Chapter 2.18 - Relation among Physical Quantities

In the previous section, we saw how to convert the units from one system to another using dimensional analysis. In this section, we will see another application of dimensional analysis, which is: Deducing relation among physical quantities

We know that the basic physical quantities like mass, length and time do not depend on other physical quantities. 
We can say that, mass, length, time etc., are independent physical quantities
But the derived physical quantities like velocity, force etc., are dependent on other physical quantities
We will want to know the 'nature of this dependence' 
A dimensional analysis can help us to achieve it
For using dimensional analysis in this regard, two conditions must be satisfied:
(i) The physical quantity must not depend on more than 3 physical quantities
That is., there must not be more than 3 quantities on the right side
(ii) The dependence must be a 'product type'
That is., the physical quantities on the right side must be multiplied together

Let us see an example:
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
Solution:
1. Let the time period be denoted as T
2. Given that T depends on l, g and m
• But we do not know the powers to which 'l, g and m' have to be raised
3. We can write the dependence as: T = k × l^x × g^y × m^z
    ♦ Where 'k' is a dimensionless constant
4. Considering the dimensions on both sides, we can write:
[M⁰L⁰T¹] = [L¹]^x [LT^-2]^y [M]^z
⇒ [M⁰L⁰T¹] = [L^x] [L^y T^-2y] [M^z]
⇒ [M⁰L⁰T¹] = [L^x+y] [T^-2y] [M^z]
5. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y = 0
(ii) -2y = 1
(iii) z = 0
• From (ii), we get: y = -¹⁄₂
• Substituting this in (i), we get: x - ¹⁄₂ = 0
⇒ x = ¹⁄₂
■ So we can write:
• x = ¹⁄₂
• y = -¹⁄₂
• z = 0
6. Substituting these in (3),we get:
$\mathbf\small{T=k\times l^{\frac{1}{2}}\times g^{-\frac{1}{2}}}$
$\mathbf\small{\Rightarrow T=k \sqrt{\frac{l}{g}}}$
7. Note that, the value of 'k' cannot be obtained by the method of dimensions
• Multiplying by a number on the right side does not affect the dimensions

We will now see some solved examples
Solved example 2.42
The centripetal force F acting on a particle moving uniformly in a circle depends upon mass (m), velocity (v) and radius (r) of the circle. Derive the expression for F using method of dimensions
Solution:
1. Given that F depends on m, v and r
• But we do not know the powers to which 'm, v and r' have to be raised
2. We can write the dependence as: F = k × m^x × v^y × r^z
    ♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M¹L¹T^-2] = [M¹]^x [LT^-1]^y [L]^z
⇒ [M¹L¹T^-2] = [M^x] [L^y T^-y] [L^z]
⇒ [M¹L¹T^-2] = [M^x] [L^y+z] [T^-y]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x = 1
(ii) y+z = 1
(iii) y = 2
Substituting 'y=2' in (ii), we get: z = -1
■ So we can write:
• x = 1
• y = 2
• z = -1
5. Substituting these in (2),we get:
$\mathbf\small{F=k\times m^1\times v^2\times r^{-1}}$
$\mathbf\small{F=k \frac{mv^2}{r}}$
• We will see the actual value of 'k' when we learn about centripetal force in a later chapter

Solved example 2.43
A gas bubble formed from an explosion under water oscillates with a period T. This period T is proportional to the pressure (p), density of water (d) and the energy of explosion (E). Derive the expression for T using method of dimensions
Solution:
1. Given that T depends on p, d and E
• But we do not know the powers to which 'p, d and E' have to be raised
2. We can write the dependence as: T = k × p^x × d^y × E^z
    ♦ Where 'k' is a dimensionless constant
3. Considering the dimensions on both sides, we can write:
[M⁰L⁰T¹] = [M¹L^-1T^-2]^x [M¹L^-3]^y [M¹L²T^-2]^z
⇒ [M⁰L⁰T¹] = [M^xL^-xT^-2x] [M^yL^-3y] [M^zL^2zT^-2z]
⇒ [M⁰L⁰T¹] = [M^x+y+z] [L^-x-3y+2z] [T^-2x-2z]
4. Equating the dimensions on both sides, we get 3 linear equations:
(i) x+y+z = 0
(ii) -x-3y+2z = 0
(iii) -2x-2z = 1
• We have 3 equations and 3 unknowns
• Solving them, we get:
x = ^-5⁄₆
y = ¹⁄₂
z= ¹⁄₃.

