Measurement:
Need of measurement:
The prime goal of physics is to study the nature by developing suitable theories based on experiment which is possible through qualitative and quantitative measurement. We need measurement to develop suitable theories and study the nature. In the measurements certain amount of matter is taken as standard and we find the number that express how many times that standard unit is in that physical quantity. In general, physical quantity is written as Q = nu,
where, n = number, u = unit of the measurement
Certain units (quality of matter) which is accepted and used throughout the world is called as standard unit.

Characteristics:
i. They are independent and well defined.
ii. They are same throughout the world.
iii. They shouldn’t affected by external factor.
iv. They must be reproducible.

S.No.	Fundamental Quantity	Fundamental Units	Symbol
1.	Mass	Kilogram	kg
2.	Length	Metre	m
3.	Time	Second	s
4.	Temperature	Kelvin	K
5.	Amount of Substance	Mole	mol
6.	Luminous intensity	Candela	Cd
7.	Electric Current	Ampere	A

Supplementary Units:

S.No.	Supplementary Quantity	Unit	Symbol
1.	Plane angle	Radian	Rd
2.	Solid angle	Steradian	Sr

Dimension:
Dimensions of a physical quantity are the powers to be raised on the fundamental units in order to obtain the desired unit of that physical quantity.
For dimension of a physical quantity mass, length, time, temperature, amount of substance, luminous intensity and electric current are represented by M, L, T, K, Mol, Cd, and A respectively.
Examples:
1) Area (A)
A = l x b
Dimensional formula of area (A) = [L] x [L]

[latex][M^oL^2T^o][/latex]
Dimensions of area are 0, 2 and 0 in mass, length and time respectively.
2) Volume (V)
V = l x b x h
= [latex][M^oL^3T^o][/latex]
Dimension of volume are 0, 3 and 0 in mass, length and time respectively.
3) Density (D) = [latex]\frac{M}{V}[/latex]
D = [latex][ML^{-3}T^o][/latex]
Dimensional formula of a physical quantity is an expression that tells us how and which fundamental units are needed in order to obtain the desired units of that physical quantity.
4) Force (F)
F = m x a
= [latex][MLT^{-2}][/latex]
The dimensional formula of force is [latex][MLT^{-2}][/latex].
Two things are clear from the expression that unit of force depends on mass, length and time.
Unit of force varies as directly as mass and length and inversely as square of time.
Dimensional formula of physical quantity:

S.No.	Physical quantities	Relations	Dimensional formula
1.	Area	A = l x b	[MoL2To]
2.	Volume	V = l x b x h	[MoL3To]
3.	Density	D = M/V	[ML-3To]
4.	Velocity/ speed	v = d/t	[MoLT-1]
5.	Acceleration	a = v/t	[MoLT-2]
6.	Force	F = m x a	[MLT-2]
7.	Linear momentum	P = mv	[MLT-1]
8.	Pressure	P = F/A	[ML-1T-2]
9.	Work or energy	W = Fd	[ML2T-2]
10.	Power	P = W/t	[ML2T-3]

Demerit:

1. It is not clear about the constant of the physical quantity.
2. It is not clear about scalar or vector product.

Dimensional equations:

If the given relation is expressed in terms of dimension, then, it is called dimensional equation.

For e.g.  s = [latex]ut+\frac{1}{2}at^2[/latex]

L = [LT^-1] [T] + [LT^-2][T²]

[M⁰LT⁰] = [M⁰LT⁰] + [M⁰LT⁰]

Dimensionally correct relation may not be physically correct relation.

Principle of homogeneity:

It states that a given relation is dimensionally correct if and only if each and every terms on both side has the same dimensions.

i. S = ut + at²

Dimension of LHS = [M⁰LT⁰]

Dimension of RHS = [LT^-1] [T] + [LT^-2] [T²]

                        = [M⁰LT⁰] + [M⁰LT⁰]

Since, the dimension of each and every terms on both side are same, the given relation is dimensionally correct.

Q. If dimensionally correct relation is physically correct, what about dimensionally incorrect/wrong relation?

No, dimensionally correct relation does not necessarily be correct relation because in the above example, the terms of LHS and RHS are correct in dimension but it is not correct relation rather the correct relation is, [latex]S = ut + \frac{1}{2}[/latex] at²

ii. v³ = u² + 2at

Dimension of LHS = [M^oLT^-1]³ = [M^oL³T^-3]

Dimension of RHS = [LT^-1]² + [LT^-2] [T]

                        = [M^oL²T^-2] + [M^oLT^-1]

Since, the dimensions of both sides are not the same. It is incorrect relation.

Uses of dimensional analysis:

A. To check the correctness of the given relation.

By using homogenetic principle, we can check the correctness of the given physical relation. If the dimension of each and every term on both side are the same, then, the given relation is dimensionally correct.

