Dimensional Analysis – Principle of Homogeneity, Applications and Limitations

Principle of Homogeneity

The Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another.

To better understand this principle, let us consider an example:

Example.1 : Check the correctness of physical equation s = ut + ½ at². In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs.

Solution:

We know that L.H.S = s and R.H.S = ut + 1/2at²

The dimensional formula for the L.H.S can be written as s = [L¹M⁰T⁰] ………..(1)

We know that R.H.S is ut + ½ at² , simplifying we can write R.H.S as [u][t] + [a] [t]²

[L¹M⁰T^-1][L⁰M⁰T^-1] +[L¹M⁰T^-2][L⁰M⁰T²]

=[L¹M⁰T⁰]………..(2)

From (1) and (2), we have [L.H.S] = [R.H.S]

Hence, by the principle of homogeneity, both sides have the same dimensions, the equation is dimensionally correct.

Applications of Dimensional Analysis

Dimensional analysis is widely used in physics for multiple purposes. The three main applications are:

1. Checking the Consistency of Equations

Using the principle of homogeneity, we can verify whether a given physical equation is dimensionally consistent.

2. Deriving Relations between Physical Quantities

By analyzing the dimensions of different quantities, we can derive expressions for unknown relationships between them.

3. Converting Units

We can use dimensional analysis to convert units from one system (e.g., SI) to another (e.g., CGS).

Limitations of Dimensional Analysis

Despite its usefulness, dimensional analysis has some limitations:

• It does not provide information about dimensionless constants (e.g., numerical factors like 1/2 in kinematic equations).
• It cannot derive formulas involving trigonometric, logarithmic, or exponential functions.
• It does not distinguish between scalar and vector quantities.

Frequently Asked Questions (FAQs)

Q1: What is the principle of homogeneity?

Answer: The principle of homogeneity states that the dimensions of all terms in a physical equation must be the same on both sides.

Q2: What is dimensional analysis?

Dimensional analysis is the study of the relationship between physical quantities with the help of dimensions and units of measurement.

Q3: State the principle of homogeneity of dimensions

The principle of homogeneity of dimensions states that an equation is dimensionally correct if the dimensions of the various terms on either side of the equation are the same.

Q4: Why do we use dimensional analysis?

We make use of dimensional analysis for three prominent reasons:

• To check the consistency of a dimensional equation
• To derive the relation between physical quantities in physical phenomena
• To change units from one system to another.

Q5: What are the limitations of dimensional analysis?

Some limitations of dimensional analysis are:

• It doesn’t give information about the dimensional constant.
• The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.
• It gives no information about whether a physical quantity is a scalar or vector.

Multiple Choice Questions (MCQs)

Q1: The principle of homogeneity states that:

A) The numerical values on both sides of an equation must be equal.
B) The dimensions of all terms in a physical equation must be the same.
C) The physical equation must be correct mathematically.
D) A physical quantity can have different dimensions in different systems.

Answer: B) The dimensions of all terms in a physical equation must be the same.

Explanation: The principle ensures that equations remain consistent in all unit systems.

Conceptual Questions

Q1: Why does dimensional analysis fail for trigonometric equations?

Answer: Trigonometric functions (e.g., sine, cosine) are dimensionless and involve angles, which do not have physical dimensions.

Q2: Can dimensional analysis be used to determine the value of universal constants?

Answer: No, dimensional analysis cannot predict numerical constants like π or 1/2.

Q3: What Is Meant By Dimensional Formula?

The Dimensional Formula of any quantity serves as an expression that shows the powers by which fundamental units must be raised to yield a single unit of that derived quantity. These Dimensional Formula play an important role in establishing relationships between variables in nearly every dimensional equation.

Q4: How do you Find the Dimensional Formula?

The Dimensional Formula of any quantity can be given by expressing the formula for it and breaking it down in terms of the base dimensions. Using these base dimensions, we can evaluate the dimensional formula for any given quantity.

Q5: Write down the Dimensional Formula of Frequency.

The Dimensional Formula for frequency is [T^–1]. Hertz is the unit of frequency. 

Q6: What Are the Uses of Dimensional Formula?

• To verify whether a formula is dimensionally correct or not.
• Conversion of units from one system to another for any given quantity.
• To establish derivation between physical quantities based on mutual relationships.
• Dimensional Formulae express every physical quantity in terms of fundamental units.

Q7: Write down the difference between Dimensional Quantities and Dimensional Formula?

The equations resulting from equating a physical quantity to its dimensional formula are termed Dimensional Equations. These equations are an important tool for representing physical quantities in terms of fundamental units. Dimensional formulas for specific quantities used as a foundation for establishing relationships between those quantities within any given dimensional equation.

Q8: What is Dimension in mathematics?

In mathematics, Dimension refers to the measurement of an object’s size, extent, or distance in a specific direction, such as length, width, or height. As per the definition of Dimension, it represents the extent of a point or a line in a specific direction.

Q9: What are the Important Dimensional Formula?

Length = L Mass = M Time are the important Dimensional Formula.