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We have completed this discussion on units and measurements. In the next chapter, we will see Motion in a straight line

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Chapter 2.17 - Unit Conversion

In the previous section, we saw how to check whether an equation is dimensionally correct or not. In this section, we will see another application of dimensional analysis, which is: Unit conversion

Let us see an example. We will write it in steps:
1. An engineer measures the length of a wall and notes it down as 3.05 m
• Another engineer measures the length of the wall and notes it down as 10 feet
■ Doing a dimensional analysis will help us to understand why the difference occurred:
2. The first engineer used the SI system
• We know that, in this system, the basic unit for length is 1 m
• So we can write: [L] = [1 m]
3. The second engineer use the Imperial system
• In this system, the basic unit for length is 1 foot
• Let L₁ represent the basic unit for length in that system
• From the data book, we have: 1 foot = 0.3048 m
(This information is available online also)
• So we can write: [L₁] = [0.3048 m] 
4. Let 'x' be the reading obtained by the second engineer
• This implies that, 'x' number of his units will make up the 'length of the wall'
■ The length of the wall is a constant regardless of the system used for measuring it's length
5. So we can write:
3.05 × [L] = x × [L₁]
⇒ 3.05 × [1 m] = x × [0.3048 m]
$\mathbf\small{\Rightarrow x = \frac{3.05 \times 1}{0.3048}=10.00}$
• So the reading obtained by the second engineer will be 10 feet

Let us see an example in volume:
1. An engineer calculates the volume of a rectangular prism as 5.67 m³.
• Another engineer calculates the volume of the prism as 200.23 cubic feet
2. Let us do a dimensional analysis and see if the same 200.23 can be obtained:
• We can write:
5.67 × [L³] = x × [L₁³]
⇒ 5.05 × [1 m³] = x × [0.3048³ m³]
$\mathbf\small{\Rightarrow x = \frac{5.67 \times 1}{0.3048^3}=200.23}$
• So the reading obtained by the second engineer will be 200.23 cubic feet

An example in speed:
In an experiment, the speed of a body was obtained as 10 ms^-1. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for length is km
• Basic unit for time is hours
■ What would be the result of the experiment?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{10[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. L₁ is the basic unit for length in the new system. It is given as 1 km
    ♦ So we get: L₁ = 1 km = 1000 m
• T₁ is the basic unit for time in the new system. It is given as 1 hour
    ♦ So we get: T₁ = 1 hour = (60 × 60) s
4. Substituting the values, we get:
$\mathbf\small{\frac{10[1\,m]}{[1\,s]}=\frac{x[1000\,m]}{[(60\times 60)s]}}$
$\mathbf\small{\Rightarrow x=\frac{10\times(60\times 60)}{1000}=36}$
• So the result would be 36 km hr^-1

Now we will see some solved examples
Solved example 2.37
In an experiment which uses the SI system, the energy was measured as 100 J. Suppose that the measurements were made using another system in which, the basic units are as follows:
• Basic unit for mass is 250 g
• Basic unit for length is 20 cm
• Basic unit for time is half a minute
■ What would be the result of the experiment?
Solution:
1. The dimensions of energy are: $\mathbf\small{[ML^2T^{-2}]}$
• This can be written as: $\mathbf\small{\frac{[M][L^2]}{[T^{2}]}}$
2. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{100[M][L^2]}{[T^{2}]}=\frac{x[M_1][L_1^2]}{[T_1^{2}]}}$
3. M₁ is the basic unit for mass in the new system. It is given as 250 g
    ♦ So we get: M₁ = 250 g = 0.250 kg
• L₁ is the basic unit for length in the new system. It is given as 20 cm
    ♦ So we get: L₁ = 20 cm = 0.20 m
• T₁ is the basic unit for time in the new system. It is given as half a minute
    ♦ So we get: T₁ = half a minute = 30 s
4. Substituting the values, we get:
$\mathbf\small{\frac{100[1\,kg][1^2\,m^2]}{[1^2\,s^2]}=\frac{x[0.250\,kg][0.20^2\,m^2]}{[30^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=\frac{100\times 30^2}{0.250\times 0.20^2}=9 \times 10^6}$
• So the result would be 9 × 10⁶ new units.

