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# Homogineous Equations

advertisement ```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).
1.
2.
Examples
[𝐴𝑟𝑒𝑎] = [𝑙𝑒𝑛𝑔𝑡ℎ × 𝑏𝑟𝑒𝑎𝑑𝑡ℎ]
= 𝐿 × 𝐿 = 𝐿2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑎𝑟𝑒𝑎 = 𝑚2
[𝑉𝑒𝑙𝑜𝑐𝑖𝑡𝑦] = [
𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡
𝑡𝑖𝑚𝑒
𝐿
] = 𝑇 = 𝐿𝑇 −1
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦 = 𝑚𝑠 −1
𝑐ℎ𝑎𝑛𝑔𝑒 𝑖𝑛 𝑣𝑒𝑙𝑜𝑐𝑖𝑡𝑦
𝐿𝑇 −1
3.
[𝐴𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛] = [
4.
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑎𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛 = 𝑚/𝑠 2
[𝐹𝑜𝑟𝑐𝑒] = [𝑚𝑎𝑠𝑠 × 𝑎𝑐𝑐𝑒𝑙𝑒𝑟𝑎𝑡𝑖𝑜𝑛] = 𝑀𝐿𝑇 −2
𝑡𝑖𝑚𝑒
]=
𝑇
= 𝐿𝑇 −2
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑓𝑜𝑟𝑐𝑒 = 𝑁 = 𝑘𝑔𝑚𝑠 −2
5.
[𝑊𝑜𝑟𝑘] = [𝑓𝑜𝑟𝑐𝑒 × 𝑑𝑖𝑠𝑝𝑙𝑎𝑐𝑒𝑚𝑒𝑛𝑡] = 𝑀𝐿2 𝑇 −2
6.
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑤𝑜𝑟𝑘 = 𝐽 = 𝑘𝑔𝑚2 𝑠 −2
[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:
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 𝑟 with linear
speed 𝑣.
Solution:
Given the equation
𝐹=
m𝑣 2
r
𝐿. 𝐻. 𝑆. = 𝐹, ℎ𝑒𝑛𝑐𝑒 [𝐿. 𝐻. 𝑆. ] = [𝐹] = [𝐿1 𝑀1 𝑇 −2 ] . . . . . . . . . . . . . . . . … . . . . (1)
𝑅. 𝐻. 𝑆 =
=
m𝑣 2
, 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 ]
= [𝐿1 𝑀1 𝑇 −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 = 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:
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 = 𝐴𝑙√𝐸
iii).
𝑇 = 𝜌 √𝑔
𝐸
√𝑔𝑙 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
√𝑔𝑙 3
𝑇=
𝐸
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑇 =
𝐴𝜌
𝑘𝑔𝑚−3
1
√𝑔𝑙 3 =
(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
3
Equation (ii) is homogeneous and can be used.
For equation (iii)
𝑇=
𝐴𝐸 𝑙
√
𝜌 𝑔
𝐴𝐸 𝑙 𝑘𝑔𝑚−1 𝑠 −2 𝑚
√ =
𝑈𝑛𝑖𝑡 𝑓𝑜𝑟 𝑇 =
(
) = 𝑚2 𝑠 −1
𝜌 𝑔
𝑘𝑔𝑚−3 𝑚𝑠 −2
Which is different from the unit of ‘T’, so it cannot be used.
Example - 05:
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.
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
4
[𝑀0 𝐿3 𝑇 0 ]
[𝑑𝑉] [𝑑𝑉]
𝐿. 𝐻. 𝑆. =
=
=
= [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 ]
=
[𝑀1 𝐿−1 𝑇 −2 ][𝑀0 𝐿4 𝑇 0 ]
[𝑀1 𝐿0 𝑇 −1 ]
=
[𝑀1 𝐿3 𝑇 2 ]
= [𝑀0 𝐿3 𝑇 −1 ] … … … … … … … … … … . … … … … (2)
[𝑀1 𝐿0 𝑇 −1 ]
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
2𝑔𝑎2 (𝜎 − 𝜌)
𝑣=
9𝜂
𝐿. 𝐻. 𝑆. = [𝑣] = [𝑀0 𝐿1 𝑇 −1 ] … … … … … … … … … … … . . . . . . . . . . . . . . . (1)
2𝑔𝑎2 (𝜎 − 𝜌) [𝑔][𝑎]2 ([𝜎] − [𝜌])
𝑅. 𝐻. 𝑆 =
=
9𝜂
[𝜂]
𝑅. 𝐻. 𝑆 =
[𝑀0 𝐿1 𝑇 −2 ] [𝑀0 𝐿1 𝑇 0 ]2 ([𝑀1 𝐿−3 𝑇 0 ] − [𝑀1 𝐿−3 𝑇 0 ])
[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]
5
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 ]
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
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
∴𝛾=
∴ 𝑣2 =
∴𝛾=
[𝑣 2 ][𝑑] [𝐿1 𝑀0 𝑇 −1 ][𝐿−3 𝑀1 𝑇 0 ]
=
[𝐿−1 𝑀1 𝑇 −2 ]
[𝑃]
∴𝛾=
[𝐿−1 𝑀1 𝑇 −2 ]
= [𝐿0 𝑀0 𝑇 0 ]
[𝐿−1 𝑀1 𝑇 −2 ]
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]
6
[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
[𝑏] =
[𝐿1 𝑀1 𝑇 −2 ]
[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?
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 ]
Now the dimensions of
[a] [𝐿5 𝑀1 𝑇 −2 ]
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
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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:
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𝑝 + ℎ 𝜌 𝑔 + 𝜌𝑣 2 = 𝑘
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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.
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