Dimensions of Physical Quantities and Its Applications
•
Dimensions
They are the powers (or exponents) to which the units of base quantities are raised for
representing a derived unit of that quantity.
Examples: Dimensional formula of volume [M0L3T0]
Dimensional formula of velocity [M0LT−1]
Dimensional formula of acceleration [M0LT−2]
Principle Of Homogeneity
According to this principle, the dimensions of two physical relations must be same.
[MaLbTc] = [MxLy Tz]
e.g. let us have the relation
Then, a = x, b = y, c = z
S. No.
1
Physical quantity
Pressure
DIMENSIONAL FORMULAS
Relation
Dimensional
formula
Units
Force/area
[M1L-1T –2]
Nm-2
2
Impulse
Force * time
[M1 L1T –1]
Nm
3
Power
Work/time
[M1L2T –3]
Js-1
4
Velocity of light
Velocity
[M0L1T –1]
ms-1
5
Frequency
I/time period
[T-1]
s-1
6
Velocity gradient
Velocity/distance
[M0L0T –1]
s-1
7
Wavelength
Velocity/frequency
[M0L1T0]
m
8
9
10
11
12
Gravitational
constant
Force × ( distance )
Planck’s constant
Energy *frequency
( mass )
2
Stress
14
Coefficient
of viscosity
–1
[M L T ]
Strain
Coefficien
t
of
elasticity
Js
1 2
Change in length
Original length
13
Nm2kg-2
2
(h)
Rate of flow
[M-1L3 T-2]
Dimensionless
No units
Force /area
[M1L-1T-2]
Nm-2
Volume/time
[M0L3T-1]
ms-1
Stress/strain
[M1L-1 T-2]
Nm-2
Force
[M1L-1T-1]
Nm–2s
Area *velo.grad.
M1L0T-2
Nm-1
Energy/area
[M1L0T-2]
Jm-2
Length of arc/ Radius
DIMENSIONLES No units
S
15
Surface tension
16
Surface energy
17
Angle
18
Angular velocity
Angle /time
[M0L0T-1]
s-1
19
Angular
acceleration
Angular velocity
time
[M0L0T-2]
s-2
Force /length
20
Torque
Force*distance
[M1L2T-2]
Nm
21
Angular
momentum
Momentum*distance
[M1L2T-1]
Kgm2s-1
22
Moment of inertia
Mass*(radius)2
[M1L2T0]
Kgm2
23
Radius of gyration
Distance
[M0L1T0]
M
24
Coefficient
friction
Force
Noraml reaction
Dimensionless
No unit
25
Temperature
Fundamental quantity [M0L0T0K1]
26
Heat
Energy
Gas constant
Pressure × volume
Temperature
27
28
of
Boltzmann
constant
K
[M1L2T-2]
J
[M1L2T-2K-1]
JK-1
[M1L2T-2K-1]
JK-1
R /N
Coefficient of
thermal
conductivity
Qd
A(θ 2 − θ1 ) t
[M1L1T-3K-1]
Wm-1K-1
30
Spring constant
Force
Displacement
[M1L0T-2]
Nm-1
31
Solar constant
Or Wave intensity
Energy received
Time × Area
[M1L0T-3]
Kgs-3
32
Specific heat
S = Q /(mθ)
[M0L2T-2K-1]
m2s-2K-1
29
33
Latent heat
L = Q/m
[M0L2T-2]
m2s-2
34
Poission ratio
Lateral strain
Longitudinal strain
Dimensionless
No unit
35
Specific gravity
Density of body
Density of water ( 40 C )
Dimensionless
No unit
Types Of Physical Quantities
The physical quantities are of two types:
(i)Variables (ii) constants.
Both these may be dimensional or non-dimensional in nature.
(i) Dimensional variables. These are the quantities, which arevariable and have dimensions
(i.e. units) e.g. acceleration, force,density etc.
.
(ii) Dimensional Constants. These are quantities which have constant values and yet have
dimensions e.g., gravitational constant, gas constant, Planck's constant etc.
(iii) Non-dimensional variables. These are the quantities which are variable and yet have no
dimension e.g., specific gravity, angle, strain etc.
(iv) Non dimensional constants. These are mere numbers like 1, 2, 3, π etc. Their values are
constant and they do not possess dimensions.
•
Applications of Dimensional Analysis
o Checking the dimensional consistency of equations
o Deducing relation among the physical quantities.
o Conversion from one system to another system.
•
Checking the Dimensional Consistency of Equations
Based on the principle of homogeneity of dimensions
o According to this principle, only that formula is correct in which the dimensions
of the various terms on one side of the relation are equal to the respective
dimensions of these terms on the other side of the relation.
Example:
Check the correctness of the relation, t = 2π
l
where l is length and t is time
g
period of a simple pendulum; g is acceleration due to gravity.
Solution:
Dimension of L.H.S = t = [T]
Dimension of R.H.S =
( 2π is a constant)
Dimensionally, L.H.S = R.H.S; therefore, the given relation is correct.
•
Deducing Relation Among Various Physical Quantities
Based on the principle of homogeneity of dimensions
Example: The centripetal force, F acting on a particle moving uniformly in a circle may
depend upon the mass (m), velocity (v) and radius (r) of the circle. Derive the formula
for F using the method of dimensions.
Solution: Let F = kmavbrc … (i)
Where, k is the dimensionless constant of proportionality, and a, b, c are the powers of
m, v, r respectively.
On writing the dimensions of various quantities in (i), we get
[M1L1T−2] = Ma [LT−1]b [Lc]
=[ MaLbT−bLc]
]M1L1T−2 ]=[MaLb + cT−b]
On applying the principle of homogeneity of dimensions, we get
a = 1, b= 2, b + c = 1 …(ii)
From (ii), c = 1 − b = 1 − 2 = −1
On putting these values in (i), we get
F = km1v2r−1
OR
This is the required relation for centripetal force.
Conversion from one system to another system
Convert of power of one mega watt on a system whose fundamental units are 10kg , 1 dm & 1 minute.
Q.
Ans.
P = 1 MW
6W 106  J 
=
Givensystem =
P 10
Re quired systen
 s 
6.
n 10
=
P [ M 1L2T −3 ] =
n
?
1
2
M =
a=
2, c =
−3
M =
10kg .
1Kg
1, b =
1
2
L 1=
M
L 1dm
1
2
=
=
= 60 s
T 1sec.
T 1min
1
2
a
b
c
M 
L 
T 
 1
 1
n =n  1
2 1M 
L 

