Home | | Physics 11th std | Dimensional Analysis

# Dimensional Analysis

All the derived physical quantities can be expressed in terms of some combination of the seven fundamental or base quantities.

DIMENSIONAL ANALYSIS

## Dimension of Physical Quantities

All the derived physical quantities can be expressed in terms of some combination of the seven fundamental or base quantities. These base quantities are known as dimensions of the physical world, and are denoted with square bracket [ ]. The three dimensions in mechanics are [L] for length, [m] for mass and [T] for time. In electricity, [A] is the dimension of electric current. In thermodynamics, [K] is the dimension for the temperature. In optics [cd] or [Φ] is the dimension for luminous intensity. The dimension of amount of substance is [mol].

The dimensions of a physical quantity are the powers to which the units of base quantities are raised to represent a derived unit of that quantity.

For example,

Hence the dimensions of velocity are 0 in mass, 1 in length and -1 in time.

### Dimensional formula and equation

Dimensional formula is an expression which shows how and which of the fundamental units are required to represent the unit of a physical quantity.

For example, [M0LT−2] is the dimensional formula of acceleration.

When the dimensional formula of a physical quantity is expressed in the form of an equation, such an equation is known as the dimensional equation.

Example, acceleration = [M0LT−2]. The dimensional formula of various physical quantities are tabulated in Table 1.11.

## Dimensional Quantities, Dimensionless Quantities, Principle of Homogeneity

On the basis of dimension, we can classify quantities into four categories.

### 1. Dimensional variables

Physical quantities, which possess dimensions and have variable values are called dimensional variables. Examples are length, velocity, and acceleration etc.

### 2. Dimensionless variables

Physical quantities which have no dimensions, but have variable values are called dimensionless variables. Examples are specific gravity, strain, refractive index etc.

### 3. Dimensional Constant

Physical quantities which possess dimensions and have constant values are called dimensional constants. Examples are Gravitational constant, Planck’s constant etc.

### 4. Dimensionless Constant

Quantities which have constant values and also have no dimensions are called dimensionless constants. Examples are π, e, numbers etc.

### Principle of homogeneity of dimensions

The principle of homogeneity of dimensions states that the dimensions of all the terms in a physical expression should be the same. For example, in the physical expression v2 = u2 + 2as, the dimensions of v2, u2 and 2 as are the same and equal to [L2T−2].

## Application and Limitations of the Method of Dimensional Analysis.

This method is used to

i. Convert a physical quantity from one system of units to another.

ii. Check the dimensional correctness of a given physical equation.

iii. Establish relations among various physical quantities.

### (i) To convert a physical quantity from one system of units to another

This is based on the fact that the product of the numerical values (n) and its corresponding unit (u) is a constant. i.e, n [u] = constant (or) n1[u1] = n2[u2].

Consider a physical quantity which has dimension ‘a’ in mass, ‘b’ in length and ‘c’ in time. If the fundamental units in one system are M1, L1 and T1 and the other system are M2, L2 and T2 respectively, then we can write, n1 [M1a L1b T1c] = n2 [M2a L2b T2c]

We have thus converted the numerical value of physical quantity from one system of units into the other system.

### (ii) To check the dimensional correctness of a given physical equation

Let us take the equation of motion

v = u + at

Apply dimensional formula on both sides

[LT−1] = [LT−1] + [LT−2] [T]

[LT−1] = [LT−1] + [LT−1]

(Quantities of same dimension only can be added)

We see that the dimensions of both sides are same. Hence the equation is dimensionally correct.

### (iii) To establish the relation among various physical quantities

If the physical quantity Q depends upon the quantities Q1, Q2 and Q3 ie. Q is proportional to Q1, Q2 and Q3.

Then,

Q α Q1a Q2b Q3c

Q = k Q1a Q2b Q3c

where k is a dimensionless constant. When the dimensional formula of Q, Q1, Q2 and Q3 are substituted, then according to the principle of homogeneity, the powers of M, L, T are made equal on both sides of the equation. From this, we get the values of a, b, c

## Limitations of Dimensional analysis

This method gives no information about the dimensionless constants in the formula like 1, 2, ……..π,e, etc.

This method cannot decide whether the given quantity is a vector or a scalar.

This method is not suitable to derive relations involving trigonometric, exponential and logarithmic functions.

It cannot be applied to an equation involving more than three physical quantities.

It can only check on whether a physical relation is dimensionally correct but not the correctness of the  relation.  For  example  using dimensional analysis, s = ut + 1/3 at2 is dimensionally correct whereas the correct relation is s = ut + 1/2 at2.

## Solved Example Problems

(i) To convert a physical quantity from one system of units to another

Example 1.12

Convert 76 cm of mercury pressure into Nm−2 using the method of dimensions.

Solution

In cgs system 76 cm of mercury pressure = 76 × 13.6 × 980 dyne cm−2

The dimensional formula of pressure P is [ML−1T−2]

Example 1.13

If the value of universal gravitational constant in SI is 6.6x10−11 Nm2 kg−2, then find its value in CGS System?

Solution

Let GSI be the gravitational constant in the SI system and Gcgs in the cgs system. Then

The dimensional formula for G is M−1 L3T −2

(ii) To check the dimensional correctness of a given physical equation

Example 1.14

Check the correctness of the equation

using dimensional analysis method

Solution

Both sides are dimensionally the same, hence the equations

is dimensionally correct.

(iii) To establish the relation among various physical quantities

Example 1.15

Obtain an expression for the time period T of a simple pendulum. The time period T depends on (i) mass ‘m’ of the bob (ii) length ‘l’ of the pendulum and (iii) acceleration due to gravity g at the place where the pendulum is suspended. (Constant k = 2π) i.e

Solution

Here k is the dimensionless constant. Rewriting the above equation with dimensions

Comparing the powers of M, L and T on both sides, a=0, b+c=0, -2c=1

Solving for a,b and c a = 0, b = 1/2, and c = −1/2

From the above equation T = k. m0  l1/2 g−1/2

Example 1.16

The force F acting on a body moving in a circular path depends on mass of the body (m), velocity (v) and radius (r) of the circular path. Obtain the expression for the force by dimensional analysis method. (Take the value of k=1)

where k is a dimensionless constant of proportionality. Rewriting above equation in terms of dimensions and taking k = 1, we have

Comparing the powers of M, L and T on both sides

From the above equation we get

Tags : Dimension of Physical Quantities, Application and Limitations, Solved Example Problems , 11th Physics : UNIT 1 : Nature of Physical World and Measurement
Study Material, Lecturing Notes, Assignment, Reference, Wiki description explanation, brief detail
11th Physics : UNIT 1 : Nature of Physical World and Measurement : Dimensional Analysis | Dimension of Physical Quantities, Application and Limitations, Solved Example Problems