Homogeneity of equations
This is a method of checking if an equation is correct by looking at the units. An equation is
homogeneous if, when the base units of all the quantities are written, they are the same on
both sides of the equation.
Dimension of physical quantity
Dimension of a physical quantity is the relationship between the physical quantity and the
basic physical quantity such as mass, M, length L, time T, electric current A, temperature 𝜽
and amount of substance N.
Square brackets are used to indicate the dimensions of M, L and T are used to denote mass,
length and time when we are dealing with dimensions. Thus, the dimensions of force are;
[𝐹] = [𝑀𝐿𝑇 −2 ]
Important point note:
Two physical quantities can be added or subtracted if and only if their dimensions are the
same. e.g. 𝐹 = 𝑎𝑡 + 𝑏𝑡 2
From the equation (𝐹 − 𝑎𝑡) it means (at) has the dimensions of (F) as well (𝑏𝑡 2 ) has the
dimensions of (F).
Examples
1.
[𝐴𝑟𝑒𝑎] = [𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ]
= 𝐿 × 𝐿 = 𝐿2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑎𝑟𝑒𝑎 = 𝑚2
2.
[𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦] = [
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
𝐿
] = 𝑇 = 𝐿𝑇 −1
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑚𝑠 −1
3.
[𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛] = [
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝑡𝑖𝑚𝑒
]=
𝐿𝑇 −1
𝑇
= 𝐿𝑇 −2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑚/𝑠 2
4.
[𝐹𝑜𝑟𝑐𝑒] = [𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛] = 𝑀𝐿𝑇 −2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑓𝑜𝑟𝑐𝑒 = 𝑁 = 𝑘𝑔𝑚𝑠 −2
5.
[𝑊𝑜𝑟𝑘] = [𝑓𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡] = 𝑀𝐿2 𝑇 −2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑤𝑜𝑟𝑘 = 𝐽 = 𝑘𝑔𝑚2 𝑠 −2
6.
[𝐸𝑙𝑒𝑐𝑡𝑟𝑖𝑐 𝑐ℎ𝑎𝑟𝑔𝑒] = [𝑐𝑢𝑟𝑟𝑒𝑛𝑡 × 𝑡𝑖𝑚𝑒)= AT
Unit for electric charge = C = A s
1
7.
[𝐹𝑟𝑒𝑞𝑢𝑒𝑛𝑐𝑦] =
8.
𝑒𝑥𝑡𝑒𝑛𝑠𝑖𝑜𝑛
Strain =
𝑜𝑟𝑖𝑔𝑖𝑛𝑎𝑙 𝑙𝑒𝑛𝑔ℎ𝑡
𝑝𝑒𝑟𝑖𝑜𝑑
= 𝑇 −1
=
𝐿
𝐿
which is dimensionless
More examples
Example - 01:
1
Check the correctness of physical equation, 𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2 , Where 𝑢 is the initial velocity, vis
the final velocity, 𝑎 is the acceleration, 𝑠 is the displacement and 𝑡 is the time in which the
change occurs.
Solution:
1
Given equation is, 𝑠 = 𝑢𝑡 + 2 𝑎𝑡 2
𝐿. 𝐻. 𝑆. = 𝑣, ℎ𝑒𝑛𝑐𝑒 [𝐿. 𝐻. 𝑆. ] = [𝑠] = [𝐿1 𝑀0 𝑇 0 ] . . . . . . . . . . . . . . . . . . . (1)
𝑅. 𝐻. 𝑆 = 𝑢𝑡 +
1 2
𝑎𝑡 , ℎ𝑒𝑛𝑐𝑒 [𝑅. 𝐻. 𝑆] = [𝑢][𝑡] + [𝑎] [𝑡]2
2
= [𝐿1 𝑀0 𝑇 −1 ][𝐿0 𝑀0 𝑇 1 ] + [𝐿1 𝑀0 𝑇 −2 ][𝐿0 𝑀0 𝑇 1 ]2
= [𝐿1 𝑀0 𝑇 0 ] + [𝐿1 𝑀0 𝑇 −2 ][𝐿0 𝑀0 𝑇 2 ]
= [𝐿1 𝑀0 𝑇 0 ] + [𝐿1 𝑀0 𝑇 0 ]
= [𝐿1 𝑀0 𝑇 0 ]
. . . . . . … … … … … … … … … … … … … … . . . . . . . . . . . . . (2)
(1) and (2) we have 𝐿. 𝐻. 𝑆. = [𝑅. 𝐻. 𝑆. ]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
Example - 02:
Check the correctness of physical equation,𝐹 =
m𝑣 2
r
, Where 𝐹 is the centripetal force acting
on a body of mass m performing uniform circular motion along a circle of radius 𝑟 with
linear speed 𝑣.