Solved example 2.38
The value of gravitational constant is G = 6.67 × 10^-11 N m² kg^-2 in SI units. Convert it into CGS system of units
Solution:
1. In the CGS system,
• Basic unit for mass is 1 g
• Basic unit for length is 1 cm
• Basic unit for time is 1 s
2. The unit of G is given as N m² kg^-2.
• N is the unit of force. Force has dimensions $\mathbf\small{[MLT^{-2}]}$
• m is the unit of length. Length has dimension [L]
• kg is the unit for mass. Mass has the dimension [M]
3. So the dimensions of G are: $\mathbf\small{\frac{[MLT^{-2}][L^2]}{[M^{2}]}}$   
• This can be written as: $\mathbf\small{\frac{[L^3]}{[M][T^{2}]}}$
4. Let 'x' be the new result
• So we can write: $\mathbf\small{\frac{6.67 \times 10^{-11}[L^3]}{[M][T^{2}]}=\frac{x[L_1^3]}{[M_1][T_1^{2}]}}$
5. M₁ is the basic unit for mass in the new system. It is 1 g
    ♦ So we get: M₁ = 1 g = 0.001 kg
• L₁ is the basic unit for length in the new system. It is 1 cm
    ♦ So we get: L₁ = 1 cm = 0.01 m
• T₁ is the basic unit for time in the new system. It is the same 1 s
    ♦ So we get: T₁ = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{6.67 \times 10^{-11}[1^3\,m^3]}{[1\,kg][1^{2}\,s^2]}=\frac{x[0.01^3\,m^3]}{[0.001\,kg][1^{2}\,s^2]}}$
$\mathbf\small{\Rightarrow x =\frac{6.67 \times 10^{-11}\times 0.001}{0.01^3}=6.67 \times 10^{-8}}$

Solved example 2.39
A calorie is a unit of heat (energy in transit) and it equals about 4.2 J where 1 J = 1 kg m² s^-2. Suppose we employ a system of units in which the unit of mass equals α kg, the unit of length equals β m, the unit of time is γ s. Show that a calorie has a magnitude 4.2 α^-1 β^-2 γ² in terms of the new units
Solution:
1. Given that: 1 J = 1 kg m² s^-2 
• kg is the unit for mass. Mass has the dimension [M]
• m is the unit of length. Length has dimension [L]
• s is the unit for time. Time has the dimension [T]
2. So the dimensions of energy are: $\mathbf\small{\frac{[M][L^2]}{[T^2]}}$
3. Let 'x' be the new result
• So we can write:
$\mathbf\small{\frac{4.2[M][L^2]}{[T^2]}=\frac{x[M_1][L_1^2]}{[T_1^2]}}$
4. M₁ is the basic unit for mass in the new system. It is α kg
• L₁ is the basic unit for length in the new system. It is β m
• T₁ is the basic unit for time in the new system. It is γ s
5. Substituting the values, we get:
$\mathbf\small{\frac{4.2[1\,kg][1^2\;m^2]}{[1^2\,s^2]}=\frac{x[\alpha\,kg][\beta^2\;m^2]}{[\gamma^2\,s^2]}}$
$\mathbf\small{\Rightarrow x=4.2\,\alpha^{-1}\beta^{-2}\gamma^2}$

Solved example 2.40
A new unit of length is chosen such that the speed of light in vacuum is unity. What is the distance between the Sun and the Earth in terms of the new unit if light takes 8 min and 20 s to cover this distance?
Solution:
1. The dimensions of speed are: $\mathbf\small{[LT^{-1}]}$
• This can be written as: $\mathbf\small{\frac{[L]}{[T]}}$
2. When we employ the SI system, the speed of light is 3 × 10⁸ ms^-1 
• In the 'new system', let the speed of light be 'x'
• So we can write:
$\mathbf\small{\frac{3\times 10^8[L]}{[T]}=\frac{x[L_1]}{[T_1]}}$
3. But 'x' is given to us as 'unity'
• That means, in the new system, the light travels '1 new unit' in one second
4. L₁ is the basic unit for length in the new system. We have to find it
5. T₁ is the basic unit for time in the new system
• It is clear that: T₁ = 1 s
6. Substituting the values, we get:
$\mathbf\small{\frac{3\times 10^8[1\,m]}{[1\,s]}=\frac{1[L_1]}{[1\,s]}}$
$\mathbf\small{\Rightarrow L_1=3\times 10^8\,m}$
• That means, 'the basic unit for length (L₁) in the new system' is equal to 3 × 10⁸ m
7. Now we apply the method in reverse:
(The word 'reverse' can be used because, first we calculated L₁ using x. Now we are going to calculate x using L₁)
(i) Distance in the SI system 
= Velocity in the SI system × Time in the SI system
= 3 × 10⁸ ms^-1 × [(8 × 60) + 20] s
= 3 × 10⁸ ms^-1 × [500] s
= 3 × 500 × 10⁸ m.
(ii) The dimension of distance is: [L]
(iii). Let 'x' be the new result
• So we can write:
3 × 10⁸[L] = x[L₁]
(iv) L₁ is the basic unit for length in the new system. We obtained it as 3 × 10⁸ m 
    ♦ So we get: L₁ = 3 × 10⁸ m
(v) Substituting the values in (iii), we get:
3 × 500 × 10⁸[1 m] = x[3 × 10⁸ m]
⇒ x = 500
• So we can write:
Distance between Sun and Earth = 500 'new units'  

