physicalquantities
TRANSCRIPT
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 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
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
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
Compare the graph with the equation
a)
a) t =Aρ E
gl3
y = mxGeneral equation
c ≠ 0
c = 0
Therefore the equation
is incorrect
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.