Q10: Write down the Limitation of Dimensional Formula.

• It’s important to note that quantities like trigonometric functions, plane angles, and solid angles do not possess defined dimensional formulae since they are inherently dimensionless in nature.
• The applicability of Dimensional Formulas is confined to a specific set of physical quantities.
• They are unable to determine proportionality constants, which can be a drawback in certain situations.
• Dimensional formulas are primarily suitable for addition and subtraction operations, limiting their use in other mathematical operations.

Do You Know?

• The concept of dimensions was first introduced by Joseph Fourier.
• Dimensional analysis is used in fluid mechanics to derive relationships like the Reynolds number.
• Buckingham π-theorem is an advanced application of dimensional analysis.

Worksheet

Questions:

1. State the principle of homogeneity.
2. Use dimensional analysis to check the validity of the equation T=2π√(l/g).
3. Convert a force of 10 Newtons to the CGS system.
4. Why can’t dimensional analysis determine the direction of a vector quantity?
5. Identify a limitation of dimensional analysis with an example.

Test Paper (Total Marks: 10)

Important Points for Quick Revision

• The principle of homogeneity ensures the correctness of equations.
• Dimensional analysis helps in checking equations, deriving relations, and converting units.

Additional Solved Problems

Problem.1. : Check the correctness of the physical equation v² = u² + 2as².
Solution:

The computations made on the L.H.S and R.H.S are as follows:

L.H.S: v² = [v²] = [ L¹M⁰T^–1]² = [ L¹M⁰T^–2] ……………(1)

R.H.S: u² + 2as²

Hence, [R.H.S] = [u]² + 2[a][s]²

[R.H.S] = [L¹M⁰T^–1]² + [L¹M⁰T^–2][L¹M⁰T⁰]²

[R.H.S] = [L²M⁰T^–2] + [L¹M⁰T^–2][L²M⁰T⁰]

[R.H.S] = [L²M⁰T^–2] + [L¹M⁰T^–2][L²M⁰T⁰]

[R.H.S] = [L²M⁰T^–2] + [L³M⁰T^–2]…………………(2)

From (1) and (2), we have [L.H.S] ≠ [R.H.S]

Hence, by the principle of homogeneity, the equation is not dimensionally correct.

Problem.2 : Evaluate the homogeneity of the equation when the rate flow of a liquid has a coefficient of viscosity η through a capillary tube of length ‘l’ and radius ‘a’ under pressure head ‘p’ given as

\(\begin{array}{l}\frac{dV}{dt}=\frac{\pi p a^4}{8l\eta}\end{array} \)

Solution:

\(\begin{array}{l} \frac{ dV }{ dt } = \frac{ \pi p ^ { 4 }}{ 8 l \eta } \end{array} \)

\(\begin{array}{l} [\textup{L.H.S}] = \frac{ [dV] }{[dt]} = \frac{[M^{0} L^{3} T^ {0}]}{[M^{0} T^{0} T^{1}]} = [M^{0}L^{3}T^{-1}] \textup{ …..(1)} \end{array} \)

\(\begin{array}{l} [\textup{R.H.S}] = \frac{[p] [a] ^ {4}}{[l] [\eta]} \end{array} \)

\(\begin{array}{l} \therefore [\textup{R.H.S}] = \frac{ [M ^ { 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 } L ^ { 1 } T ^ { 0 }]^ { 4 }} { [M ^ { 0 } L ^ { 1 } T ^ { 0 }] [M ^ { 1 } L ^ { -1 } T ^{ -1 }] } \end{array} \)

\(\begin{array}{l} = \frac{ [M^{ 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 }L ^ { 4 } T ^ { 0 } ] }{ [M^{ 1 }L^ { 0 }T^ { -1 }] } \end{array} \)

\(\begin{array}{l} = \frac{ [M^ { 1 } L ^ { 3 } T^ { 2 }] }{ [M^ { 1 } L^ { 0 } T ^ { -1 }] } = [M^ { 0 }L ^ { 3 }T^{ -1 }] \textup{ …….(2)} \end{array} \)

From (1) and (2), we have [L.H.S] = [R.H.S]

Hence, by the principle of homogeneity, the given equation is homogenous.

Problem.3 : Using Dimensional Formula, X= M^aL^bT^c, find the values of a, b, and c for density.

Solution:     

To find: Values for a, b, and c

Given:

Quantity = Density

Using the Dimensional Formula,

X = M^aL^bT^c

We know,

Density = (mass/length³)

= M/L³

= M¹L^-3T⁰

Comparing with Dimensional Formula, we get,

a = 1, b = -3, c = 0

Answer: a = 1, b = -3, c = 0

Practice Questions on Dimensional Formula

Q1. Using Dimensional Formula, X= M^aL^bT^c, find the values of a, b, and c for Energy.

Q2. Using Dimensional Formula, X= M^aL^bT^c, find the values of a, b, and c for Acceleration.

Q3. Determine the Dimensional Formula of Power.

Q4. Determine the Dimensional Formula of Time period of wave.

Test Your Knowledge (Quiz)

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