Solution:
Given the equation
m𝑣 2
𝐹=
r
𝐿. 𝐻. 𝑆. = 𝐹, ℎ𝑒𝑛𝑐𝑒 [𝐿. 𝐻. 𝑆. ] = [𝐹] = [𝐿1 𝑀1 𝑇 −2 ] . . . . . . . . . . . . . . . . … . . . . (1)
m𝑣 2
𝑅. 𝐻. 𝑆 =
, ℎ𝑒𝑛𝑐𝑒 [𝑅. 𝐻. 𝑆] = [𝑚][𝑣]2 /[𝑟]
r
=
[𝐿0 𝑀1 𝑇 0 ][𝐿1 𝑀0 𝑇 −1 ]2
[𝐿1 𝑀0 𝑇 0 ]
= [𝐿0 𝑀1 𝑇 0 ][𝐿2 𝑀0 𝑇 −2 ] [𝐿−1 𝑀0 𝑇 0 ]
2
= [𝐿1 𝑀1 𝑇 −2 ] … … … … … … … . … … … … … … … … … . . . . . . . . . . . . . (2)
Hence by the principle of homogeneity, the given equation is dimensionally correct.
Example - 03:
Check the correctness of physical equation.
𝑣 2 = 𝑢2 + 2𝑎𝑠 2
Solution:
Given equation is
𝑣 2 = 𝑢2 + 2𝑎𝑠 2
𝐿. 𝐻. 𝑆. = 𝑣2 , ℎ𝑒𝑛𝑐𝑒 [𝐿. 𝐻. 𝑆. ] = [𝑣2 ] = [𝐿1 𝑀0 𝑇 −1 ]2 = [𝐿2 𝑀0 𝑇 −2 ] . . . . . . . . . . . . . . . (1)
𝑅. 𝐻. 𝑆 = 𝑢2 + 2𝑎𝑠 2
= [𝐿1 𝑀0 𝑇 −1 ] + [𝐿1 𝑀0 𝑇 −2 ][𝐿1 𝑀0 𝑇 0 ]2
= [𝐿2 𝑀0 𝑇 −2 ] + [𝐿1 𝑀0 𝑇 −2 ][𝐿2 𝑀0 𝑇 0 ]
= [𝐿2 𝑀0 𝑇 −2 ] + [𝐿3 𝑀0 𝑇 −2 ]. … … … … … … … … … … … … … … . . . . . . . . . . (2)
𝐹𝑟𝑜𝑚 (1) 𝑎𝑛𝑑 (2) 𝑤𝑒 ℎ𝑎𝑣𝑒 [𝐿. 𝐻. 𝑆. ] ≠ [𝑅. 𝐻. 𝑆. ]
Example - 04:
Check the homogeneity of equation, when the periodic time, “T” of vibration of magnet of
moment of inertia 'I', magnetic moment 'M' vibrating in magnetic induction 'B' is given by
𝐼
𝑇 = 2π√
𝑀𝐵
𝜇
0𝐼
Where 𝐼 = mr 2 , 𝑀 = and 𝐵 = 2𝜋𝑟
Given the equation;
𝐼
𝑇 = 2π√
𝑀𝐵
𝐿. 𝐻. 𝑆 = [𝑇] = [𝐿0 𝑀0 𝑇1 ] . . . . . . . . . . . . . . . (1)
[𝐼]
𝑅. 𝐻. 𝑆 = √
[𝑀][𝐵]
[𝑀1 𝐿2 𝑇 0 ]
= √ 2 0 0 1 0 1 −2 −1
[𝐿 𝑀 𝑇 𝐼 ][𝐿 𝑀 𝑇 𝐼 ]
[𝑀1 𝐿2 𝑇 0 ]
= √ 2 1 −2 0
[𝐿 𝑀 𝑇 𝐼 ]
𝑅. 𝐻. 𝑆 = √[𝐿0 𝑀0 𝑇 2 ] = [𝐿0 𝑀0 𝑇 1 ] … … … … … … … . . (2)
𝐹𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2)𝑤𝑒 ℎ𝑎𝑣𝑒 [𝐿. 𝐻. 𝑆. ] = [𝑅. 𝐻. 𝑆. ]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
3
Example - 05:
a) The period T of a tuning fork depends on the density 𝜌, Young modulus E and length 𝑙 of
the tuning fork. Which of the following equations can be used to relate T with the
quantities mentioned?
Nb: E has the units of pascals per second.
𝐴𝜌
√𝑔𝑙 3
i).
𝑇=
ii).
𝑇 = 𝐴𝑙√𝐸
iii).
𝑇 = 𝜌 √𝑔
𝐸
𝜌
𝐴𝐸
𝑙
Where A is a dimensionless constant and g is the acceleration due to gravity.
b) The table below is obtained using various steel tuning forks which are geometrically
identical in an experiment.
Frequency /cycle 𝑠 −1 (Hz)
256
288
320
384
480
Length I/cm
12.0
10.6
9.6
8.0
6.4
Use the above data to confirm the equations you have chosen. Hence determine the value
of the constant A, if, for steel, E = 2·0 x 1011 N𝑚−2 p = 8 500 kg 𝑚−3. [Unit for E = kg
𝑚−1 𝑆 −2]
Solution
a) Unit for period T = s
For equation (i)
𝐴𝜌
√𝑔𝑙 3
𝑇=
𝐸
𝐴𝜌
𝑘𝑔𝑚−3
1
3
√𝑔𝑙 =
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑇 =
(𝑚𝑠 −2 𝑚3 )− ⁄2
−1
−2
𝐸
𝑘𝑔𝑚 𝑠
= 𝑠
Equation (i) is homogeneous and can be used.
For equation (ii)
𝜌
𝑇 = 𝐴𝑙 √
𝐸
1⁄
2
𝑘𝑔𝑚−3
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑇 = 𝐴𝑙 √ = 𝑚 (
)
𝐸
𝑘𝑔𝑚−1 𝑠 −2
𝜌
=𝑠
Equation (ii) is homogeneous and can be used.
4
For equation (iii)
𝑇=
𝐴𝐸 𝑙
√
𝜌 𝑔
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑇 =
𝐴𝐸 𝑙 𝑘𝑔𝑚−1 𝑠 −2 𝑚
√ =
(
) = 𝑚2 𝑠 −1
𝜌 𝑔
𝑘𝑔𝑚−3 𝑚𝑠 −2
Which is different from the unit of ‘T’, so it cannot be used.
b)
Frequency f /cycle 𝒔−𝟏
256
288
320
384
480
Period T/(x 𝟏𝟎−𝟑 s)
3·91
3.47
3·13
2·60
2·08
Length I/cm
12.0
10.6
9.6
8.0
6.4
𝟑
𝟑
𝒍 ⁄𝟐 /𝒄𝒎 ⁄𝟐
41.6
34.5
29.7
22.6
16.2
(Hz)
From the graphs above, the graph of 𝑇 against 𝑙 is a straight line passing through the origin.
