SlideShare a Scribd company logo
experiment
Careful observations
precise
Accurate measurement




Quantities
that can be
measured
Quantities
that can be
measured
Examples:
Length
Mass
Time
Weight
Electric current
Force
Velocity
energy
50 kg
Numericalvalue
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 feet
In inches
In 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
Physicalquantities
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
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
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=
= L
T
= 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=
= 1
T
= T-1
Unit of frequency = s-1
Example 8 :
Dimension of strain = Strain
Extension
Original length=
= L
L
= 1
Unit of frequency = no unit
dimensionless
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-2
L
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-2
L
[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-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.
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
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 =
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
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 =
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
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
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
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
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
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
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
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
4.0
tx10-3
/s
2 4 6 8 10 12 14 16 18 20
l /cm
b) t = Al
ρ
E
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 = 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
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
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-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
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 =
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.
Solution :
a) T =
h
mpc2
h = Tmpc2
h = Tmpc2[ ] [ ]
= T M (L T–1
)2
= M L2
T–1
Unit = kg m2
s-1

More Related Content

PDF
Physics f4 chapter2
PDF
2.0 forces and motion
PPS
2.jarak dan sesaran
PPTX
Lower Secondary Grade 8 Measuring Motion PART 1 distance speed and measuremen...
PDF
Bölüm 6
PPTX
Homogeneity of an equation
PDF
Measurements and Dimensional Analysis.pdf
PPT
Chapter 13-1219584195577289-8
Physics f4 chapter2
2.0 forces and motion
2.jarak dan sesaran
Lower Secondary Grade 8 Measuring Motion PART 1 distance speed and measuremen...
Bölüm 6
Homogeneity of an equation
Measurements and Dimensional Analysis.pdf
Chapter 13-1219584195577289-8

Similar to Physicalquantities (20)

PPT
Lecture 1.ppt
PDF
1 Units Dimension- Fall 2024 (4) general eng.pdf
PPT
Chapter 1(3)DIMENSIONAL ANALYSIS
PPTX
1_PHYSICAL Quantities to define the laws of physics
PDF
Solucionario de fluidos_white
PPTX
Physical Quantities and Measurements
PPTX
Quarter 1 Lesson 1 - Physical Quantities and Measurements
PPT
Basic Quantities,Prefixes&Dimensions.ppt
PPTX
Reaction Kinetics
PDF
Allen Physics JEE Module 1st Edition Allen Experts Faculty
DOC
Problem and solution i ph o 31
PDF
Thermal diffusivity
PDF
units and dimensions
DOCX
Revision notes on unit and dimensions
PDF
Mit2 092 f09_lec20
PDF
Es 2017 18
PDF
1.2 solutions serway physics 6th edition
PDF
Serway, raymond a physics for scientists and engineers (6e) solutions
PPT
Physics .. An introduction
PPTX
NS 6141 - Physical quantities.pptx
Lecture 1.ppt
1 Units Dimension- Fall 2024 (4) general eng.pdf
Chapter 1(3)DIMENSIONAL ANALYSIS
1_PHYSICAL Quantities to define the laws of physics
Solucionario de fluidos_white
Physical Quantities and Measurements
Quarter 1 Lesson 1 - Physical Quantities and Measurements
Basic Quantities,Prefixes&Dimensions.ppt
Reaction Kinetics
Allen Physics JEE Module 1st Edition Allen Experts Faculty
Problem and solution i ph o 31
Thermal diffusivity
units and dimensions
Revision notes on unit and dimensions
Mit2 092 f09_lec20
Es 2017 18
1.2 solutions serway physics 6th edition
Serway, raymond a physics for scientists and engineers (6e) solutions
Physics .. An introduction
NS 6141 - Physical quantities.pptx
Ad

Physicalquantities

  • 2. Quantities that can be measured Examples: Length Mass Time Weight Electric current Force Velocity energy 50 kg Numericalvalue unit Describing a physical quantity:
  • 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
  • 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
  • 25. A physical equation whose dimensions are consistent need not necessary be correct :
  • 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
  • 38. Compare the graph with the equation a) a) t = Aρ E gl3 y = mxGeneral equation c ≠ 0 c = 0 Therefore the equation is incorrect
  • 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