3. Different units can be used to describe
the same quantity
Example : The height of a person
can be expressed
in feet
In inches
In metres
……
4. unit
Is the standard use to compare
different magnitudes of the same
physical quantity
5. 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
6.
7. 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
8. 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
…………………
9. Prefix Factor Symbol
pico 10-12
p
nano 10-9
n
micro 10-6
µ
Milli 10-3
m
centi 10-2
c
deci 10-1
d
kilo 103
k
Mega 106
M
Giga 109
G
Tera 1012
T
10. 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
11. 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
12. 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
13. 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
14. Example 2 :
Dimension of velocity = Velocity
Displacement
Time=
= L
T
= L T-1
Unit of velocity = m s-1
15. Example 3 :
Dimension of acceleration = Acceleration
Change of velocity
Time=
= L T-1
T
= L T-2
Unit of acceleration = m s-2
16. 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
17. 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
18. 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
19. Example 6 :
Dimension of electric charge = Charge
Current x Time=
= A x T
= A T
Unit of area = A s
20. Example 7 :
Dimension of frequency = Frequency
1 .
Period=
= 1
T
= T-1
Unit of frequency = s-1
21. Example 8 :
Dimension of strain = Strain
Extension
Original length=
= L
L
= 1
Unit of frequency = no unit
dimensionless
22. Uses of Dimensions
Dimensional homogeneity of a physical
equation
1st
Physical
quantity
2nd
Physical
quantity+
_=
Both have same dimensions
23. 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-2
L
Every
term has
the same
dimension
The equation is dimensionally consistent
Case 1
24. Example :
v = u + 2as
v = L T-1
= L T-1
u = L T-1
= L T-1
2as = L T-2
= L2
T-2
L
[v]
=
[u]
≠
[2as]
The dimension is not consistent
The
equation is
incorrect.
Case 2
26. 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-2
L
Every
term has
the same
dimension
The equation is
dimensionally consistent
Case 3
The constant of
proportionality is wrong
The
equation is
incorrect.
27. 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-2
L
Every
term has
the same
dimension
The equation is
dimensionally consistent
Case 4
Has extra term
The
equation is
incorrect.
s2
t2
s2
t2
=
L2
T2 = L2
T-2
28. The truth of a physical
equation can be
confirmed
experimentally :
29. 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 =
ρ
EAl c) t =
AE
ρ
l
g
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
30. 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
31. 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 =
ρ
EAl c) t =
AE
ρ
l
g
[t] = T
Aρ
E
gl3 =
M L-1
T-2
½
(L T-2
) L3M L-3
[ρ] =
Mass
volume
=
L-3
L-1
T-2
L2
T-1
= T
+ xx
The equation is
dimensionally consistent
32. Solution :
a) t =
Aρ
E
gl3
b) t =
ρ
EAl c) t =
AE
ρ
l
g
[t] = T
ρ
EAl = L
½
[ρ] =
Mass
volume
M L-3
M L-1
T-2
= L
½
L-2
T-2
= L
L-1
T-1
= T
Theequationis
dimensionallyconsistent
33. Solution :
a) t =
Aρ
E
gl3
b) t =
ρ
EAl c) t =
AE
ρ
l
g
[t] = T
AE
ρ
l
g
=
½M L-1
T-2
M L-3
L
L T-2
= L2
T-2 1
T-1
= L2
T-1
The dimension
is not consistent
Hence,
the equation
is incorrect
34. Solution :
a) t =
Aρ
E
gl3
b) t =
ρ
EAl c) t =
AE
ρ
l
g
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 l
3
2
t against l
35. Frequency
/Hz
256 288 320 384 480
Length l
/cm
12.0 10.6 9.6 8.0 6.4
period, t, s
1
256
3.91x10-3
period = 1
frequency
1
288
3.47x10-3 1
320
3.13x10-3 1
384
2.60x10-3 1
480
2.08x10-3
36. l
3
2
,s
3
2, cm
( l )3
( 12)3
41.6 ( 10.6)3
34.5 ( 9.6 )3
29.7 ( 8.0 )3
22.6 ( 6.4 )3
16.2
37. t against l
3
2
Graph of
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
tx10-3
/s
5 10 15 20 25 30 35 40 45
39. t against l
Graph of
0.5
1.0
1.5
2.0
2.5
3.0
3.5
4.0
tx10-3
/s
2 4 6 8 10 12 14 16 18 20
l /cm
40. b) t = Al
ρ
E
y = mxGeneral equation
c = 0
c = 0
Therefore
the equation
is correct
Compare
the graph
with the
equation
b)
41. 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 = Al
ρ
E
y = mx
m = A
ρ
E
m =
3.91x10-3
s
12 cm
= 3.91x10-3
s
12 m
100
= 0.03258333
0.03258333 = A
ρ
E
42. 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 = Al
ρ
E
y = mx
m = A
ρ
E
0.03258333 = A
ρ
E
0.03258333 = A
8500 .
2.0 x 1011
A = 1.58 x 102
43. 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?
44. Solution :
C = αT + βT3
All the terms have same unit
Unit for C = J K-1
= kg m s-1 K-1
2
= 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
45. 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
46. Example :
A recent theory suggests that time may be
quantized. The quantum or elementary
amount of time is given by the equation :
T =
h
mpc2
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.
47. Solution :
a) T =
h
mpc2
h = Tmpc2
h = Tmpc2[ ] [ ]
= T M (L T–1
)2
= M L2
T–1
Unit = kg m2
s-1