Solved example 2.41
The photograph of a house occupies an area of 1.75 cm² on a 35 mm slide. The slide is projected on to a screen, and the area of the house on the screen is 1.55 m². What is the linear magnification of the projector-screen arrangement.
Solution:
1. Consider the two rectangles in fig.2.26 below:

Fig.2.26

• The larger rectangle is a 'proportional enlargement' of the smaller rectangle
(Details about proportionality can be seen here)
2. Length l₂ of the larger rectangle is obtained by multiplying l₁ by a factor 'k'
    ♦ Where l₁ is the length of the smaller rectangle
• Width b₂ of the larger rectangle is obtained by multiplying b₁ by the same factor 'k'
    ♦ Where b₁ is the width of the smaller rectangle
3. Note that, the same factor 'k' must be used for both length and width. otherwise, there will be distortion
• Thus we get: (l₁ × b₁) = 1.75 cm² 
(l₂ × b₂) = 1.55 m² = 1.55 × 10⁴ cm² 
⇒ (kl₁ × kb₁) = 1.55 × 10⁴ cm² 
⇒ k² × (l₁ × b₁) = 1.55 × 10⁴ cm².
4. But (l₁ × b₁) = 1.75 cm²
• So we get: k²(1.75) = 1.55 × 10⁴ cm²  
⇒ k² = (^1.55⁄_1.75) × 10⁴ = 0.885714286 × 10⁴ = 8857.14
⇒ k = √(8857.14) = 94.11
• So the linear magnification (k) = 94.11

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In the next section, we will see how to deduce relation among physical quantities

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Chapter 2.16 - Dimensional Analysis

In the previous section, we saw dimensions of physical quantities. In this section, we will see dimensional analysis and it's applications

Consider the arithmetic operations of addition and subtraction. Let us see some examples involving those operations
Example 1:
• Suppose we have two masses
    ♦ We can add the 'magnitude of one mass' to the 'magnitude of the other mass'
    ♦ We can subtract the 'magnitude of the smaller mass' from the 'magnitude of the larger mass' 
• Note that, both the quantities involved in addition/subtraction:
    ♦ have the same unit kg
    ♦ have the same dimension [M]

Example 2:
• Suppose we have two forces
    ♦ We can add the 'magnitude of one force' to the 'magnitude of the other force'
    ♦ We can subtract the 'magnitude of the smaller force' from the 'magnitude of the larger force'
• Note that, both the quantities involved in addition/subtraction:
    ♦ have the same unit N
    ♦ have the same dimension $\mathbf\small{\left[MLT^{-2}\right]}$ 

Example 3:
• Suppose we have two volumes
    ♦ We can add the 'magnitude of one volume' to the 'magnitude of the other volume'
    ♦ We can subtract the 'magnitude of the smaller volume' from the 'magnitude of the larger volume'
• Note that, both the quantities involved in addition/subtraction:
    ♦ have the same unit m³
    ♦ have the same dimension [L³]
■ It is clear that, addition or subtraction can be carried out only between those quantities which have the same dimensions
• Addition or subtractions like 'volume to mass',  'force to velocity, etc., are not possible

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Next we will consider multiplication and division. Let us see some examples involving those operations:

Example 1:
(i) Consider the multiplication between mass and velocity
• We can multiply their magnitudes
• We then specify an appropriate unit for the product
(ii) We know that product of mass and velocity is momentum
• Unit of mass is kg
• Unit of velocity is ms^-1.
• So unit of momentum is kg ms^-1.
(iii) In the same way, dimensions of momentum can also be calculated:
• Dimension of mass is [M]
• Dimensions of velocity is $\mathbf\small{\left[LT^{-1}\right]}$
• So the dimensions of momentum is $\mathbf\small{\left[MLT^{-1}\right]}$
■ Note that, we are combining dimensions as if they are ordinary algebraic symbols
eg: a × b = ab

Example 2:
(i) Consider the division of force by area
• We can divide their magnitudes
• We then specify an appropriate unit for the quotient
(ii) We know that force divided by area gives pressure
• Unit of force is N
• Unit of area is m².
• Force is in the numerator and area is in the denominator
• So unit of pressure is N m^-2
(iii) In the same way, dimensions of pressure can also be calculated:
• Dimension of force is $\mathbf\small{\left[MLT^{-2}\right]}$
• Dimensions of area is $\mathbf\small{\left[L^2\right]}$
• So the dimensions of pressure is $\mathbf\small{\frac{\left[MLT^{-2}\right]}{\left[L^2\right]}=\left[ML^{-1}T^{-2}\right]}$
■ One [L] in the numerator gets cancelled by the [L] in the denominator. We are combining dimensions as if they are ordinary algebraic symbols
eg: $\mathbf\small{\frac{abc^{-2}}{b^2}=ab^{-1}c^{-2}}$
■ So we can write a conclusion about multiplication/division:
For multiplication/division, the two physical quantities need not have the same dimensions. The dimensions can be combined as if they are ordinary algebraic symbols

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Based on the above discussion, we can now learn how to check the dimensional consistency of equations. We will write it in steps:

1. We have seen that, while doing addition or subtraction, the physical quantities must be of the same dimensions
2. An equation will contain many terms. An example is shown below:
$\mathbf\small{a=bc^2+df^3g}$ 
3. There are 2 terms on the right side. We are adding those two terms 
• This addition will be possible only if they have the same dimensions
4. At a first look, they may appear to have different dimensions
• But on simplification, it is quite possible that, they may end up having the same dimensions
An example:
(i) After simplification, '$\mathbf\small{\left[MLT^{-2}\right]\left[L\right]}$' will become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
(ii) After simplification, '$\mathbf\small{\left[M\right]\left[LT^{-1}\right]^2}$' will also become '$\mathbf\small{\left[ML^2T^{-2}\right]}$'
• So we can do addition/subtraction on them
5. For doing addition or subtraction, two physical quantities must have the same dimensions. This is called the principle of homogeneity of dimensions. 
• This principle can be used for testing the correctness of an equation in the 'initial stage'
6. If the two terms on the right side of the equation in (2) are not of the same dimensions, the equation itself is wrong
• Not only the two terms, the term 'a' on the left side also must have the very same dimensions
7. Let us see an example:
• Test the dimensional consistency or homogeneity of the equation: $\mathbf\small{x=x_0+v_0t+\frac{1}{2}at^2}$
    ♦ x is the total distance traveled by the object in time t
    ♦ x₀ is the distance already traveled when the stop watch was turned on
    ♦ v₀ is the initial velocity at the instant when the stop watch was turned on
    ♦ a is the acceleration with which the object traveled
Solution:
(i) There are a total of 4 terms. Let us write the dimensions of each:
    ♦ [x] = [L]
    ♦ [x₀] = [L]
    ♦ [v₀t] = [LT^-1][T] = [L] 
    ♦ [¹⁄₂at²] = [LT^-2][T²] = [L]
(ii) The dimensions of all the three terms on the right side of the equation is [L]
• The dimension of the term on the left side of the equation is also [L]
■ So the given equation is dimensionally correct
8. But it is important to remember that, 'being dimensionally correct' is not the final word
• An equation may be dimensionally correct
    ♦ But as a whole, it may be wrong
• This is because, dimensional analysis do not take 'factors' into account  
    ♦ While deriving an equation, we may wrongly obtain a factor as (𝛑) instead of (0.5𝛑)
    ♦ Dimensional analysis cannot detect such mistakes
• We will come across more such shortcomings in the later chapters
9. So we can write a conclusion
The conclusion can be written in two steps:
(i) If an equation fails in a 'dimensional consistency test', then it is guaranteed that, the equation is wrong
(ii) Even if an equation passes in a 'dimensional consistency test', it is not guaranteed that, the equation is correct