𝜌
Therefore, the equation 𝑇 = 𝐴𝑙√𝐸 is correct.
The graph or T against is a straight line but does not pass through the origin. Therefore, the
equation 𝑇 =
𝐴𝜌
𝐸
√𝑔𝑙 3 is wrong.
5
𝜌
From the equa1ion 𝑇 = 𝐴𝑙√𝐸 and from the table. when l = 12·0 cm. T = 3 · 91 × 10− 3 𝑠,
𝐴=
𝑇 𝐸
√
𝑙 𝜌
𝐴=
𝑇 𝐸
√
𝑙 𝜌
(3 · 91 × 10− 3 ) 2.0 × 1011
√(
)
0.12
8 500
= 158
Example - 06:
Check the homogeneity of equation, when the rate of flow of a liquid having coefficient of
viscosity '𝜂' through a capillary tube of length '𝑙' and radius '𝑎' under pressure head ′𝑝' is
given by
𝑑𝑉 𝜋𝑝𝑎4
=
𝑑𝑡
8𝑙𝜂
Nb: the unit of ′ 𝜂′ is the pascal per second.
Solution
Given the equation
𝑑𝑉 𝜋𝑝𝑎4
=
𝑑𝑡
8𝑙𝜂
[𝑀0 𝐿3 𝑇 0 ]
[𝑑𝑉] [𝑑𝑉]
=
=
= [𝑀0 𝐿3 𝑇 −1 ] . . . . . . . . . . . . . . . (1)
[𝑑𝑡]
[𝑑𝑡]
[𝑀0 𝐿0 𝑇 1 ]
[𝑀1 𝐿−1 𝑇 −2 ] [𝑀0 𝐿1 𝑇 0 ]4
[𝑝][𝑎]4
𝑅. 𝐻. 𝑆 =
=
[𝑙][𝜂]
[𝑀0 𝐿1 𝑇 0 ] [𝑀1 𝐿−1 𝑇 −1 ]
𝐿. 𝐻. 𝑆. =
=
[𝑀1 𝐿−1 𝑇 −2 ][𝑀0 𝐿4 𝑇 0 ]
[𝑀1 𝐿0 𝑇 −1 ]
[𝑀1 𝐿3 𝑇 2 ]
= 1 0 −1 = [𝑀0 𝐿3 𝑇 −1 ] … … … … … … … … … … . … … … … (2)
[𝑀 𝐿 𝑇 ]
𝐹𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2)𝑤𝑒 ℎ𝑎𝑣𝑒 [𝐿. 𝐻. 𝑆. ] = [𝑅. 𝐻. 𝑆. ]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
Example - 07:
Check the homogeneity of equation, when the terminal velocity 'v' of a small sphere of radius
'a' and density ' 𝜌 ' falling through a liquid of density ' 𝜎 ' and coefficient of viscosity ' 𝜂 ' is
given by
2𝑔𝑎2 (𝜎 − 𝜌)
𝑣=
9𝜂
Given the equation
6
2𝑔𝑎2 (𝜎 − 𝜌)
9𝜂
0 1 −1 ]
[𝑣]
[𝑀
𝐿. 𝐻. 𝑆. =
=
𝐿𝑇
… … … … … … … … … … … . . . . . . . . . . . . . . . (1)
𝑣=
𝑅. 𝐻. 𝑆 =
2𝑔𝑎2 (𝜎 − 𝜌) [𝑔][𝑎]2 ([𝜎] − [𝜌])
=
9𝜂
[𝜂]
[𝑀0 𝐿1 𝑇 −2 ] [𝑀0 𝐿1 𝑇 0 ]2 ([𝑀1 𝐿−3 𝑇 0 ] − [𝑀1 𝐿−3 𝑇 0 ])
𝑅. 𝐻. 𝑆 =
[𝑀1 𝐿−1 𝑇 −1 ]
𝑅. 𝐻. 𝑆 =
[𝑀0 𝐿1 𝑇 −2 ][𝑀0 𝐿2 𝑇 0 ]([𝑀1 𝐿−3 𝑇 0 ])
[𝑀1 𝐿−1 𝑇 −1 ]
[𝑀1 𝐿0 𝑇 −2 ]
𝑅. 𝐻. 𝑆 = 1 −1 −1 = [𝑀0 𝐿1 𝑇 −2 ] … … … … … … … … … … … … … … . . (2)
[𝑀 𝐿 𝑇 ]
𝐹𝑟𝑜𝑚 (1)𝑎𝑛𝑑 (2)𝑤𝑒 ℎ𝑎𝑣𝑒 [𝐿. 𝐻. 𝑆. ] = [𝑅. 𝐻. 𝑆. ]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
Example - 08:
The distance covered by a body in time tis given by the relation 𝑆 = 𝑎 + 𝑏𝑡 + 𝑐𝑡 2 . What are
the dimension of a,b,c? Also write the quantities they represent.
