The principle of dimensional homogeneity states that the dimensions of all terms in a physical equation must be the same. This principle is widely used to check the correctness of physical relations and equations in physics. By comparing the dimensions on both sides of an equation, we can verify whether a formula is dimensionally consistent, even without knowing its exact derivation. This concept is an important application of dimensional analysis and plays a key role in solving numerical problems in Class 11 Physics and competitive exams.
- State the Principle of Homogeneity of Dimensions
- Applications (Uses) of Dimensional Analysis
- To Check Correctness of Physical Relation and The Principle of Homogeniety of Dimensions
- Solved Numerical Problems
- Limitations of Dimensional Analysis
- Frequently Asked Questions (FAQs) Very Short Answer Type Questions and Answers
State the Principle of Homogeneity of Dimensions
It states that the dimensions of the fundamental quantities (mass, length and time) are same in each and every term on either side of a physical relation.
The Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same.
Applications (Uses) 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).
To Check Correctness of Physical Relation and The Principle of Homogeniety of Dimensions
Checking the correctness of a physical relation (or equation) is based on the principle of homogeneity of dimensions.
According to this principle, the dimensions of the fundamental quantities (mass, length and time) are same in each and every term on either side of the physical relation.
To check the correctness of a given physical equation, the physical quantities on the two sides of the equation are expressed in terms of fundamental units of mass, length and time. If the powers of M, L and T on two sides of the equation are same, then the physical equation is correct and otherwise not.
Solved Numerical Problems
Problem. 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.
Problem. Check the accuracy of the relation :
$$
\frac{1}{2}mv^2 = mgh
$$
using dimensional analysis, where m is the mass of the body, v its velocity, g is the acceleration due to gravity and h is the height.
Solution. Here, $\frac{1}{2}mv^2 = mgh$
$$
\text{[L.H.S]} = \left[ \frac{1}{2}mv^2 \right] = [M][L T^{-1}]^2 = [M L^2 T^{-2}]
$$
$$
\text{[R.H.S]} = [mgh] = [M][L T^{-2}][L] = [M L^2 T^{-2}]
$$
Since dimensions of L.H.S and R.H.S are the same, the given relation is correct.
Problem. Check the dimensional consistency of the equation:
$$\mathbf{F}\mathbf{S} = \frac{1}{2} m\nu^2 -\frac{1}{2} m u^2,$$
where $S$ is distance moved, $u$ and $\nu$ are the initial and final velocities of a body of mass $m$ and $\mathbf{F}$ is the force acting on it.
Solution. We have, $\mathrm{F}\mathrm{S} = \frac{1}{2} m\nu^2 -\frac{1}{2} m u^2$
Dimensional formula of $\mathrm{F} \mathrm{S} = [\mathrm{M}\mathrm{L}\mathrm{T}^{-2}]\times [\mathrm{L}] = [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]$
Dimensional formula of $\frac{1}{2} m\nu^2 = [\mathrm{M}][\mathrm{L}\mathrm{T}^{-1}]^2 = [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]$
Dimensional formula of $\frac{1}{2} m u^2 = [\mathrm{M}][\mathrm{L}\mathrm{T}^{-1}]^2 = [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]$
As the dimensional formula of all the terms in the given equation is same, the equation is a correct one.
Problem. In vander Waals’ equation:
$$\left(P + \frac{a}{V^2}\right) (V – b) = RT,$$
what are the dimensions of constants ‘$a$’ and ‘$b$’?
Solution. Here, $\left(P + \dfrac{a}{V^2}\right)(V – b) = RT$
According to the principle of homogeneity of dimension, $\dfrac{a}{V^2}$ and $b$ should have dimensions of $P$ (pressure) and $V$ (volume) respectively.
Thus, $\dfrac{a}{V^2} =$ pressure
or
$a =$ pressure $\times V^2 = [\mathrm{M}\mathrm{L}^{-1}\mathrm{T}^{-2}][\mathrm{L}^3]^2 = [\mathrm{M}\mathrm{L}^5 \mathrm{T}^{-2}]$
Also, $b =$ volume $= [\mathrm{L}^3] = [\mathrm{M}^0 \mathrm{L}^3 \mathrm{T}^0]$
Problem. Check the correctness of the relation by dimensional analysis: $\lambda = \dfrac{h}{mv}$ where the letters have their usual meanings.