 T2 
 2
 2
1
−3
 1kg   1m  2
1sec 
= 106 
 

 60 s 
10kg  1dm 
2
−3
1
 1  1  

6
n = 10   

2
10  10−1   60 × 60 × 60 
1
= 106   ×100 × 60 × 60 × 60
10 
=216 ×1010 =2.16 × 1012
As n u = n u
11 2 2
J 
 ergs 
106 =
2.16 ×1012 

s
 s 
Like Quantities are added and subtracted
Q.
Ans.
 b 
what are dimensions of a, b &c.
 t + c 
The velocity ‘V’ of particle is given by V= at + 
 b 
V= at + 
 t + c 
t + c ⇒ t Should have some dim ensions as that of ' t '.
b
V= at +  
T 
Acc. to dim ension analsis.
b
=
V at=
,   V
T 
−1
V   LT 
a =
b V=
,
=
=
[T ]  LT −1  [T ] ,



t   T 
−2
a [ LT
b [ L]
=
]
Limitations of Dimensional Analysis:*Dimensional analysis does not give any information about pure numerical values.
*Dimensional formula of more than two quantities may be same i.e. Dimensions of pressure
and stress are same.
*The formulae involving logarithmic, exponential and trigonometric functions can not be
derived with the help of dimensional analysis nor their validity can be tested.
*This relation is limited only to power relations.
*They don’t give any information whether the physical quantity is a scalar or vector.