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Homogineous Equations

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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 βˆ’2 ]
Important point note:
Two physical quantities can be added or subtracted if and only if their dimensions are the
same. e.g. 𝐹 = π‘Žπ‘‘ + 𝑏𝑑 2
From the equation (𝐹 βˆ’ π‘Žπ‘‘) it means (at) has the dimensions of (F) as well (𝑏𝑑 2 ) has the
dimensions of (F).
1.
2.
Examples
[π΄π‘Ÿπ‘’π‘Ž] = [π‘™π‘’π‘›π‘”π‘‘β„Ž × π‘π‘Ÿπ‘’π‘Žπ‘‘π‘‘β„Ž]
= 𝐿 × πΏ = 𝐿2
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘Žπ‘Ÿπ‘’π‘Ž = π‘š2
[π‘‰π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦] = [
π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘
π‘‘π‘–π‘šπ‘’
𝐿
] = 𝑇 = 𝐿𝑇 βˆ’1
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦ = π‘šπ‘  βˆ’1
π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦
𝐿𝑇 βˆ’1
3.
[π΄π‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›] = [
4.
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘Žπ‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› = π‘š/𝑠 2
[πΉπ‘œπ‘Ÿπ‘π‘’] = [π‘šπ‘Žπ‘ π‘  × π‘Žπ‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›] = 𝑀𝐿𝑇 βˆ’2
π‘‘π‘–π‘šπ‘’
]=
𝑇
= 𝐿𝑇 βˆ’2
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘“π‘œπ‘Ÿπ‘π‘’ = 𝑁 = π‘˜π‘”π‘šπ‘  βˆ’2
5.
[π‘Šπ‘œπ‘Ÿπ‘˜] = [π‘“π‘œπ‘Ÿπ‘π‘’ × π‘‘π‘–π‘ π‘π‘™π‘Žπ‘π‘’π‘šπ‘’π‘›π‘‘] = 𝑀𝐿2 𝑇 βˆ’2
6.
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘€π‘œπ‘Ÿπ‘˜ = 𝐽 = π‘˜π‘”π‘š2 𝑠 βˆ’2
[πΈπ‘™π‘’π‘π‘‘π‘Ÿπ‘–π‘ π‘β„Žπ‘Žπ‘Ÿπ‘”π‘’] = [π‘π‘’π‘Ÿπ‘Ÿπ‘’π‘›π‘‘ × π‘‘π‘–π‘šπ‘’)= AT
Unit for electric charge = C = A s
1
7.
[πΉπ‘Ÿπ‘’π‘žπ‘’π‘’π‘›π‘π‘¦] =
8.
𝑒π‘₯π‘‘π‘’π‘›π‘ π‘–π‘œπ‘›
Strain =
π‘œπ‘Ÿπ‘–π‘”π‘–π‘›π‘Žπ‘™ π‘™π‘’π‘›π‘”β„Žπ‘‘
π‘π‘’π‘Ÿπ‘–π‘œπ‘‘
= 𝑇 βˆ’1
=
𝐿
𝐿
which is dimensionless
More examples
Example - 01:
1
Check the correctness of physical equation, 𝑠 = 𝑒𝑑 + 2 π‘Žπ‘‘ 2 , Where 𝑒 is the initial velocity, vis
the final velocity, π‘Ž is the acceleration, 𝑠 is the displacement and 𝑑 is the time in which the
change occurs.