Solution:
By the principle of homogeneity, the dimensions on either side of the physical equation
must be the same. Now two physical quantities can be added or subtracted if and only if
their dimensions are the same.
Dimensions of S = Dimensions of a
[𝐿1 𝑀0 𝑇 0 ] = [𝑎]
Dimensions of 'a' are that of displacement. Hence 'a' represent the displacement.
Dimensions of S = Dimensions of 𝑏𝑡
[𝐿1 𝑀0 𝑇 0 ] = [𝑏][𝑡]
[𝐿1 𝑀0 𝑇 0 ]
[𝑏] =
[𝑇]
[𝑏] = [𝐿1 𝑀0 𝑇 −1 ]
Dimensions of 'a' are that of velocity. Hence 'b' represents the velocity.
Dimensions of S = Dimensions of 𝑐𝑡 2 .
[𝐿1 𝑀0 𝑇 0 ] = [𝑐]
[𝐿1 𝑀0 𝑇 0 ]
[𝑐] =
[𝑇]2
[𝑐] = [𝐿1 𝑀0 𝑇 −2 ]
7
Dimensions of 'c' are that of acceleration. Hence 'c' represents the acceleration.
Ans: The dimensional formula of 'a', 'b' and 'c' are [𝐿1 𝑀0 𝑇 0 ], [𝐿1 𝑀0 𝑇 −1 ], and [𝐿1 𝑀0 𝑇 −2 ].
The quantities represented by a, b and c are displacement, velocity and acceleration.
Example - 09:
𝛾𝑃⁄ 1⁄2
𝑑) ,
where P is the pressure, d is the density of a gas. What is the dimensional formula for 𝛾?
According to Laplace's formula, the velocity (V) of sound in a gas is given by 𝑣 = (
Solution:
𝛾𝑃⁄
𝑑
2
𝑣
𝑑
⁄𝑃
∴𝛾=
∴ 𝑣2 =
∴𝛾=
[𝑣 2 ][𝑑] [𝐿1 𝑀0 𝑇 −1 ][𝐿−3 𝑀1 𝑇 0 ]
=
[𝐿−1 𝑀1 𝑇 −2 ]
[𝑃]
[𝐿−1 𝑀1 𝑇 −2 ]
∴ 𝛾 = −1 1 −2 = [𝐿0 𝑀0 𝑇 0 ]
[𝐿 𝑀 𝑇 ]
Example - 10:
A force F is given by 𝐹 = 𝑎𝑡 + 𝑏𝑡 2 , where 𝑡 is the time. Find the dimensional formula of '𝑎'
and 'b'.
Solution:
By the principle of homogeneity, the dimensions on either side of the physical equation must
be the same. Now two physical quantities can be added or subtracted if and only if their
dimensions are the same.