Solution. Here, $\lambda = \dfrac{h}{mv}$ Dimensional formula of wavelength $\lambda = [\mathrm{L}]$
Therefore, dimensional formula of L.H.S.
$$= [\mathrm{L}] = [\mathrm{M}^0 \mathrm{L}^1 \mathrm{T}^0]$$
Dimensional formula of Planck’s constant $h$
$$\left[\frac{\text{energy}}{\text{frequency}}\right] = \frac{[\mathrm{M}\mathrm{L}^2\mathrm{T}^{-2}]}{[\mathrm{T}^{-1}]} = [\mathrm{M}\mathrm{L}^2\mathrm{T}^{-1}]$$
Dimensional formula of mass $m = [\mathrm{M}]$
Dimensional formula of velocity $v = [\mathrm{L}\mathrm{T}^{-1}]$
Therefore, dimensional formula of R.H.S.
$$= \frac{[\mathrm{M}\mathrm{L}^2\mathrm{T}^{-1}]}{[\mathrm{M}]\times [\mathrm{L}\mathrm{T}^{-1}]} = [\mathrm{M}^0 \mathrm{L}^1 \mathrm{T}^0]$$
As dimensional formula of L.H.S. is the same as that of R.H.S., the given relation is correct.
Problem. Check the correctness of the relation $\tau = I \alpha$, where $\tau$ is the torque acting on a body, $I$ is the moment of inertia and $\alpha$ is the angular acceleration.
Solution. We have, $\tau = I \alpha$
Now, $\tau = \text{force} \times \text{distance} = [\mathrm{M}\mathrm{L}\mathrm{T}^{-2}][\mathrm{L}] = [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]$
Therefore, dimensional formula of L.H.S. $= [\mathrm{M}\mathrm{L}^{2}\mathrm{T}^{-2}]$
$I = \text{mass} \times \text{distance}^2 = [\mathrm{M}][\mathrm{L}]^2 = [\mathrm{M}\mathrm{L}^2]$
$\alpha = \text{angular acceleration} = \dfrac{\text{angle}}{(\text{time})^{2}} = [\mathrm{T}^{-2}]$
Therefore, dimensional formula of R.H.S.
$$= [\mathrm{M}\mathrm{L}^2][\mathrm{T}^{-2}] = [\mathrm{M}\mathrm{L}^2 \mathrm{T}^{-2}]$$
Since the dimensional formulae of L.H.S. and R.H.S. are the same, the given relation is correct.
Problem. Test by using dimensional analysis, the accuracy of the relation:
$$v_{c} = \dfrac{k \eta}{r \rho}$$
Solution. Here, $v_{c} = \dfrac{k \eta}{r \rho}$
Dimensional formula of $v_{c}$ (critical velocity) $= [\mathrm{L} \mathrm{T}^{-1}]$
Therefore, dimensional formula of L.H.S.
$$= [\mathrm{L} \mathrm{T}^{-1}] = [\mathrm{M}^{0} \mathrm{L} \mathrm{T}^{-1}]$$
Dimensional formula of $\eta$ (coefficient of viscosity)
$$= [\mathrm{M} \mathrm{L}^{-1} \mathrm{T}^{-1}] $$
Dimensional formula of $r$ (radius) $= [\mathrm{L}]$
Dimensional formula of $\rho$ (density) $= [M L^{-3}]$
The constant $k$ is dimensionless.
Therefore, dimensional formula of R.H.S.
$$= \frac{[M L^{-1} T^{-1}]}{[L][M L^{-3}]} = [M^0 L T^{-1}]$$
As the dimensional formula of L.H.S. is the same as that of R.H.S., the given relation is correct.
Problem. 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. 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
\(\begin{array}{l}\dfrac{dV}{dt}=\dfrac{\pi p a^4}{8l\eta}\end{array} \)
Solution:
\(\begin{array}{l} \dfrac{ dV }{ dt } = \dfrac{ \pi p ^ { 4 }}{ 8 l \eta } \end{array} \)
\(\begin{array}{l} [\textup{L.H.S}] = \dfrac{ [dV] }{[dt]} = \dfrac{[M^{0} L^{3} T^ {0}]}{[M^{0} T^{0} T^{1}]} = [M^{0}L^{3}T^{-1}] \textup{ …..(1)} \end{array} \)
\(\begin{array}{l} [\textup{R.H.S}] = \dfrac{[p] [a] ^ {4}}{[l] [\eta]} \end{array} \)
\(\begin{array}{l} \therefore [\textup{R.H.S}] = \dfrac{ [M ^ { 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 } L ^ { 1 } T ^ { 0 }]^ { 4 }} { [M ^ { 0 } L ^ { 1 } T ^ { 0 }] [M ^ { 1 } L ^ { -1 } T ^{ -1 }] } \end{array} \)
\(\begin{array}{l} = \dfrac{ [M^{ 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 }L ^ { 4 } T ^ { 0 } ] }{ [M^{ 1 }L^ { 0 }T^ { -1 }] } \end{array} \)
\(\begin{array}{l} = \dfrac{ [M^ { 1 } L ^ { 3 } T^ { 2 }] }{ [M^ { 1 } L^ { 0 } T ^ { -1 }] } = [M^ { 0 }L ^ { 3 }T^{ -1 }] \textup{ …….(2)} \end{array} \)
From (1) and (2), we have [L.H.S] = [R.H.S]
Hence, by the principle of homogeneity, the given equation is homogenous.
