Dimensional Analysis - Meaning, Examples, FAQs

What Is Principle of Homogeneity Of Dimensions:

The principle of homogeneity of dimensions says that “ In any physical mathematical equation the dimensions of each term appearing in the equation are the same on each side of that equation”. This is called the principle of homogeneity.

What is Meant by Dimension?

In physics, any physical quantity can be expressed in terms of fundamental units, and the representation of a physical quantity in terms of fundamental units is called the dimension of the physical quantity.

Following are the symbols for fundamental units used in Dimensional Analysis class 11.

What is Dimensional Analysis Class 11?

When we represent each physical quantity of a mathematical equation in its dimensional form then analysis of dimensions to determine whether a given equation is correct or not dimensionally is known as dimensional analysis.

Also read -

• NCERT Solutions for All Subjects
• NCERT Notes for all subject
• NCERT Exemplar Solutions for All Subjects

Application of Dimensional Analysis:

Application of dimensional analysis in various fields are:

1. Dimensional Analysis is applicable in the physics field for uses like unit conversion, and checking the correctness of equations.
2. In thermodynamics for deriving dimensionless numbers like Reynold's number.
3. In environmental science by helping researchers develop equations to predict natural occurrences like cyclones.

Uses Of Dimensional Analysis Class 11

The most widely uses of dimensional analysis are mentioned as:

1. The validity of a physical equation can be checked using dimensional analysis.
2. Dimensional analysis is used to determine the dimensions of any unknown variable’s dimension in a given physical equation.
3. Units can be converted from one system to another using dimensional analysis.

Examples of Dimensional Analysis:

When we analyze the physical equation by using their dimensions such as Distance, velocity, and Time relation.

We know that the Dimension of physical quantity Velocity is $\left[\mathrm{LT}^{-1}\right]$ while the Dimensions of quantity Time is [T] and from the relation, we know that Distance=Velocity×time so we can find the dimension of quantity distance by multiplying the dimensions of Velocity and Time and we get, Dimension of Distance as $\mathrm{LT}^{-1} \mathrm{~T}=[\mathrm{L}]$. Hence, this is a simple example of dimensional analysis showing a valid physical equation can be checked by using dimensional analysis.

Limitations Of Dimensional Analysis

Some of the limitations of dimensional analysis are:

1. Cannot determine dimensionless quantities or constants
2. It depends on known variables only during the analysis.
3. Not able to determine the addition and multiplication of terms with the same dimensions

Problem Solving Strategy of Dimensional Analysis

1. Check the correctness of the equation $\mathrm{f}=\frac{\mathrm{mv}^2 }{ \mathrm{r}}$

We will use the dimensional analysis and principle of homogeneity which can be used If the dimension of quantity ‘$f$’ represents force and the dimension of quantity $\frac{\mathrm{mv}^2 }{ \mathrm{r}}$ where $m$ represents mass, $v$ represents velocity and $r$ represents radius are the same then the given equation will be correct dimensionally.

The force has a dimension of mass×acceleration so, the dimension of ‘$f$’ can be written as $M\left[\mathrm{LT}^{-2}\right]$ or Dimension of force $\mathrm{f}=\left[\mathrm{MLT}^{-2}\right]$

Now, the dimension of radius which is simply the distance will be the Dimension of $r=[L]$, and for mass Dimension of mass $m=[M]$, and velocity Dimension of velocity $\mathrm{v}=\left[\mathrm{LT}^{-1}\right]$ Now, on putting these dimensions on the right-handed part of the equation which is $\frac {\mathrm{mv}^2 }{ \mathrm{r}}$ we get, $\frac{[M]\left[L T^{-1}\right]^2 }{[L]}$ on solving we get, $\left[\mathrm{MLT}^{-2}\right]$

Hence, the dimension of quantity f is the Dimension of force $\mathrm{f}=\left[\mathrm{MLT}^{-2}\right]$ and the dimension of quantity $\frac {\mathrm{mv}^2 }{ \mathrm{r}}$ is, $\left[\mathrm{MLT}^{-2}\right]$ which are the same, so from the principle of homogeneity this physical equation has dimensions the same on both sides so, this is a correct equation.

2. Check the correctness of the equation $\mathrm{v}^2-\mathrm{u}^2=2 \mathrm{aS}$

We will use the dimensional analysis and principle of homogeneity which can be used If the dimension of quantity $\mathrm{v}^2-\mathrm{u}^2$ where $v$, $u$ represents velocity and has a dimension of $\left[\mathrm{LT}^{-1}\right]$ and the dimension of quantity $2aS$ where a represents acceleration and has a dimension of $\left[\mathrm{LT}^{-2}\right]$ and $S$ represent distance which has a dimension of [L] is the same, then the equation will be correct dimensionally. Now, using these dimensions let us find the dimension of quantity $\mathrm{v}^2-\mathrm{u}^2$ as $\left[\mathrm{LT}^{-1}\right]^2=\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$ since u and v are both velocities so their difference is also a velocity.

Now, let us find the dimension of quantity 2aS as 2 is a dimensionless constant, and the multiplication of $a$ and $S$ will have the dimension of $\mathrm{LT}^{-2} \mathrm{~L}=\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$ so, we see that both parts have the same dimension of $\left[\mathrm{L}^2 \mathrm{~T}^{-2}\right]$ so, according to the principle of homogeneity, both parts have the same dimension which shows, the given equation is correct.

3. Dimensional analysis of $\mathrm{S}=\mathrm{ut}+0.5 \mathrm{a} \mathrm{t}^2$

We will do the Dimensional analysis of $\mathrm{S}=\mathrm{ut}+ \frac{1}{ 2} \mathrm{at}{ }^2$ by using the principle of homogeneity to check the correctness of the following equation $\mathrm{S}=\mathrm{ut}+ \frac{1}{ 2} \mathrm{at}{ }^2$ where $S$ represents distance having a dimension of [L] which is on the left side of the equation. Now coming to the right side, we have u which is velocity having the dimension of $\left[\mathrm{LT}^{-1}\right]$ and $t$ is time having the dimension of [T] so the net dimension of the product of velocity $u$ and time t will be $\mathrm{LT}^{-1} \mathrm{~T}=[\mathrm{L}]$ and similarly another part of right side equation is $0.5 \mathrm{at}^2$ where a is acceleration having the dimension of $\left[\mathrm{LT}^{-2}\right]$ and the dimension of the square of t is $\left[\mathrm{T}^2\right]$ so the net dimension of term $0.5 a t^2$ will be [L]. Hence, the net right-sided equation has a dimension of $\mathrm{L}+\mathrm{L}=[\mathrm{L}]$ since the addition of two dimensions is the same dimension. Hence, the left side and right side of the equation have the same dimension of [L] so, by the principle of homogeneity, the equation $S=u t+0.5 a t^2$ is correct.