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# Dimensional Analysis and Its Applications

Dimensional analysis answers some very interesting questions. Which is more, one meter or one second? Are they even comparable? Can you add a kilogram and a dozen eggs to each other? No, right? After studying this section, you will be able to understand how dimensional analysis answers such questions. Let’s begin!

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## Dimensional Analysis

Dimensional analysis is the use of dimensions and the dimensional formula of physical quantities to find interrelations between them. It is based on the following facts:

### The Physical laws

The physical laws are independent of the units in which a quantity is measured.If   $$n_1 a_1$$ is the measured value of a physical quantity in one system of units and $$n_2 a_2$$ is the value in another system of units then, from the above reasoning, these two must be equal.

$$n_1 a_1$$ = $$n_2 a_2$$    …… (1)

### The principle of Homogeneity

The equations depicting physical situations must have the same dimensions throughout. If two sides of an equation have different dimensions, that equation can’t represent any physical situation. This is known as the Principle of Homogeneity. For example, if

$$[M]^{a}$$ $$[L]^{b}$$ $$[T]^{c}$$ = $$[M]^{x}$$ $$[L]^{y}$$ $$[T]^{z}$$

then from the principle of Homogeneity, we have:

a = x; b = y; c = z

## Applications of the Dimensional Analysis

### Conversion of units

The dimensions of a physical quantity are independent of the system of units used to measure the quantity in. Let us suppose that  $$M_1$$ , $$L_1$$ and $$T_1$$ and $$M_2$$ , $$L_2$$ and $$T_2$$ are the fundamental quantities in two different systems of units. We will measure a quantity Q (say) in both these systems of units. Suppose, a, b, c be the dimensions of the quantity respectively.

In the first system of units, Q = $$n_1$$ $$u_1$$ = $$n_1$$ [ $$M_1^a$$ $$L_1^b$$ $$T_1^c$$ ]      …. (2)

In the second system of units, Q = $$n_2$$ $$u_2$$ = $$n_1$$ [ $$M_2^a$$ $$L_2^b$$ $$T_2^c$$ ] …. (3)

$$n_1$$ [$$M_1^a$$ $$L_1^b$$ $$T_1^c$$] = $$n_2$$ [ $$M_2^a$$ $$L_2^b$$ $$T_2^c$$ ]   … (4)  [using (1)]

Substitution of the respective values will give the value of $$n_1$$ or $$n_2$$

### Checking the consistency of an equation

All the physical equations must be consistent. The reverse may not be true. For example, the following equations are not consistent because of the dimensions of the L.H.S. ≠ Dimensions of the R.H.S.

F = $$m^2$$× a

F = m × $$a^2$$

### Finding relations between physical quantities in a physical phenomenon

The principle of  Homogeneity can be used to derive the relations between various physical quantities in a physical phenomenon. Let us see it with the help of an example.

## Solved Example for You

Example 1: The Time period (T) of a simple pendulum is observed to depend on the  following factors:

• length of the pendulum (L),
• mass of the Bob (m)
• acceleration due to gravity (g)

Sol. Let T ∼  Lαmβgγ or …… (5)

[T] = [L]α[M]β[L]γ[T]-2γ or

Solving the above for α, β and γ we have:

β = 0; α + γ = 0;

-2γ = 1 or γ= -1/2

Using these in equation (5), we have:

T ∼ $$\sqrt[]{L/g}$$

Example 2: Convert 1 J to erg.

Sol. Joule is the S.I. unit of work. Let this be the first system if units. Also erg is the unit of work in the cgs system of units. This will be the second system of units. Also $$n_1$$ = 1J and we have to find the value of $$n_2$$

From equation (4), we have:

[$$M_1^a$$ $$L_1^b$$ $$T_1^c$$] = $$n_2$$ [ $$M_2^a$$ $$L_2^b$$ $$T_2^c$$ ]

The dimensional formula of work is  [$$M^1$$ $$L^2$$ $$T^{-2}$$]. As a result a = 1, b = 2 and c = -2.

[$$M_1^1$$ $$L_1^2$$ $$T_1^-2$$] = $$n_2$$ [ $$M_2^1$$ $$L_2^2$$ $$T_2^{-2}$$ ]

Also $$M_1$$ = 1 kg , $$L_1$$ = 1 m,  $$T_1$$=1 s

and $$M_2$$ = 1 g , $$L_2$$ = 1 cm,  $$T_2$$=1 s

$$n_2$$ = [$$M_1^1$$ $$L_1^2$$ $$T_1^{-2}$$] / [ $$M_2^1$$ $$L_2^2$$ $$T_2^{-2}$$ ]

= [$$1kg^1$$ $$1g^2$$ $$1s^{-2}$$] /  [ $$1g^1$$ $$1cm^2$$ $$1s^{-2}$$ ]

$$n_2$$ = $$10^7$$

Hence, 1J = $$10^7$$ erg

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