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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).
Examples
1.
[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
2.
[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
3.
[π΄π‘π‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘›] = [
π‘β„Žπ‘Žπ‘›π‘”π‘’ 𝑖𝑛 π‘£π‘’π‘™π‘œπ‘π‘–π‘‘π‘¦
π‘‘π‘–π‘šπ‘’
]=
𝐿𝑇 βˆ’1
𝑇
= 𝐿𝑇 βˆ’2
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘Žπ‘π‘’π‘™π‘’π‘Ÿπ‘Žπ‘‘π‘–π‘œπ‘› = π‘š/𝑠 2
4.
[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‘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
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ π‘€π‘œπ‘Ÿπ‘˜ = 𝐽 = π‘˜π‘”π‘š2 𝑠 βˆ’2
6.
[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:
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 π‘Ÿ with
linear speed 𝑣.
Solution:
Given the equation
m𝑣 2
𝐹=
r
𝐿. 𝐻. 𝑆. = 𝐹, β„Žπ‘’π‘›π‘π‘’ [𝐿. 𝐻. 𝑆. ] = [𝐹] = [𝐿1 𝑀1 𝑇 βˆ’2 ] . . . . . . . . . . . . . . . . … . . . . (1)
m𝑣 2
𝑅. 𝐻. 𝑆 =
, 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 ]
2
= [𝐿1 𝑀1 𝑇 βˆ’2 ] … … … … … … … . … … … … … … … … … . . . . . . . . . . . . . (2)
Hence by the principle of homogeneity, the given equation is dimensionally correct.
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:
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.
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
βˆšπ‘”π‘™ 3
i).
𝑇=
ii).
𝑇 = π΄π‘™βˆšπΈ
iii).
𝑇 = 𝜌 βˆšπ‘”
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 βˆ’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 Nπ‘šβˆ’2 p = 8 500 kg π‘šβˆ’3. [Unit for E = kg
π‘šβˆ’1 𝑆 βˆ’2]
Solution
a) Unit for period T = s
For equation (i)
𝐴𝜌
βˆšπ‘”π‘™ 3
𝑇=
𝐸
𝐴𝜌
π‘˜π‘”π‘šβˆ’3
1
3
βˆšπ‘”π‘™ =
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ 𝑇 =
(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
Equation (ii) is homogeneous and can be used.
4
For equation (iii)
𝑇=
𝐴𝐸 𝑙
√
𝜌 𝑔
π‘ˆπ‘›π‘–π‘‘ π‘“π‘œπ‘Ÿ 𝑇 =
𝐴𝐸 𝑙 π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’2 π‘š
√ =
(
) = 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.
b)
Frequency f /cycle π’”βˆ’πŸ
256
288
320
384
480
Period T/(x πŸπŸŽβˆ’πŸ‘ s)
3·91
3.47
3·13
2·60
2·08
Length I/cm
12.0
10.6
9.6
8.0
6.4
πŸ‘
πŸ‘
𝒍 β„πŸ /π’„π’Ž β„πŸ
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 = π΄π‘™βˆšπΈ is correct.
The graph or T against is a straight line but does not pass through the origin. Therefore, the
equation 𝑇 =
𝐴𝜌
𝐸
βˆšπ‘”π‘™ 3 is wrong.
5
𝜌
From the equa1ion 𝑇 = π΄π‘™βˆšπΈ and from the table. when l = 12·0 cm. T = 3 · 91 × 10βˆ’ 3 𝑠,
𝐴=
𝑇 𝐸
√
𝑙 𝜌
𝐴=
𝑇 𝐸
√
𝑙 𝜌
(3 · 91 × 10βˆ’ 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 '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‚
[𝑀0 𝐿3 𝑇 0 ]
[𝑑𝑉] [𝑑𝑉]
=
=
= [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 ]
𝐿. 𝐻. 𝑆. =
=
[𝑀1 πΏβˆ’1 𝑇 βˆ’2 ][𝑀0 𝐿4 𝑇 0 ]
[𝑀1 𝐿0 𝑇 βˆ’1 ]
[𝑀1 𝐿3 𝑇 2 ]
= 1 0 βˆ’1 = [𝑀0 𝐿3 𝑇 βˆ’1 ] … … … … … … … … … … . … … … … (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 - 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
6
2π‘”π‘Ž2 (𝜎 βˆ’ 𝜌)
9πœ‚
0 1 βˆ’1 ]
[𝑣]
[𝑀
𝐿. 𝐻. 𝑆. =
=
𝐿𝑇
… … … … … … … … … … … . . . . . . . . . . . . . . . (1)
𝑣=
𝑅. 𝐻. 𝑆 =
2π‘”π‘Ž2 (𝜎 βˆ’ 𝜌) [𝑔][π‘Ž]2 ([𝜎] βˆ’ [𝜌])
=
9πœ‚
[πœ‚]
[𝑀0 𝐿1 𝑇 βˆ’2 ] [𝑀0 𝐿1 𝑇 0 ]2 ([𝑀1 πΏβˆ’3 𝑇 0 ] βˆ’ [𝑀1 πΏβˆ’3 𝑇 0 ])
𝑅. 𝐻. 𝑆 =
[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]
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 ]
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 βˆ’1 ], and [𝐿1 𝑀0 𝑇 βˆ’2 ].