Solution:
1
Given equation is, 𝑠 = 𝑒𝑑 + 2 π‘Žπ‘‘ 2
𝐿. 𝐻. 𝑆. = 𝑣, β„Žπ‘’π‘›π‘π‘’ [𝐿. 𝐻. 𝑆. ] = [𝑠] = [𝐿1 𝑀0 𝑇 0 ] . . . . . . . . . . . . . . . . . . . (1)
𝑅. 𝐻. 𝑆 = 𝑒𝑑 +
1 2
π‘Žπ‘‘ , β„Žπ‘’π‘›π‘π‘’ [𝑅. 𝐻. 𝑆] = [𝑒][𝑑] + [π‘Ž] [𝑑]2
2
= [𝐿1 𝑀0 𝑇 βˆ’1 ][𝐿0 𝑀0 𝑇 1 ] + [𝐿1 𝑀0 𝑇 βˆ’2 ][𝐿0 𝑀0 𝑇 1 ]2
= [𝐿1 𝑀0 𝑇 0 ] + [𝐿1 𝑀0 𝑇 βˆ’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:
m𝑣 2
Check the correctness of physical equation,𝐹 = r , Where 𝐹 is the centripetal force acting
on a body of mass m performing uniform circular motion along a circle of radius π‘Ÿ with linear
speed 𝑣.
Solution:
Given the equation
𝐹=
m𝑣 2
r
𝐿. 𝐻. 𝑆. = 𝐹, β„Žπ‘’π‘›π‘π‘’ [𝐿. 𝐻. 𝑆. ] = [𝐹] = [𝐿1 𝑀1 𝑇 βˆ’2 ] . . . . . . . . . . . . . . . . … . . . . (1)
𝑅. 𝐻. 𝑆 =
=
m𝑣 2
, β„Žπ‘’π‘›π‘π‘’ [𝑅. 𝐻. 𝑆] = [π‘š][𝑣]2 /[π‘Ÿ]
r
[𝐿0 𝑀1 𝑇 0 ][𝐿1 𝑀0 𝑇 βˆ’1 ]2
[𝐿1 𝑀0 𝑇 0 ]
= [𝐿0 𝑀1 𝑇 0 ][𝐿2 𝑀0 𝑇 βˆ’2 ] [πΏβˆ’1 𝑀0 𝑇 0 ]
= [𝐿1 𝑀1 𝑇 βˆ’2 ] … … … … … … … . … … … … … … … … … . . . . . . . . . . . . . (2)
Hence by the principle of homogeneity, the given equation is dimensionally correct.
2
Example - 03:
Check the correctness of physical equation.
𝑣 2 = 𝑒2 + 2π‘Žπ‘  2
Solution:
Given equation is
𝑣 2 = 𝑒2 + 2π‘Žπ‘  2
𝐿. 𝐻. 𝑆. = 𝑣2 , β„Žπ‘’π‘›π‘π‘’ [𝐿. 𝐻. 𝑆. ] = [𝑣2 ] = [𝐿1 𝑀0 𝑇 βˆ’1 ]2 = [𝐿2 𝑀0 𝑇 βˆ’2 ] . . . . . . . . . . . . . . . (1)
𝑅. 𝐻. 𝑆 = 𝑒2 + 2π‘Žπ‘  2
= [𝐿1 𝑀0 𝑇 βˆ’1 ] + [𝐿1 𝑀0 𝑇 βˆ’2 ][𝐿1 𝑀0 𝑇 0 ]2
= [𝐿2 𝑀0 𝑇 βˆ’2 ] + [𝐿1 𝑀0 𝑇 βˆ’2 ][𝐿2 𝑀0 𝑇 0 ]
= [𝐿2 𝑀0 𝑇 βˆ’2 ] + [𝐿3 𝑀0 𝑇 βˆ’2 ]. … … … … … … … … … … … … … … . . . . . . . . . . (2)
πΉπ‘Ÿπ‘œπ‘š (1) π‘Žπ‘›π‘‘ (2) 𝑀𝑒 β„Žπ‘Žπ‘£π‘’ [𝐿. 𝐻. 𝑆. ] β‰  [𝑅. 𝐻. 𝑆. ]
Example - 04:
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
i).
𝑇=
ii).
𝑇 = π΄π‘™βˆšπΈ
iii).
𝑇 = 𝜌 βˆšπ‘”
𝐸
βˆšπ‘”π‘™ 3
𝜌
𝐴𝐸
𝑙
Where A is a dimensionless constant and g is the acceleration due to gravity.