Problem. Using Dimensional Formula, X= MaLbTc, 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 = MaLbTc
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
Problem. Match the physical quantities given in column I with dimensions expressed in terms of mass (M), length (L), time (T) and charge (Q) given in column II and write the correct answer against the matched quantity in tabular form :
| I | II | I | II |
|---|---|---|---|
| Angular momentum | $M L^2 T^{-2}$ | Capacitance | $M L^3 T^{-1} Q^{-2}$ |
| Latent heat | $M L^2 Q^{-2}$ | Inductance | $M^{-1} L^{-2} T^2 Q^2$ |
| Torque | $M L^2 T^{-2}$ | Resistivity | $L^2 T^{-2}$ |
(I.I.T. 1983)
Solution. It can be proved that angular momentum $= M L^2 T^{-1}$; torque $= M L^2 T^{-2}$
Now, latent heat $= \dfrac{\text{quantity of heat}}{\text{mass}} = \dfrac{\text{energy}}{\text{mass}}$
$$= \frac{M L^2 T^{-2}}{M} = L^2 T^{-2}$$
Capacitance $= \dfrac{\text{charge}}{\text{potential}} = \dfrac{\text{charge}}{\text{work done per unit charge}}$
$$= \frac{Q}{M L^2 T^{-2} Q^{-1}} = M^{-1} L^{-2} T^2 Q^2$$
Inductance $= \dfrac{\text{induced e.m.f.}}{\text{change in current / time taken}}$
$$= \frac{\text{work done per unit charge} \times \text{time taken}}{\text{rate of flow of charge}}$$
$$= \frac{M L^2 T^{-2} Q^{-1} \times T}{Q T^{-1}} = M L^{-2} Q^{-2}$$
Resistivity $= \dfrac{\text{resistance} \times \text{area}}{\text{length}}$
$$= \frac{\text{potential difference} \times \text{area}}{\text{current} \times \text{length}}$$
$$= \frac{\text{work done per unit charge}}{\text{rate of the flow of charge}} \times \frac{\text{area}}{\text{length}}$$
$$= \frac{M L^2 T^{-2} Q^{-1} \times L^2}{Q T^{-1} \times L} = M L^3 T^{-1} Q^{-2}$$
Hence the given physical quantities and their dimensional formulae are correctly given as below :
| I | II | I | II |
|---|---|---|---|
| Angular momentum | $M L^2 T^{-1}$ | Capacitance | $M^{-1} L^{-2} T^2 Q^2$ |
| Latent heat | $L^2 T^{-2}$ | Inductance | $M L^2 Q^{-2}$ |
| Torque | $M L^2 T^{-2}$ | Resistivity | $M L^3 T^{-1} Q^{-2}$ |
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) Very Short Answer Type Questions and Answers
What is the principle of homogeneity?
The principle of homogeneity states that the dimensions of all terms in a physical equation must be the same on both sides.
What is dimensional analysis?
Dimensional analysis is the study of the relationship between physical quantities with the help of dimensions and units of measurement.
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.
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.
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.
Why does dimensional analysis fail for trigonometric equations?
Trigonometric functions (e.g., sine, cosine) are dimensionless and involve angles, which do not have physical dimensions.
Can dimensional analysis be used to determine the value of universal constants?
No, dimensional analysis cannot predict numerical constants like π or 1/2.
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.
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.
Write down the Dimensional Formula of Frequency.
The Dimensional Formula for frequency is [T–1]. Hertz is the unit of frequency.
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.
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.
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.
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.
Units and Measurements – Summary
The chapter Units and Measurements lays the foundation of physics by explaining how physical quantities are defined, measured, and expressed using standard units. It starts with physical quantities, their classification into fundamental and derived quantities, and a detailed understanding of SI units, supplementary units, and systems of units (CGS, MKS, FPS). You can interlink these topics to explore unit definitions, symbols, prefixes, and conversions in detail.
Next, the chapter focuses on measurement and errors, covering concepts like accuracy and precision, types of errors (absolute, relative, percentage error), and propagation of errors (sum, difference, multiplication, division, and powers). These topics are essential for understanding uncertainty in measurements and are frequently used in numerical problems and JEE PYQs.
The chapter also introduces significant figures and rounding off rules, which help maintain proper precision in calculations. Finally, it covers dimensional analysis, including dimensional formulae, principle of dimensional homogeneity, checking correctness of physical relations, and identifying quantities having the same dimensions, along with applications in unit conversion and equation verification. This chapter is crucial for building a strong base in physics.