be
$v = \dfrac{d}{t} = \dfrac{L}{T} = L{T^{ - 1}}$
Therefore,
${v^2} = {L^2}{T^{ - 2}}$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$\dfrac{1}{2}m{v^2} = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensions of both $\dfrac{1}{2}m{v^2}$ and $mgh\;$ are the same.
Therefore we can say that the relation $mgh = \dfrac{1}{2}m{v^2}$ is dimensionally correct.
Additional Information:
There are seven fundamental quantities are represented as mass $\left[ M \right]$, length $\left[ L \right]$, time $\left[ T \right]$, electric current $\left[ A \right]$, Thermodynamic temperature $\left[ K \right]$, luminous intensity $\left[ {cd} \right]$, and amount of substance $\left[ {mol} \right]$.
Note:
An equation connecting the physical quantity with its dimensional formula is known as the dimensional equation of that physical quantity. The dimensions of the equivalent fundamental quantity must be equal on either side of a dimensional equation.
Complete step by step solution:
Here we are given the expressions of kinetic energy and potential energy.
The expression for P.E is given as, $P.E = mgh$.
Let us first find the dimensional formula for potential energy for $mgh\;$
The mass,
$m = \left[ M \right]$
Acceleration can be written as,
$a = \dfrac{d}{{{t^2}}}$
where $d$ stands for the distance and $t$ stands for the time.
The dimensional formula for acceleration can be written as,
$a = \dfrac{L}{{{T^2}}} = L{T^{ - 2}}$
The dimensional formula for the acceleration due to gravity is, therefore,
$g = \left[ {L{T^{ - 2}}} \right]$
The dimensional formula for height
$h = \left[ L \right]$
The dimensional formula for $mgh\;$ can be written as,
$mgh = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$m = \left[ M \right]$
The formula for velocity
$v = \dfrac{d}{t}$
where $d$ stands for the displacement and $t$ stands for time.
The dimensional formula for velocity will be
$v = \dfrac{d}{t} = \dfrac{L}{T} = L{T^{ - 1}}$
Therefore,
${v^2} = {L^2}{T^{ - 2}}$
The dimensional formula for $\dfrac{1}{2}m{v^2}$ can be written as,
$\dfrac{1}{2}m{v^2} = \left[ {M{L^2}{T^{ - 2}}} \right]$
The dimensions of both $\dfrac{1}{2}m{v^2}$ and $mgh\;$ are the same.
Therefore we can say that the relation $mgh = \dfrac{1}{2}m{v^2}$ is dimensionally correct.
Additional Information:
There are seven fundamental quantities are represented as mass $\left[ M \right]$, length $\left[ L \right]$, time $\left[ T \right]$, electric current $\left[ A \right]$, Thermodynamic temperature $\left[ K \right]$, luminous intensity $\left[ {cd} \right]$, and amount of substance $\left[ {mol} \right]$.
Note:
An equation connecting the physical quantity with its dimensional formula is known as the dimensional equation of that physical quantity. The dimensions of the equivalent fundamental quantity must be equal on either side of a dimensional equation.
Recently Updated Pages
Elastic Collision in Two Dimensions Explained Simply

Elastic Collisions in One Dimension Explained

Electric Field Due to a Uniformly Charged Ring Explained

Electric Field of Infinite Line Charge and Cylinders Explained

Electric Flux and Area Vector Explained Simply

Electric Field of a Charged Spherical Shell Explained

Trending doubts
JEE Main 2026: Application Form Open, Exam Dates, Syllabus, Eligibility & Question Papers

Derivation of Equation of Trajectory Explained for Students

Hybridisation in Chemistry – Concept, Types & Applications

Understanding the Angle of Deviation in a Prism

Understanding Collisions: Types and Examples for Students

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

Units And Measurements Class 11 Physics Chapter 1 CBSE Notes - 2025-26

NCERT Solutions For Class 11 Physics Chapter 8 Mechanical Properties Of Solids

Motion in a Straight Line Class 11 Physics Chapter 2 CBSE Notes - 2025-26

NCERT Solutions for Class 11 Physics Chapter 7 Gravitation 2025-26

Understanding Atomic Structure for Beginners