Solution
Unit for period T = s
For equation (i)
𝐴𝜌
βˆšπ‘”π‘™ 3
𝑇=
𝐸
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ 𝑇 =
𝐴𝜌
π‘˜π‘”π‘šβˆ’3
1
βˆšπ‘”π‘™ 3 =
(π‘šπ‘  βˆ’2 π‘š3 )βˆ’ ⁄2
βˆ’1
βˆ’2
𝐸
π‘˜π‘”π‘š 𝑠
= 𝑠
Equation (i) is homogeneous and can be used.
For equation (ii)
𝜌
𝑇 = 𝐴𝑙 √
𝐸
1⁄
2
π‘˜π‘”π‘šβˆ’3
√
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ 𝑇 = 𝐴𝑙
= π‘š(
)
𝐸
π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’2
𝜌
=𝑠
3
Equation (ii) is homogeneous and can be used.
For equation (iii)
𝑇=
𝐴𝐸 𝑙
√
𝜌 𝑔
𝐴𝐸 𝑙 π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’2 π‘š
√ =
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ 𝑇 =
(
) = π‘š2 𝑠 βˆ’1
𝜌 𝑔
π‘˜π‘”π‘šβˆ’3 π‘šπ‘  βˆ’2
Which is different from the unit of β€˜T’, so it cannot be used.
Example - 05:
Check the homogeneity of equation, when the periodic time, β€œT” of vibration of magnet of
moment of inertia 'I', magnetic moment 'M' vibrating in magnetic induction 'B' is given by
𝐼
𝑇 = 2Ο€βˆš
𝑀𝐡
πœ‡
0𝐼
Where 𝐼 = mr 2 , 𝑀 = and 𝐡 = 2πœ‹π‘Ÿ
Given the equation;
𝐼
𝑇 = 2Ο€βˆš
𝑀𝐡
𝐿. 𝐻. 𝑆 = [𝑇] = [𝐿0 𝑀0 𝑇1 ] . . . . . . . . . . . . . . . (1)
[𝐼]
𝑅. 𝐻. 𝑆 = √
[𝑀][𝐡]
[𝑀1 𝐿2 𝑇 0 ]
= √ 2 0 0 1 0 1 βˆ’2 βˆ’1
[𝐿 𝑀 𝑇 𝐼 ][𝐿 𝑀 𝑇 𝐼 ]
[𝑀1 𝐿2 𝑇 0 ]
= √ 2 1 βˆ’2 0
[𝐿 𝑀 𝑇 𝐼 ]
𝑅. 𝐻. 𝑆 = √[𝐿0 𝑀0 𝑇 2 ] = [𝐿0 𝑀0 𝑇 1 ] … … … … … … … . . (2)
πΉπ‘Ÿπ‘œπ‘š (1)π‘Žπ‘›π‘‘ (2)𝑀𝑒 β„Žπ‘Žπ‘£π‘’ [𝐿. 𝐻. 𝑆. ] = [𝑅. 𝐻. 𝑆. ]
Hence by the principle of homogeneity, the given equation is dimensionally correct.
Example - 06:
Check the homogeneity of equation, when the rate of flow of a liquid having coefficient of
viscosity 'πœ‚' through a capillary tube of length '𝑙' and radius 'π‘Ž' under pressure head ′𝑝' is given
by
𝑑𝑉 πœ‹π‘π‘Ž4
=
𝑑𝑑
8π‘™πœ‚
Nb: the unit of β€² πœ‚β€² is the pascal per second.