Dimensions of F = Dimensions of 𝑎𝑡
[𝐿1 𝑀1 𝑇 −2 ] = [𝑎][𝑡]
[𝑎] =
[𝐿1 𝑀1 𝑇 −2 ]
[𝑇]
[𝑎] = [𝐿1 𝑀1 𝑇 −3 ]
Dimensions of F = Dimensions of 𝑏𝑡 2
[𝐿1 𝑀1 𝑇 −2 ] = [𝑏][𝑇]2
[𝐿1 𝑀1 𝑇 −2 ]
[𝑏] =
[𝑇]2
[𝑏] = [𝐿1 𝑀1 𝑇 −4 ]
The dimensional formula of a and b are [𝐿1 𝑀1 𝑇 −3 ] and [𝐿1 𝑀1 𝑇 −4 ]
Example - 11:
In an equation {𝑃 + 𝑎⁄𝑉 2 }(𝑉 − 𝑏) = 𝑅𝑇, where P is the pressure, V is the volume, T is the
temperature and a, b, R are constants. What is the dimensional formula of 𝑎 ⁄𝑏?
8
Solution:
Two physical quantities can be added or subtracted if and only if their dimensions are the
same.
Thus, the dimensions of P and 𝑎⁄𝑉 2 are same
[𝑎]
[𝑃] =
2
[𝑉]
∴ [𝑎] = [𝑃][𝑉]
2
∴ [𝑎] = [𝐿−1 𝑀1 𝑇 −2 ][𝐿3 ]
2
∴ [𝑎] = [𝐿−1 𝑀1 𝑇 −2 ][𝐿6 ]
∴ [𝑎] = [𝐿5 𝑀1 𝑇 −2 ]
Dimensions of b = Dimensions of V
[𝑏] = [𝐿3 𝑀0 𝑇 0 ]
[a] [𝐿5 𝑀1 𝑇 −2 ]
Now the dimensions of
is 3 0 0 = [𝐿2 𝑀1 𝑇 −2 ]
[𝐿 𝑀 𝑇 ]
[b]
The dimensions of 𝑎⁄𝑏 is = [𝐿2 𝑀1 𝑇 −2 ]4
Practice questions
1. What are the S. I. units of k so that the equation?
𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑘 × 𝑑𝑒𝑛𝑠𝑖𝑡𝑦
2. A sphere of radius a moving with a velocity v under streamline conditions in a viscous
fluid experiences a retarding force 𝐹 given by 𝐹 = 𝑘𝑎𝑣 where 𝑘 is a constant What are
the S.I. units of 𝑘 in terms of the base units?
3. The heat capacity C of a solid can be expressed as a function of temperature T to fit the
expression:
𝐶 = 𝛼𝑇 + 𝛽𝑇 3
a. Find the possible units of ′𝛼′ and ′𝛽′.
b. Write down the base S.I. units of specific heat capacity.
4. Bernoulli’s equation. which applies to fluid flow. states that:
1
𝑝 + ℎ 𝜌 𝑔 + 𝜌𝑣 2 = 𝑘
2
where p is pressure, h height, p density, g acceleration due to gravity, v velocity and k a
constant. Show that the equation is dimensionally consistent and state an S.I. unit for
k.
Deriving Physical equations
Determining a physical quantity usually depends on a number of other physical quantities.
For example, the period T of a simple pendulum depends on its length 𝑙 and the
acceleration due to gravity g. Using units or dimensions. an C4ua1ion can be derived to
relate the period T with / and g. Suppose that
𝑇𝛼 𝑙 𝑥 𝑔y
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where.𝑥 and 𝑦 are non-dimensional constants.
Then
𝑇 = 𝑘 𝑙 𝑥 𝑔y
where k is a dimensionless constant of proportionality.
Since a physical equation must be dimensionally consistent, units on the left side of the
equation must be the same as the units on the right side of the equation.
