experiment

Careful observations

precise

Accurate measurement

Quantities that can be measured

Quantities that can be measured

Examples:LengthMassTimeWeightElectric currentForceVelocityenergy

50 kg

Num

eric

al v

alue

unit

Describing a physical quantity:

Different units can be used to describe the same quantity

Example : The height of a person

can be expressed

in feetIn inchesIn metres……

unit

Is the standard use to compare different magnitudes of the same

physical quantity

Quantity SI unit Symbol

Length metre m

Mass kilogram kg

Time second s

Electric current Ampere A

Thermodynamic temperature

Kelvin K

Amount of substance mole mol

Light intensity Candela cd

Base quantity

standard

International Prototype Metre standard bar made of platinum-iridium. This was the standard until

1960

The metre is the length equal to 1650763.73 wavelengths in    vacuum of the radiation corresponding to the transition between the levels 2p10 and 5d5 of the krypton-86 atom

In 1960

The metre is the length of the path travelled by light in vacuum during a time interval of 1⁄299792458 of    a second.

In 1975

……

……

……

…

Prefix Factor Symbol

pico 10-12 p

nano 10-9 n

micro 10-6 µMilli 10-3 m

centi 10-2 c

deci 10-1 d

kilo 103 k

Mega 106 M

Giga 109 G

Tera 1012 T

Derived quantities

Is a combination of different base quantities

Derived unit = the unit for derived quantity

= is obtained by using the relation between the derived quantities and base quantities

Derived quantity Derived unit

Area m2

Volume m3

Frequency Hz

Density kg m-3

Velocity m s-1

Acceleration m s-2

Force N or kg m s-2

Pressure Pa

Derived quantity Derived unit

Energy or Work J or Nm

Power W or J s-1

Electric charge C or As

Electric potential V or J C-1

Electric intensity V m-1

Electric resistance Ω Or V A-1

Capacitance F or C V-1

Heat capacity J K-1

Specific heat capacity J kg-1 K-1

Dimensions of Physical quantities

Is the relation between the physical quantities and the base physical quantities

Example 1 :

Dimension of area = Area

Length x Breadth =

= L x L= L2

Unit of area = m2

Example 2 :

Dimension of velocity = Velocity

Displacement Time =

= LT

= L T-1

Unit of velocity = m s-1

Example 3 :

Dimension of acceleration = Acceleration

Change of velocity Time =

= L T-1

T= L T-2

Unit of acceleration = m s-2

Example 4 :

Dimension of force = Force

Mass x Acceleration =

= M x L T-2

= M L T-2

Unit of force = kg m s-2 or N

Example 5 :

Dimension of energy = Energy

Force x Displacement =

= M L T-2 x L= M L2 T-2

Unit of energy = kg m2 s-2 or J

Example 5 :

Dimension of energy = Energy

Mass x Velocity x Velocity =

= M x L T-1 x L T-1

= M L2 T-2

Unit of energy = kg m2 s-2 or J

OR

Example 6 :

Dimension of electric charge = Charge

Current x Time =

= A x T= A T

Unit of area = A s

Example 7 :

Dimension of frequency = Frequency

1 .Period =

= 1T

= T-1

Unit of frequency = s-1

Example 8 :

Dimension of strain = Strain

Extension Original length=

= LL

= 1

Unit of frequency = no unitdimensionless

Uses of Dimensions

Dimensional homogeneity of a physical equation

1st Physical quantity

2nd Physical quantity+_=

Both have same dimensions

Example :

v2 = u2 + 2as

v2 = L T-1 2 = L2 T-2

u2 = L T-1 2 = L2 T-2

2as = L T-2 = L2 T-2L

Every term has the same dimension

The equation is dimensionally consistent

Case 1

Example :

v = u + 2as

v = L T-1 = L T-1

u = L T-1 = L T-1

2as = L T-2 = L2 T-2L

[v]=[u]≠

[2as]

The dimension is not consistent

The equation is incorrect.

Case 2

A physical equation whose dimensions are

consistent need not necessary be correct :

Example :

v2 = u2 + as

v2 = L T-1 2 = L2 T-2

u2 = L T-1 2 = L2 T-2

as = L T-2 = L2 T-2L

Every term has the same dimension

The equation is dimensionally consistent

Case 3The constant of

proportionality is wrong

The equation is incorrect.

Example :

v2 = u2 + 2as +

v2 = L T-1 2 = L2 T-2

u2 = L T-1 2 = L2 T-2

2as = L T-2 = L2 T-2L

Every term has the same dimension

The equation is dimensionally consistent

Case 4Has extra term

The equation is incorrect.

s2

t2

s2

t2 =

L2

T2 = L2 T-2

The truth of a physical equation can be

confirmed experimentally :

Example :The period of vibration t of a tuning fork depends on

the density ρ, Young modulus E and length l of the tuning fork. Which of the following equation may be used to relate t with the quantities mentioned?

a) t =Aρ E

gl3 b) t =ρEAl c) t =

AE ρ

lg

Where A is a dimensionless constant and g is the acceleration due to gravity. The table below shows the data obtained from various tuning forks made of steel and are geometrically identical.

