Dimensional analysis and its application

While using addition and subtraction for two or more physical quantities, we notice that only those quantities can be added or subtracted that have similar dimension. On the other hand, in the case of multiplication and division, when two quantities are multiplied, along with their magnitudes, the dimensions are also treated accordingly, in order to find the dimension of the product. In mathematical equations, the physical quantities on either side of the equation are required to be of the same dimension. The dimensional analysis is thus very important when dealing with these physical quantities. In this section, we will learn about some more applications of the dimensional analysis.

Checking the dimensional consistency

As we know, only similar physical quantities can be added or subtracted, thus two quantities having different dimensions cannot be added together. For example, we cannot add mass and force or electric potential and resistance. For any given equation, the principle of homogeneity of dimensions is used to check the correctness and consistency of the equation. The dimensions of each component on either side of the sign of equality are checked, and if they are not the same, the equation is considered wrong.

Let us consider the equation given below,

Dimensional Analysis

The dimensions of the LHS and the RHS are calculated

Dimensional Analysis

As we can see the dimensions of the LHS and the RHS are the same, hence, the equation is consistent.

Deducing the relation among physical quantities

Dimensional analysis is also used to deduce the relation between two or more physical quantities. If we know the degree of dependence of a physical quantity on another, that is the degree to which one quantity changes with the change in another, we can use the principle of consistency of two expressions to find the equation relating these two quantities. This can be understood more easily through the following illustration.

Example: Derive the formula for centripetal force F acting on a particle moving in a uniform circle.

As we know, the centripetal force acting on a particle moving in a uniform circle depends on its mass m, velocity v and the radius r of the circle. Hence, we can write



Writing the dimensions of these quantities,


As per the principle of homogeneity, we can write,

a = 1, b+c = 1 and b = 2

Solving the above three equations we get, a = 1, b = 2 and c = -1.

Hence, the centripetal force F can be represented as,


Commonly Asked Questions on Dimension Analysis

Some of the FAQ’s related to this topics are discussed below.

1) What is the meaning of dimension in physics?

Ans – It is an expression that relates derived quantity to fundamental quantities. But it is not related to the magnitude of the derived quantity.

2) What is the dimension of force?

Ans – We know, \(F\) = \(ma\) —– (1)

Mass is fundamental quantity but acceleration is a derived quantity and can be represented in terms of fundamental quantities.

\(a\) = \([LT^{-2}]\) —– (2)

Using (1) and (2),

\(F\) = \([M LT^{-2}]\)

This is the dimension of force.

3) What is dimensional analysis?

Ans – Dimensional analysis is based on the principle that two quantities having same dimensions can only be compared with one another. For example, I can compare kinetic energy with potential energy and say they equal or one is greater than another because they have the same dimension. But I cannot kinetic energy with force or acceleration as their dimensions are not same.

4) How to do demonstrate dimensional analysis with an example?

Ans – Suppose I have the following equation,

\(F\) = \(E^a~ . ~V^b~ .~ T^c\)

Where, \(F\) = Force; \(E\) = Energy; \(V\) = Velocity; \(M\) = Mass

We need to find the value of \(a\), \(b\) and \(c\).

Following are the dimensions of the given quantities,

\(F\) = \([MLT^{-2}]\), \(E\) = \([ML^2T^{-2}]\), \(V\) = \([LT^{-1}]\)

According to dimensional analysis the dimension of RHS should be equal to LHS hence,

\([MLT^{-2}]\) = \([ML^2T^{-2}]^a~ . ~[LT-1]^b~ . ~[T]^c\)

\([MLT^{-2}]\) = \([M^a~ L^{2a+b}~ T^{-2a-b+c}]\)

Now we have three equations,

\(a\) = \(1\)

\(2a+b\) = \(1\)

\(-2a~-~b~+~c\) = \(-2\)

Solving the three equations we get,

\(a\) = \(1\), \(b\) = \(-1\) and \(c\) = \(-1\).

5) What are some of the best books to refer for unit and dimensions in physics?

Ans – When it comes to books for JEE preparation H.C. Verma is definitely one of the best books. Apart from that, books by Arihant Publication also have good dimensional analysis practice problems.

To learn more about dimensional analysis and its applications, download Byju’s The Learning App.

Practise This Question

A dimensionless quantity