For e.g. s = ut + [latex]\frac{1}{2}at^2[/latex]

L = [LT^-1] [T] + [LT^-2][T²]

[M⁰LT⁰] = [M⁰LT⁰] + [M⁰LT⁰]

B. To obtain dimension of constant.

By using homogenetic principle, we can find dimensional constant from the given relation.

a. Dimension of G

We have,

F = [latex]\frac{GM_1M_2}{R^2}[/latex]

G = [latex]\frac{FR^2}{M_1M_2}[/latex]

Now, G = [latex]\frac{[MLT^{-2}][L^2]}{[M][M]}[/latex] = [M^-1L³T^-2]

b. Dimension of coefficient of viscosity ([latex]\eta[/latex]):

We have, F = 6π[latex]\eta[/latex]rv where,

F = viscous force

[latex]\eta[/latex] = coefficient of viscosity

r = radius

v = velocity

Or, [latex]\eta = \frac{F}{6\pi rv}[/latex]

            = [latex]\frac{[MLT^{-2}]}{[L][LT^{-1}]}[/latex]

            = [ML^-1T^-1]

C. To obtain relation between different physical quantities.

By using homogeneity principle, we can obtain the relation between different physical quantities from the given statement.

Example: The time taken by a freely falling body from a certain height depends on its mass (m), height (h) and acceleration due to gravity (g). Mathematically,

t ∝ m^ah^bg^c

Or, t = K m^ah^bg^c ………….. (1)

Where, K is dimensionally dimensionless proportionality constant. a, b and c are unknowns to be determined. Dimensionally eqⁿ(1) must be correct, i.e.

Dimension of LHS = Dimension of RHS

[T] = [M]^a[L]^b[M^oLT^-2]^c

Or, [M^oL^oT] = [M^aL^b+cT^-2c]

Equating value of M, L and T

a = 0, b + c = 0

Or, b = – c

And, c = -1/2

So, a = 0, b = ½ and c = -1/2

Substituting the values of a, b and c in equation (1), we get,

t = Km^oh^1/2g^-1/2

            = k [latex]\sqrt{\frac{h}{g}}[/latex]

Experimentally, value of k is [latex]\sqrt{2}[/latex]. Therefore, t = [latex]\sqrt{\frac{2h}{g}}[/latex]

D. To convert one system of measurement into another.

Generally, a physical quantity is written as Q = n.u ……………… (1)

Where, n is numeric value

U is unit of measurement

Let, n₁ and n₂ be the numerical value of physical quantity in two system of measurements.

Where, u₁ and u₂ are units respectively.

Q = n₁u₁ = n₂u₂………….. (2)

Let, a, b and c be the dimensions of physical quantity in terms of MLT. Then,

[u₁] = [M₁^aL₁^bT₁^c]

[u₂] = [M₂^aL₂^bT₂^c]

Then, from eq^n. (2), we get,

n₁ [M₁^aL₁^bT₁^c] = n₂ [M₂^aL₂^bT₂^c]

E.g.: Convert 1 Newton into dynes:

In SI system

n₁= 1

M₁ = 1 Kg = 1000 g

L₁ = 1m = 100cm

T₁ = 1s

In CGS system

n₂ = ?

M₂ = 1g

L₂ = 1cm

T₂ = 1s

Dimensional formula of force is

[F] = [MLT^-2]

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

We have,

n₂= n₁ [latex][\frac{M_1}{M_2}]^a[\frac{L_1}{L_2}]^b[\frac{T_1}{T_2}]^c[/latex]

            = 1 [latex][\frac{1000g}{1g}]^1[\frac{100cm}{1cm}]^1[\frac{1s}{1s}]^{-2}[/latex]

            = 10⁵

∴ 1N = 10⁵ dynes.

Limitations of dimensional analysis:

i. It doesn’t tell us the value of constants.

ii. It does not tell us whether the physical quantity is scalar or vector.

iii. We cannot obtain the relation between physical quantities by dimensional analysis if the proportionality constant has dimensions.

iv. By dimensional analysis, we cannot obtain the relation between different physical quantities, if the relation contains trigonometric ratios.

v. By dimensional analysis, we cannot obtain the relation between different physical quantities if the relation contains more than four different physical quantities.

Classification of physical quantities:

On the basis of dimensional analysis, all physical quantities are classified into four group. i.e.

1. Dimensional variables: Those physical quantities which have fixed dimension but not fixed value. For e.g. velocity, force, acceleration, energy, mass, etc.

2. Dimensionless variable: Those physical quantities which do not have dimension and no fixed value. E.g. ratio, angle, strain.

3. Dimensionless constant: Those physical quantities which have no dimension and fixed value is called dimensionless constant. E.g., relative permittivity of water.