Solution
Given the equation
𝑑𝑉 πœ‹π‘π‘Ž4
=
𝑑𝑑
8π‘™πœ‚
4
[𝑀0 𝐿3 𝑇 0 ]
[𝑑𝑉] [𝑑𝑉]
𝐿. 𝐻. 𝑆. =
=
=
= [𝑀0 𝐿3 𝑇 βˆ’1 ] . . . . . . . . . . . . . . . (1)
[𝑑𝑑]
[𝑑𝑑]
[𝑀0 𝐿0 𝑇 1 ]
[𝑀1 πΏβˆ’1 𝑇 βˆ’2 ] [𝑀0 𝐿1 𝑇 0 ]4
[𝑝][π‘Ž]4
𝑅. 𝐻. 𝑆 =
=
[𝑙][πœ‚]
[𝑀0 𝐿1 𝑇 0 ] [𝑀1 πΏβˆ’1 𝑇 βˆ’1 ]
=
[𝑀1 πΏβˆ’1 𝑇 βˆ’2 ][𝑀0 𝐿4 𝑇 0 ]
[𝑀1 𝐿0 𝑇 βˆ’1 ]
=
[𝑀1 𝐿3 𝑇 2 ]
= [𝑀0 𝐿3 𝑇 βˆ’1 ] … … … … … … … … … … . … … … … (2)
[𝑀1 𝐿0 𝑇 βˆ’1 ]
πΉπ‘Ÿπ‘œπ‘š (1)π‘Žπ‘›π‘‘ (2)𝑀𝑒 β„Žπ‘Žπ‘£π‘’ [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 ' πœ‚ ' is given
by
2π‘”π‘Ž2 (𝜎 βˆ’ 𝜌)
𝑣=
9πœ‚
Given the equation
2π‘”π‘Ž2 (𝜎 βˆ’ 𝜌)
𝑣=
9πœ‚
𝐿. 𝐻. 𝑆. = [𝑣] = [𝑀0 𝐿1 𝑇 βˆ’1 ] … … … … … … … … … … … . . . . . . . . . . . . . . . (1)
2π‘”π‘Ž2 (𝜎 βˆ’ 𝜌) [𝑔][π‘Ž]2 ([𝜎] βˆ’ [𝜌])
𝑅. 𝐻. 𝑆 =
=
9πœ‚
[πœ‚]
𝑅. 𝐻. 𝑆 =
[𝑀0 𝐿1 𝑇 βˆ’2 ] [𝑀0 𝐿1 𝑇 0 ]2 ([𝑀1 πΏβˆ’3 𝑇 0 ] βˆ’ [𝑀1 πΏβˆ’3 𝑇 0 ])
[𝑀1 πΏβˆ’1 𝑇 βˆ’1 ]
𝑅. 𝐻. 𝑆 =
[𝑀0 𝐿1 𝑇 βˆ’2 ][𝑀0 𝐿2 𝑇 0 ]([𝑀1 πΏβˆ’3 𝑇 0 ])
[𝑀1 πΏβˆ’1 𝑇 βˆ’1 ]
[𝑀1 𝐿0 𝑇 βˆ’2 ]
𝑅. 𝐻. 𝑆 = 1 βˆ’1 βˆ’1 = [𝑀0 𝐿1 𝑇 βˆ’2 ] … … … … … … … … … … … … … … . . (2)
[𝑀 𝐿 𝑇 ]
πΉπ‘Ÿπ‘œπ‘š (1)π‘Žπ‘›π‘‘ (2)𝑀𝑒 β„Žπ‘Žπ‘£π‘’ [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 = π‘Ž + 𝑏𝑑 + 𝑐𝑑 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 ] = [π‘Ž]
5
Dimensions of 'a' are that of displacement. Hence 'a' represent the displacement.
Dimensions of S = Dimensions of 𝑏𝑑
[𝐿1 𝑀0 𝑇 0 ] = [𝑏][𝑑]
[𝐿1 𝑀0 𝑇 0 ]
[𝑏] =
[𝑇]
[𝑏] = [𝐿1 𝑀0 𝑇 βˆ’1 ]
Dimensions of 'a' are that of velocity. Hence 'b' represents the velocity.
Dimensions of S = Dimensions of 𝑐𝑑 2 .
[𝐿1 𝑀0 𝑇 0 ] = [𝑐]
[𝐿1 𝑀0 𝑇 0 ]
[𝑐] =
[𝑇]2
[𝑐] = [𝐿1 𝑀0 𝑇 βˆ’2 ]
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 βˆ’1 ], and [𝐿1 𝑀0 𝑇 βˆ’2 ].