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Now we will see some solved examples

Solved example 2.30
• Check whether the following equation is dimensionally correct:
$\mathbf\small{\frac{1}{2}mv^2=mgh}$
    ♦ m is the mass of the object
    ♦ v is the velocity of the object
    ♦ g is the acceleration due to gravity
    ♦ h is the height
Solution:
1. There are a total of 4 items. Let us write the dimensions of each:
    ♦ [m] = [M]
    ♦ [v] = [LT^-1]
    ♦ [g] = [LT^-2]
    ♦ [h] = [L]
2. So the dimensions of the left side are:
[M] × [LT^-1]² = [ML²T^-2]  
3. Dimensions of the right side are:
[M] × [LT^-2] × [L] = [ML²T^-2]  
4. So both sides have the same dimensions. The equation is dimensionally correct

Solved example 2.31
The time period T of a simple pendulum is given by $\mathbf\small{T=2\pi\sqrt{\frac{l}{g}}}$. Show that this equation is dimensionally correct
Solution:
1. There are 3 items in the equation. let us write the dimensions of each:
    ♦ [T] = [T]
    ♦ [l] = [L]
    ♦ [g] = [LT^-2]
2. Dimensions of the left side are: [T]
3. Dimensions of the right side are:
$\mathbf\small{\frac{[L]^{\frac{1}{2}}}{[LT^{-2}]^{\frac{1}{2}}}=\frac{1}{[T^{-1}]}=[T]}$
3. So both sides have the same dimension. The equation is dimensionally correct

Solved example 2.32
Write the dimensions of a and b in the relation $\mathbf\small{P=\frac{b-x^2}{at}}$
P is the power, x is the distance and t is the time
Solution:
1. 'Power' is work done in unit time. So we have: $\mathbf\small{[P]=\frac{[ML^2T^{-2}]}{[T]}=[ML^2T^{-3}]}$
2. [x] = [L]
3. [t] = [T]
4. In the given expression, we see that x² is being subtracted from b
• So they must be having the same dimensions
• We have: [x²] = [L²]
• Thus we get: [b] = [L²]
5. A 'change in a physical quantity' will have the same dimensions as the 'physical quantity'
• So we get: [b-x²] = [M⁰L²T⁰]  
6. Substituting the known dimensions in the given expression, we get:
$\mathbf\small{[ML^2T^{-3}]=\frac{[L^2]}{[a][T]}}$
$\mathbf\small{\Rightarrow[a]=\frac{[L^2]}{[ML^2T^{-3}][T]}=[M^{-1}L^0T^{2}]}$

Solved example 2.33
The velocity v of a particle depends upon the time t according to the equation $\mathbf\small{v=a+bt+\frac{c}{d+t}}$. Write the dimensions of a, b, c and d
Solution:
1. The dimension of the quantity on the left side = [v] = [LT^-1]
2. On the right side, 3 terms are being added
• Dimensions of each of them must be [LT^-1]
3. So we get: [a] = [LT^-1]
4. [bt] = [LT^-1]
⇒ [b][T] = [LT^-1]
⇒ [b] = [LT^-2]
5. In the last term, we have: $\mathbf\small{\frac{[c]}{[d+t]}=[LT^{-1}]}$
• We see that, d and t are being added
• So we get: [d] = [T]
6. Sum of two quantities will have the same dimension as the quantities being added
• Thus the last term in the given equation can be expressed dimensionally as: $\mathbf\small{\frac{[c]}{[T]}=[LT^{-1}]}$
$\mathbf\small{\Rightarrow [c]=[M^0LT^0]}$
6. So the final answer is:
• [a] = [M⁰LT^-1]
• [b] = [M⁰LT^-2]
• [c] = [M⁰LT⁰]
• [d] = [M⁰L⁰T]