Units of T = units of [𝑘 𝑙 𝑥 𝑔y ]
𝑠 = 𝑚 𝑥 (𝑚𝑠 −2 )𝑦
= 𝑚 𝑥+𝑦 𝑠 −2
𝑦
Equating indices of s:
1 = −2𝑦. ∴ 𝑦 = −
Equating indices of m:
0=𝑥+𝑦
1
𝑥 = −𝑦 =
2
1
2
∴𝑇=𝑘𝑙
1⁄
2
𝑔−
1⁄
2
𝑙
∴𝑇=𝑘√
𝑔
Example
The speed of sound 𝑣 in a medium depends on its wavelength 𝜆. the Young modulus E and
the density 𝜌 of the medium. Use the method of base units to derive a formula for the
speed of sound in a medium. Unit of E: 𝑘𝑔𝑚−1 𝑠 −2.
Solution:
Suppose:
𝑣 = 𝑘 𝜆𝑥 𝐸 y 𝜌 z
where 𝑘, 𝑥, 𝑦 and 𝑧 are non-dimensional constants.
Units of v = units of [𝑘 𝜆𝑥 𝐸 y 𝜌z ]
𝑚𝑠 −1 = 𝑚 𝑥 (𝑘𝑔𝑚−1 𝑠 −2 )𝑦 (𝑘𝑔𝑚−3 )z
= 𝑚 𝑥−𝑦−3𝑧 𝑠 −2y 𝑘𝑔y+z
Equating indices of s:
−1 = −2𝑦. ∴ 𝑦 =
Equating indices of kg:
0=𝑦+𝑧
1
2
10
𝑥 = −𝑦 =
1
1
∴ 𝑧 = −𝑦 = −
2
2
Equating indices of m:
1 = 𝑥 − 𝑦 − 3𝑧
1 3
=𝑥− −
2 2
∴𝑥=0
∴𝑇=𝑘𝐸
1⁄
2
𝜌−
1⁄
2
𝐸
∴𝑣=𝑘√
𝜌
Example
Poiseuille assumed that the rate or now or a liquid through a horizontal tube under
streamline flow depends
a) a. radius of the tube.
b) 𝜂 viscosity of the liquid. and
c)
𝑝
𝑙
, the pressure gradient along the tube,
where 𝑝 = pressure difference across the length or the tube
𝑙 = length of tube.
Using Poiseuille's assumption. derive an expression for the rate or now of a liquid through a
horizontal tube in terms of 𝑎, 𝑙, 𝑝 and 𝜂.
Solution
𝑠𝑢𝑝𝑝𝑜𝑠𝑒 𝑡ℎ𝑒 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑓𝑙𝑜𝑤,
𝑑𝑉
𝑝 𝑦
= 𝑘𝑎 𝑥 ( ) 𝜂 𝑧
𝑑𝑡
𝑙
𝑤ℎ𝑒𝑟𝑒 𝑘, 𝑥 𝑦 𝑎𝑛𝑑 𝑧 𝑎𝑟𝑒 𝑑𝑖𝑚𝑒𝑛𝑠𝑖𝑜𝑛𝑙𝑒𝑠𝑠 𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡𝑠.
𝑈𝑛𝑖𝑡𝑠 𝑜𝑓
𝑑𝑉
𝑝 𝑦
= 𝑈𝑛𝑖𝑡𝑠 𝑜𝑓 [𝑘𝑎 𝑥 ( ) 𝜂 𝑧 ]
𝑑𝑡
𝑙
𝑦
3 −1
𝑚 𝑠
𝑘𝑔𝑚𝑠 −2 1
= 𝑚 ×(
× ) (𝑘𝑔𝑚−1 𝑠 −1 ) 𝑧
𝑚2
𝑚
𝑥
= 𝑘𝑔y+z 𝑚 𝑥−2𝑦−𝑧 𝑠 −2y−z
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