Basic quantity is defined as a quantity which cannot be derived from any physical quantities

Quantity Symbol SI Unit Symbol

Length l metre m

Mass m kilogram kg

Time t second s

Temperature T/ kelvin K

Electric current I ampere A

Amount of substance N mole mol

Luminous Intensity candela cd

Derived quantity is defined as a quantity which

can be expressed in term of base quantity

PHYSICS CHAPTER 1

d. 29 cm = ? in

e. 12 mi h-1 = ? m s-1

cm 1

in cm 29cm 29 2.54

1

in .411cm 29

s 3600

h 1

km 1

m 1000

mi 1

km 1.609

h 1

mi 12h mi 12 1

11 s m365h mi 12 .

Learning Outcome:At the end of this chapter, students should be able to:

Use dimensional analysis to check homogeneity and construct equation of physics.

Dimensional AnalysisDimension is defined as a technique or method technique or method which the physical quantityphysical quantity can be expressed in terms of expressed in terms of combination of basic quantitiescombination of basic quantities.It can be written as

[physical quantity or its symbol]

Table 1.5 shows the dimension of basic quantities.

[Basic Quantity] Symbol Unit

[mass] or [m] M kg

[length] or [l] L m

[time] or [t] T s

[electric current] or [I] A @ I A

[temperature] or [T] K

[amount of substance] or [N]

N mole

PHYSICS CHAPTER 1

Dimension can be treated as algebraic quantities through the procedure called dimensional analysis.

The uses of dimensional analysis are to determine the unit of the physical quantitydetermine the unit of the physical quantity. to determine whether a physical equation is dimensionally determine whether a physical equation is dimensionally

correct or not by using the principle of homogeneitycorrect or not by using the principle of homogeneity.

to derive/construct a physical equationderive/construct a physical equation. Note:

Dimension of dimensionless constant is 1Dimension of dimensionless constant is 1,

e.g. [2] = 1, [refractive index] = 1 Dimensions cannot be added or subtracted.cannot be added or subtracted. The validity of an equation cannot determined by dimensional

analysis. The validity of an equation can only be determined by experiment.

Dimension on the L.H.S. = Dimension on the R.H.SDimension on the L.H.S. = Dimension on the R.H.S

PHYSICS CHAPTER 1

Determine a dimension and the S.I. unit for the following quantities:

a. Velocity b. Acceleration c. Linear momentum

d. Density e. Force

Solution :Solution :

a.

The S.I. unit of velocity is m sm s11.

Example 1.2 :

interval time

ntdisplacemein changeVelocity

t

sv

or

1LTT

L v

PHYSICS CHAPTER 1

b.

Its unit is m sm s22.

d.

S.I. unit : kg mkg m33.

t

va

2LT a

T

LT 1

a

vmp

1MLT p

1LTM p

c.

S.I. unit : kg m skg m s11.

V

mρ

3MLρ

hwl

mρ

LLL

M

ρ

amF

2MLT F

2LTM F

e.

S.I. unit : kg m skg m s22.

PHYSICS CHAPTER 1

Determine Whether the following expressions are dimensionally correct or not.

a. where s, u, a and t represent the displacement, initial velocity, acceleration and the time of an object respectively.

b. where t, u, v and g represent the time, initial velocity, final velocity and the gravitational acceleration respectively.

c. where f, l and g represent the frequency of a

simple pendulum , length of the simple pendulum and the gravitational acceleration respectively.

Example 1.3 :

222 atuts

gtuv 222

l

g

πf

2

1

PHYSICS CHAPTER 1

Solution :Solution :

a. Dimension on the LHS :

Dimension on the RHS :

Dimension on the LHS = dimension on the RHS

Hence the equation above is homogeneoushomogeneous or dimensionally correct.dimensionally correct.

b. Dimension on the LHS :

Dimension on the RHS :

Thus

Therefore the equation above is not homogeneous not homogeneous or dimensionally dimensionally

incorrect.incorrect.

L22 ss

LTLT122 1 tuut

LTLT 2222 taat

and

22212 TLLT v

22212 TLLT u

12 LTTLT122 tggtand

gtuv 222

PHYSICS CHAPTER 1

Solution :Solution :

c. Dimension on the LHS :

Dimension on the RHS :

Therefore the equation above is homogeneoushomogeneous or dimensionally dimensionally

correct.correct.

1T1

T

f

21

21

2

1

2

1

lg

πl

g

π

12 TLLT1 21

21

l

g

πf

2

1

PHYSICS CHAPTER 1

The period, T of a simple pendulum depends on its length l, acceleration due to gravity, g and mass, m. By using dimensional analysis, obtain an equation for period of the simple pendulum.

Solution :Solution :Suppose that :

Then

where k, x, y and z are dimensionless constants.

Example 1.4 :

zyx mglT zyx mgklT

zyx mglkT

zyx MLTL1T 2zyyx MTLT 2zyyx MTLMTL 2010

……………………………………(1)(1)

PHYSICS CHAPTER 1

By equating the indices on the left and right sides of the equation, thus

By substituting eq. (3) into eq. (2), thus

Replace the value of x, y and z in eq. (1), therefore

The value of k can be determined experimentally.

0 yx12 y

21y

0z

021 x

21x

……………………………………(2)(2)

……………………………………(3)(3)

021

21

mgklT

g

lkT

PHYSICS CHAPTER 1

Determine the unit of in term of basic unit by using the equation below:

where Pi and Po are pressures of the air bubble and R is the radius of

the bubble.

Solution :Solution :

Example 1.5 :

212

2

TMLL

MLT

A

am

A

FP

LR

RPP oi

2

PHYSICS CHAPTER 1

Since thus

Therefore the unit of is kg skg s-2-2

oi PPRγ 2

1

PPP oi

oi PPRγ

2

1

21TMLL1 γ

PRγ

2

1

2MT γ

PHYSICS CHAPTER 1

1. Deduce the unit of (eta) in term of basic unit for the equation below:

where F is the force, A is the area, v is the change in velocity

and l is the change in distance.

ANS. : kg mANS. : kg m-1-1 s s-1-1

A sphere of radius r and density s falls in a liquid of density f. It

achieved a terminal velocity vT given by the following expression:

where k is a constant and g is acceleration due to gravity.

Determine the dimension of k.

ANS. : M LANS. : M L-1-1 T T-1-1

Exercise 1.1 :

Δl

Δvη

A

F

fs

2

T ρρk

grv

9

2

PHYSICS CHAPTER 1

3. a. What is meant by homogeneity of a physical equation?

b. The escape velocity, v for a tomahawk missile which escape the gravitational attraction of the earth is depend on the

radius of the earth, r and the acceleration due to gravity, g. By using dimensional analysis, obtain an expression for escape

velocity, v.

ANS. :ANS. :

4. Show that the equation below is dimensionally correct.

Where R is the inside radius of the tube, L is its length, P1-P2 is

the pressure difference between the ends, is the coefficient of

viscosity ( N s m-2) and Q is the volume rate of flow ( m3 s-1).

Exercise 1.1 :

grkv

ηL8

PPπRQ 21

4

PHYSICS CHAPTER 1

5. Acceleration is related to velocity and time by the following expression

Determine the x and y values if the expression is dimensionally consistent.

ANS. : x= 1; y= ANS. : x= 1; y= 11

6. Bernoulli’s equation relating pressure P and velocity v of a fluid moving in a horizontal plane is given as

where is the density of the fluid and k is a constant. Determine

the dimension of the constant k and its unit in terms of basic units.

ANS. : M LANS. : M L11 T T22; kg m; kg m11 s s22

Exercise 1.1 :

kvP 2

2

1

yxtva