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Principle of Homogeneity - JEE Important Topic

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Dimensional Analysis and Principle of Homogeneity

Before we talk about the principle of homogeneity, you should have a clear idea about what dimensional analysis is! Dimensional analysis refers to ‘the study of the relationship between physical quantities with the help of dimensions and the units of measurement’. What makes dimensional analysis a necessity in the world of physical quantities is that through this technique, mathematical calculations become a lot simpler, as there is no need to convert the units of the physical quantities in question. In this theory, we implement the concept of conversion factors to further simplify the calculations. 


Let us take the following example. According to the metric system, 1 kg = 1000 gm. Therefore, if we have to convert 5kg in to gm, then the procedure is as follows:

$\because 1\,kg=1000\,gm$

$\therefore 5\,kg=5\times 1000\,gm=5000\,gm$

As you can see here, 1000 is the conversion factor (c.f.) and all conversions from kilograms to grams can be done by simply multiplying the larger value with the c.f., making the calculations easy to understand and perform.


The principle of homogeneity of dimensions essentially deals with the conversion of units from one quantity to another, easily and effectively, and is one of the most important concepts of dimensional analysis.


Definition and Explanation of the Principle of Homogeneity of Dimensions

A very valid question here would be ‘what is dimensional homogeneity’! To state the principle of homogeneity, we must understand the basic concept that in case of dimensional equations, all terms must be equal on both sides of the equation. Thus, the principle of homogeneity states that, ‘An equation is dimensionally correct if the dimensions of the various terms on either side of the equal sign  (=) must be the same’.


Consider the following principle of homogeneity of dimensions,


Example: $s=ut+\dfrac{1}{2}a{{t}^{2}}$

Here, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t stands for time. 


To check the accuracy of the above given example, we implement the principle of homogeneity of dimensions.


Solution: 

Given, $L.H.S=s$ and $R.H.S.=ut+\dfrac{1}{2}a{{t}^{2}}$.

Dimensional formula of the $L.H.S.=s={{L}^{1}}{{M}^{0}}{{T}^{0}}\,\ldots \ldots \left( 1 \right)$

Dimensional formula of the $R.H.S$ $=ut+a{{t}^{2}}$

$=[{{L}^{1}}{{M}^{0}}{{T}^{-1}}]\left[ {{L}^{0}}{{M}^{0}}{{T}^{1}}\left] + \right[{{L}^{1}}{{M}^{0}}{{T}^{-2}} \right]\left[ {{L}^{0}}{{M}^{0}}{{T}^{2}} \right]$

After simplifying, we get: $R.H.S.=[{{L}^{1}}{{M}^{0}}{{T}^{0}}]\,\,\ldots \ldots \left( 2 \right)$

Therefore, from equations (1) and (2), we see that $L.H.S.=R.H.S.=[{{L}^{1}}{{M}^{0}}{{T}^{0}}]$

Let us consider the dimensions on the right-hand side to be A and B respectively, and C be the dimensions on the left and side.


This means that, for A+B=C, the equation only holds true when the dimensions on the left-hand side are equivalent to the dimensions on the right-hand side.

Application of the Principle of Homogeneity

There are certain points of application of the principle of dimensional homogeneity. These key areas are as follows:

  • In the case of an equation, where the dimensions of a certain quantity are unknown, the principle of homogeneity is applied to find out the unknown value.

  • Principles of homogeneity and dimensional analysis are extensively used to check the accuracy of the relation between two or more quantities.

  • This method of analysis is extensively used in cases where there is a requirement of converting units from one system to another.

  • Lastly, this technique is also implemented to find out the dimensions of a particular physical quantity, when the factors or the quantities it depends on, are known. 


Some Physical Quantities and Their Dimensions

The dimensions of some of the most common physical quantities are as follows:ML

Sr. No.

Physical Quantity

Dimensions


Energy

$\left[ML^2T^{-2}\right]$


Angular momentum

$\left[ML^{2}T^{-1}\right]$


Pressure

$\left[ML^{-1}T^{-2}\right]$


Power

$\left[ML^{2}T^{-3}\right]$


Electric charge

$\left[IT\right]$


Electric field

$\left[MLT^{-3}I^{-1}\right]$


Magnetic Field

$\left[MT^{-2}I^{-1}\right]$


Principle of Homogeneity of Dimensions Examples

Here, let us take some solved principle of homogeneity examples.