The quantities represented by a, b and c are displacement, velocity and acceleration.
Example - 09:
𝛾𝑃⁄ 1⁄2
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
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]
[πΏβˆ’1 𝑀1 𝑇 βˆ’2 ]
∴ 𝛾 = βˆ’1 1 βˆ’2 = [𝐿0 𝑀0 𝑇 0 ]
[𝐿 𝑀 𝑇 ]
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]
[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
[𝐿1 𝑀1 𝑇 βˆ’2 ]
[𝑏] =
[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?
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 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 ]
[a] [𝐿5 𝑀1 𝑇 βˆ’2 ]
Now the dimensions of
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‘œπ‘π‘–π‘‘π‘¦ = π‘˜ × π‘‘π‘’π‘›π‘ π‘–π‘‘π‘¦
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
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.
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 π‘₯ 𝑔y
9
where.π‘₯ and 𝑦 are non-dimensional constants.
Then
𝑇 = π‘˜ 𝑙 π‘₯ 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 [π‘˜ 𝑙 π‘₯ 𝑔y ]
𝑠 = π‘š π‘₯ (π‘šπ‘  βˆ’2 )𝑦
= π‘š π‘₯+𝑦 𝑠 βˆ’2
𝑦
Equating indices of s:
1 = βˆ’2𝑦. ∴ 𝑦 = βˆ’
Equating indices of m:
0=π‘₯+𝑦
1
π‘₯ = βˆ’π‘¦ =
2
1
2
βˆ΄π‘‡=π‘˜π‘™
1⁄
2
π‘”βˆ’
1⁄
2
𝑙
βˆ΄π‘‡=π‘˜βˆš
𝑔
Example
The speed of sound 𝑣 in a medium depends on its wavelength πœ†. 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: π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’2.
Solution:
Suppose:
𝑣 = π‘˜ πœ†π‘₯ 𝐸 y 𝜌 z
where π‘˜, π‘₯, 𝑦 and 𝑧 are non-dimensional constants.
Units of v = units of [π‘˜ πœ†π‘₯ 𝐸 y 𝜌z ]
π‘šπ‘  βˆ’1 = π‘š π‘₯ (π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’2 )𝑦 (π‘˜π‘”π‘šβˆ’3 )z
= π‘š π‘₯βˆ’π‘¦βˆ’3𝑧 𝑠 βˆ’2y π‘˜π‘”y+z
Equating indices of s:
βˆ’1 = βˆ’2𝑦. ∴ 𝑦 =
Equating indices of kg:
0=𝑦+𝑧
1
2
10
π‘₯ = βˆ’π‘¦ =
1
1
∴ 𝑧 = βˆ’π‘¦ = βˆ’
2
2
Equating indices of m:
1 = π‘₯ βˆ’ 𝑦 βˆ’ 3𝑧
1 3
=π‘₯βˆ’ βˆ’
2 2
∴π‘₯=0
βˆ΄π‘‡=π‘˜πΈ
1⁄
2
πœŒβˆ’
1⁄
2
𝐸
βˆ΄π‘£=π‘˜βˆš
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) πœ‚ 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 π‘Ž, 𝑙, 𝑝 and πœ‚.
Solution
π‘ π‘’π‘π‘π‘œπ‘ π‘’ π‘‘β„Žπ‘’ π‘Ÿπ‘Žπ‘‘π‘’ π‘œπ‘“ π‘“π‘™π‘œπ‘€,
𝑑𝑉
𝑝 𝑦
= π‘˜π‘Ž π‘₯ ( ) πœ‚ 𝑧
𝑑𝑑
𝑙
π‘€β„Žπ‘’π‘Ÿπ‘’ π‘˜, π‘₯ 𝑦 π‘Žπ‘›π‘‘ 𝑧 π‘Žπ‘Ÿπ‘’ π‘‘π‘–π‘šπ‘’π‘›π‘ π‘–π‘œπ‘›π‘™π‘’π‘ π‘  π‘π‘œπ‘›π‘ π‘‘π‘Žπ‘›π‘‘π‘ .
π‘ˆπ‘›π‘–π‘‘π‘  π‘œπ‘“
𝑑𝑉
𝑝 𝑦
= π‘ˆπ‘›π‘–π‘‘π‘  π‘œπ‘“ [π‘˜π‘Ž π‘₯ ( ) πœ‚ 𝑧 ]
𝑑𝑑
𝑙
𝑦
3 βˆ’1
π‘š 𝑠
π‘˜π‘”π‘šπ‘  βˆ’2 1
= π‘š ×(
× ) (π‘˜π‘”π‘šβˆ’1 𝑠 βˆ’1 ) 𝑧
π‘š2
π‘š
π‘₯
= π‘˜π‘”y+z π‘š π‘₯βˆ’2π‘¦βˆ’π‘§ 𝑠 βˆ’2yβˆ’z
11
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