Solved example 2.34
In the equation $\mathbf\small{p=\frac{a-t^2}{bx}}$, P is the pressure, x is the distance and t is the time. What is the dimension of $\mathbf\small{\frac{a}{b}}$?
Solution:
1. Pressure is force per unit area. So it's dimensions is given by:
$\mathbf\small{[P]=\frac{[MLT^{-2}]}{[L^2]}=[ML^{-1}T^{-2}]}$
2. We see that, t² is being subtracted from a. So we get: [a] = [t²] = [T²]
• Also, since 'change in magnitude' does not alter the dimension, we have: [a-t²] = [a] = [t²] = [T²] 
3. Substituting all the known dimensions, we get: $\mathbf\small{[ML^{-1}T^{-2}]=\frac{[T^2]}{[b][L]}}$
$\mathbf\small{\Rightarrow[b]=\frac{[T^2]}{[ML^{-1}T^{-2}][L]}=[M^{-1}L^{0}T^{4}]}$
4. So the dimension of $\mathbf\small{\frac{a}{b}}$ 
= $\mathbf\small{=\left[\frac{a}{b}\right]=\frac{[a]}{[b]}=\frac{[T^2]}{[M^{-1}L^{0}T^{4}]}=[ML^{0}T^{-2}]}$ 

Solved example 2.35
The SI unit of energy is J = kg m² s^-2; that of speed v is ms^-1 and of acceleration a is ms^-2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) :
(a) $\mathbf\small{K=m^2v^3}$
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
(c) $\mathbf\small{K=ma}$
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
Solution:
The unit of energy is given as J = kg m² s^-2
So the dimensions of kinetic energy will be: [ML²T^-2]  
(a) $\mathbf\small{K=m^2v^3}$
• Dimensions on the left side are: [ML²T^-2] 
• Dimensions on the right side are: [M]² × [LT^-1]³ = [M²L³T^-3]
• The two sides are not equal. So this formula can be ruled out 
(b) $\mathbf\small{K=\frac{1}{2}mv^2}$
• Dimensions on the left side are: [ML²T^-2] 
• Dimensions on the right side are: [M] × [LT^-1]² = [ML²T^-2]
• The two sides are equal. So this formula can not be ruled out 
(c) $\mathbf\small{K=ma}$
• Dimensions on the left side are: [ML²T^-2] 
• Dimensions on the right side are: [M] × [LT^-2] = [MLT^-2]
• The two sides are not equal. So this formula can be ruled out 
(d) $\mathbf\small{K=\frac{3}{16}mv^2}$
• Dimensions on the left side are: [ML²T^-2] 
• Dimensions on the right side are: [M] × [LT^-1]² = [ML²T^-2]
• The two sides are equal. So this formula can not be ruled out
(e) $\mathbf\small{K=\frac{1}{2}mv^2+ma}$
• Dimensions on the left side are: [ML²T^-2] 
• Dimensions on the right side are: [M] × [LT^-1]² + [M] × [LT^-2] 
=  [ML²T^-2]+[MLT^-2]
• We cannot even add the two terms on the right side.
• So this formula can be ruled out
■ We can write a conclusion. It can be done in 4 steps:
(i) We see that, (a), (c) and (e) can be ruled out
(ii) (b) and (d) cannot be ruled out
• But that doesn't mean that, (b) and (d) are correct
(iii) To find the correct formula, we need to do further investigation. We need to look for the correct definition of kinetic energy
(iv) The advantage of doing dimensional analysis is that, no such further investigation is required to be done for (a), (c) and (e)

Solved example 2.36
A famous relation in physics relates ‘moving mass’ m to the ‘rest mass’ m₀ of a particle in terms of its speed v and the speed of light, c. (This relation first arose as a consequence of special relativity due to Albert Einstein). A boy recalls the relation almost correctly but forgets where to put the constant c. He writes: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
Guess where to put the missing c
Solution:
1. The boy writes the equation as: $\mathbf\small{m=\frac{m_0}{(1-v^2)^{\frac{1}{2}}}}$
2. In the denominator, 'v²' is being subtracted from '1'
IF 'v²' is to be subtracted from 'something',
THEN that 'something' must have the dimensions [LT^-1]²
BECAUSE, 'v²' has the dimensions [LT^-1]²
3. Here the 'something' is '1'
• '1' is just a number. It has no dimensions
• So the subtraction (1-v²) is not possible
4. If we divide 'v' by 'c', then we get a dimensionless quantity
• Because, 'c' is also a velocity
5. So we must divide 'v²' by 'c²'
• Thus the correct equation is: $\mathbf\small{m=\frac{m_0}{(1-\frac{v^2}{c^2})^{\frac{1}{2}}}}$

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In the next section, we will see unit conversion

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