Example 1: Derive the unit of force by applying the principle of homogeneity

Solution: As we know, Force,

$F=m\times a$

$\left(\text{Here, m= mass and a= acceleration, as per Newton’s second law of motion} \right)$

As we know, Dimension of mass = M and the dimension of a = LT-2

$\therefore F=m\times a=ML{{T}^{-2}}$


From the dimension of force, as derived, we can easily find out the units of force which is = kg.m.s-2. Thus, the S.I. the unit of force is $\dfrac{kg.m}{{s}^{2}}$ or $\text{Newton}$.


1 Newton refers to the force which when applied to a body weighing $1kg$, produces an acceleration of $1\,\dfrac{m}{{s}^{2}}$.


Example 2: Evaluate the homogeneity of the equation when the rate of flow of a liquid has a coefficient of viscosity, $\eta $ through a capillary tube of length l and radius r. Let p denote the pressure.

Solution:  According to the formula of viscosity, $\dfrac{dV}{dt}=\dfrac{\pi p{{a}^{4}}}{8l\eta }$

Taking the dimensional analysis of the L.H.S we get, 

$L.H.S.=\dfrac{dV}{dt}=\dfrac{[{{M}^{0}}{{L}^{3}}{{T}^{0}}]}{[{{M}^{0}}{{L}^{0}}{{T}^{-1}}]}$

$\therefore L.H.S.=[{{M}^{0}}{{L}^{3}}{{T}^{-1}}]\,\,\ldots \ldots \left( 1 \right)$

Taking the dimensional analysis of the R.H.S. we get, 

$R.H.S.=\dfrac{[p]{{[a]}^{4}}}{[l][\eta]}=\dfrac{[{{M}^{1}}{{L}^{-1}}{{T}^{-2}}]{{[{{M}^{0}}{{L}^{1}}{{T}^{0}}]}^{4}}}{[{{M}^{0}}{{L}^{1}}{{T}^{0}}][{{M}^{1}}{{L}^{-1}}{{T}^{-1}}]}$

Further simplifying, we get, $R.H.S.=\dfrac{[{{M}^{1}}{{L}^{3}}{{T}^{2}}]}{[{{M}^{1}}{{L}^{0}}{{T}^{-1}}]}=[{{M}^{0}}{{L}^{3}}{{T}^{-1}}]\,\,\ldots \ldots \left( 2 \right)$ 

From equations (1) and  (2) we get that L.H.S = R.H.S.


Therefore, the equation is homogeneous as proved by applying the principle of homogeneity. 


Conclusion

Dimensional analysis is the method through which we can derive the dimension of an unknown quantity and establish a relationship between two or more physical quantities. To state the principle of homogeneity, we can enunciate that an equation is dimensionally correct when the dimensions of the L.H.S and the R.H.S are equal. 


The chief regions of application of this theory are in cases when you need to convert the units of a physical quantity from one system to another, so as to find out the dimensions of an unknown quantity when the other factors are given. Principle of homogeneity is also implemented to find out the units of an unknown quantity.

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FAQs on Principle of Homogeneity - JEE Important Topic

1. What are the drawbacks of dimensional analysis?

Firstly, dimensional analysis cannot be used to find the values of dimensionless constants. Secondly, it does not give any hint regarding whether the quantity in question is a scalar quantity or a vector with both magnitude and direction. Dimensional analysis cannot be applied or used to derive expressions containing trigonometric functions, logarithmic functions, etc. If a physical quantity is depending upon more than three factors that are having dimensions, then deriving a formula or a relation would become very difficult.

2. What are the three major areas of application of dimensional analysis?

Dimensional analysis and the principle of homogeneity have three major areas of application. They are as follows:

  • Dimensional analysis is used to check the accuracy and the preciseness of a dimensional equation.

  • This technique is used to convert one system of units to another.

  • Dimensional analysis is extensively used to find out the dimensions of unknown quantities and establish a relationship between two or more physical quantities.  

This basically helps a person to look into any object in a mathematical way.