The quantities represented by a, b and c are displacement, velocity and acceleration.
Example - 09:
𝛾𝑃 1
According to Laplace's formula, the velocity (V) of sound in a gas is given by 𝑣 = ( ⁄𝑑 ) ⁄2 ,
where P is the pressure, d is the density of a gas. What is the dimensional formula for 𝛾?
Solution:
𝛾𝑃⁄
𝑑
2
𝑣
𝑑
⁄𝑃
βˆ΄π›Ύ=
∴ 𝑣2 =
βˆ΄π›Ύ=
[𝑣 2 ][𝑑] [𝐿1 𝑀0 𝑇 βˆ’1 ][πΏβˆ’3 𝑀1 𝑇 0 ]
=
[πΏβˆ’1 𝑀1 𝑇 βˆ’2 ]
[𝑃]
βˆ΄π›Ύ=
[πΏβˆ’1 𝑀1 𝑇 βˆ’2 ]
= [𝐿0 𝑀0 𝑇 0 ]
[πΏβˆ’1 𝑀1 𝑇 βˆ’2 ]
Example - 10:
A force F is given by 𝐹 = π‘Žπ‘‘ + 𝑏𝑑 2 , where 𝑑 is the time. Find the dimensional formula of 'π‘Ž' 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 π‘Žπ‘‘
[𝐿1 𝑀1 𝑇 βˆ’2 ] = [π‘Ž][𝑑]
6
[π‘Ž] =
[𝐿1 𝑀1 𝑇 βˆ’2 ]
[𝑇]
[π‘Ž] = [𝐿1 𝑀1 𝑇 βˆ’3 ]
Dimensions of F = Dimensions of 𝑏𝑑 2
[𝐿1 𝑀1 𝑇 βˆ’2 ] = [𝑏][𝑇]2
[𝑏] =
[𝐿1 𝑀1 𝑇 βˆ’2 ]
[𝑇]2
[𝑏] = [𝐿1 𝑀1 𝑇 βˆ’4 ]
The dimensional formula of a and b are [𝐿1 𝑀1 𝑇 βˆ’3 ] and [𝐿1 𝑀1 𝑇 βˆ’4 ]
Example - 11:
In an equation {𝑃 + π‘Žβ„π‘‰ 2 }(𝑉 βˆ’ 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 π‘Ž ⁄𝑏?
Solution:
Two physical quantities can be added or subtracted if and only if their dimensions are the
same.
Thus, the dimensions of P and π‘Žβ„π‘‰ 2 are same
[π‘Ž]
[𝑃] =
2
[𝑉]
∴ [π‘Ž] = [𝑃][𝑉]
2
∴ [π‘Ž] = [πΏβˆ’1 𝑀1 𝑇 βˆ’2 ][𝐿3 ]
2
∴ [π‘Ž] = [πΏβˆ’1 𝑀1 𝑇 βˆ’2 ][𝐿6 ]
∴ [π‘Ž] = [𝐿5 𝑀1 𝑇 βˆ’2 ]
Dimensions of b = Dimensions of V
[𝑏] = [𝐿3 𝑀0 𝑇 0 ]
Now the dimensions of
[a] [𝐿5 𝑀1 𝑇 βˆ’2 ]
is 3 0 0 = [𝐿2 𝑀1 𝑇 βˆ’2 ]
[𝐿 𝑀 𝑇 ]
[b]
The dimensions of π‘Žβ„π‘ is = [𝐿2 𝑀1 𝑇 βˆ’2 ]4+
Practice questions
1. What are the S. I. units of k so that the equation?
π’—π’†π’π’π’„π’Šπ’•π’š = π‘˜ × π’…π’†π’π’”π’Šπ’•π’š
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 = π‘˜π‘Žπ‘£ where π‘˜ is a constant What are
the S.I. units of π‘˜ 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:
𝐢 = 𝛼𝑇 + 𝛽𝑇 3
7
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 + β„Ž 𝜌 𝑔 + πœŒπ‘£ 2 = π‘˜
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.
8
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