Frequency/Hz 256 288 320 384 480

Length l /cm 12.0 10.6 9.6 8.0 6.4

Use the data above to confirm the choice of the right equation. Hence, determine the value of the constant A.

For steel: density ρ = 8500 kg m-3 , E = 2.0 x 1011 Nm-2

Solution :

Unit for E = N m-2

= ( ) m-2kg m s-2

= kg m-1 s-2

[E] = M L-1 T-2

a) t =Aρ E

gl3 b) t =ρEAl c) t =

AE ρ

lg

[t] = T

Aρ E

gl3 = M L-1 T-2

½ (L T-2) L3M L-3

[ρ] = Massvolume

=L-3

L-1 T-2L2 T-1

= T

+ xx

The equation is dimensionally consistent

Solution :

a) t =Aρ E

gl3 b) t =ρEAl c) t =

AE ρ

lg

[t] = TρEAl = L

½ [ρ] =

Massvolume

M L-3

M L-1 T-2

= L½ L-2

T-2

= L L-1

T-1

= T

The equation is

dimensionally consistent

Solution :

a) t =Aρ E

gl3 b) t =ρEAl c) t =

AE ρ

lg

[t] = T

AE ρ

lg

=½ M L-1 T-2

M L-3

LL T-2

= L2 T-2 1 T-1

= L2 T-1

The dimensionis not consistent

Hence, the equation is incorrect

Solution :

a) t =Aρ E

gl3 b) t =ρEAl c) t =

AE ρ

lg

Which one is the right equation?

Frequency/Hz 256 288 320 384 480

Length l /cm 12.0 10.6 9.6 8.0 6.4

Plot a grapht against l32

t against l

Frequency/Hz

256 288 320 384 480

Length l /cm

12.0 10.6 9.6 8.0 6.4

period, t, s 1256

3.91x10-3

period = 1frequency

1288

3.47x10-3 1320

3.13x10-3 1384

2.60x10-3 1480

2.08x10-3

l32

,s32, cm

( l )3

( 12)341.6 ( 10.6)334.5 ( 9.6 )329.7 ( 8.0 )322.6 ( 6.4 )316.2

t against l32

Graph of

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0tx10-3/s

5 10 15 20 25 30 35 40 45

Compare the graph with the equation

a)

a) t =Aρ E

gl3

y = mxGeneral equation

c ≠ 0

c = 0

Therefore the equation

is incorrect

t against l

Graph of

0.5

1.0

1.5

2.0

2.5

3.0

3.5

4.0tx10-3/s

2 4 6 8 10 12 14 16 18 20

l /cm

b) t = AlρE

y = mxGeneral equation

c = 0

c = 0

Therefore the equation

is correct

Compare the graph with the equation

b)

Use the data above to confirm the choice of the right equation. Hence, determine the value of the constant A.For steel: density ρ = 8500 kg m-3 , E = 2.0 x 1011 Nm-2

b) t = AlρE

y = mx

m = AρE

m =3.91x10-3 s 12 cm

= 3.91x10-3 s 12 m 100

= 0.03258333

0.03258333 = AρE

Use the data above to confirm the choice of the right equation. Hence, determine the value of the constant A.For steel: density ρ = 8500 kg m-3 , E = 2.0 x 1011 Nm-2

b) t = AlρE

y = mx

m = AρE

0.03258333 = AρE

0.03258333 = A 8500 .2.0 x 1011

A = 1.58 x 102

Example :

The dependence of the heat capacity C of a solid on the temperature T is given by the equation :

C = αT + βT3

What are the units of α and β in terms of the base units?

Solution :C = αT + βT3

All the terms have same unit

Unit for C = J K-1

= kg m s-1 K-12

= kg m2 s-2 K-1

Unit for αT == kg m2 s-2 K-1

Unit for α = kg m2 s-2 K-1

unit for T= kg m2 s-2 K-1

K= kg m2 s-2 K-2

Solution :C = αT + βT3

All the terms have same unit

Unit for βT3 == kg m2 s-2 K-1

Unit for β = kg m2 s-2 K-1

unit for T3

kg m2 s-2 K-1

K3=

= kg m2 s-2 K-4

Example :

A recent theory suggests that time may be quantized. The quantum or elementary amount of time is given by the equation :

T = hmpc2

where h is the Planck constant, mp = mass of proton and c = speed of light.

a)What is the dimension for Planck constant?b)Write the SI unit of Planck constant.

Solution :

a) T = hmpc2

h = Tmpc2

h = Tmpc2[ ] [ ]

= T M (L T–1)2

= M L2 T–1

Unit = kg m2 s-1