1.4 Dimensions of Physical Quantities – Class 11 Physics

Understanding dimensions is a foundational skill in physics because every physical quantity can ultimately be expressed using combinations of fundamental units. Dimensions reveal the underlying structure of physical laws, showing how mass, length, time, and other base quantities are interwoven in defining measurable physical properties. A strong grasp of dimensional analysis helps students verify equations, derive new relations, convert complex units, and solve challenging numerical problems.

Every physical measurement in physics is fundamentally connected to seven base quantities: mass (M), length (L), time (T), electric current (I), temperature (Θ), amount of substance (N), and luminous intensity (J). When a physical quantity is expressed in terms of these fundamental dimensions, we call this representation its dimensional formula. The dimensional formula identifies what the quantity depends on, rather than its magnitude.

This section elaborates on the meaning, importance, and applications of dimensions, moving gradually from basic understanding to advanced problem-solving, helping Class 11 students and competitive exam aspirants build a robust conceptual foundation.

Importance of Dimensions

Dimensions are not merely mathematical expressions; they play a crucial role in interpreting and validating physical relationships. Their significance can be understood through the following key points:

• Verification of Physical Equations: Dimensions help determine whether a physical equation is dimensionally consistent. If all terms have the same dimensions, the equation is likely correct.
• Conversion Between Systems of Units: They enable systematic conversion of units between different measurement systems (e.g., CGS to SI).
• Derivation of Relations Between Quantities: In some cases, dimensional analysis can help derive working formulas when the relationship between variables is only partially known.
• Understanding the Nature of Physical Quantities: Dimensions reveal what fundamental quantities influence a measurement-for example, showing that force depends on mass, length, and time.
• Identifying Dimensionless Quantities: Dimensionless parameters such as refractive index or strain often represent ratios or pure numbers that play important roles in physical analysis.

Thus, mastery of dimensional analysis is essential for problem-solving in physics and real-world engineering applications.

Dimensional Representation

The dimensional formula of a physical quantity expresses its dependence on the base quantities M, L, T, I, Θ, N, and J. It is written in the generalized form:

[Mᵃ Lᵇ Tᶜ Iᵈ Θᵉ Nᶠ Jᵍ]

Here, a, b, c, d, e, f, g represent the powers to which each fundamental quantity is raised. For most quantities in mechanics, the dimensions involve only M, L, and T.

Basic Examples

• Velocity: Velocity = distance/time → [L T⁻¹]
• Acceleration: Acceleration = change in velocity/time → [L T⁻²]
• Force: Force = mass × acceleration → [M L T⁻²]

These simple examples build the foundation for more complex dimensional expressions.

Dimensional Expressions of Common Physical Quantities

Below is table covering frequently used physical quantities and their dimensional formulas. These expressions are essential for both board exams and competitive exams like JEE and NEET.

This table strengthens the learner’s ability to quickly analyze dimensional dependencies.

Rules for Dimensional Representation

Dimensions follow certain universal rules that ensure logical consistency across all physical equations:

1. Homogeneity Principle: All terms in a physically meaningful equation must have identical dimensions.
2. Dimensionless Quantities: Some physical quantities (e.g., angles, refractive index) have no dimensions.
3. Arguments of Mathematical Functions: Inputs of trigonometric, logarithmic, or exponential functions must be dimensionless.
4. Constants Without Units: Pure numbers such as π, e, or 1/2 have no dimensions.
5. Additivity Rule: Only quantities with identical dimensions can be added or subtracted.

Understanding these rules prevents common errors in equation manipulation.

Additional Deep-Dive Concepts

1. Dimensionless Groups in Physics

Several important physical constants and ratios are dimensionless:

• Reynolds number
• Froude number
• Coefficient of friction
• Relative density These describe physical behavior independent of units.

2. Why Different Quantities Share the Same Dimensions

Work, energy, and torque all share [M L² T⁻²] but are not identical. Dimensions show dependence, not meaning.

3. Dimensional Inconsistency in Incorrect Formulae

Dimensions help detect errors: Example incorrect formula: v = u + at² Dimensions of RHS = [L T⁻¹] + [L T⁻² × T²] = [L T⁻¹] + [L] Not consistent → incorrect.

NCERT-Style Solved Examples

Example 1:

Work = Force × distance
Force = mass × acceleration = [M L T⁻²]
Thus, Work = [M L T⁻²] × [L] = [M L² T⁻²], showing that work depends directly on mass, the square of length, and inversely on the square of time.

Example 2:

Pressure = Force/Area = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²]
This shows that pressure increases with force but decreases with the area over which it is applied.

Example 3:

Energy has the same dimensions as work → [M L² T⁻²], emphasizing their interchangeable nature in physical processes.

Example 4:

Find the dimensions of momentum. Momentum = mass × velocity = [M] × [L T⁻¹] = [M L T⁻¹]
Momentum depends directly on mass and speed.

Example 5:

What are the dimensions of power? Power = Work/Time = [M L² T⁻²] / [T] = [M L² T⁻³]
This indicates that power depends on how quickly work is done.

Example 6:

Find the dimensions of kinetic energy using E = 1/2 mv². Dimensions = [M] × [L² T⁻²] = [M L² T⁻²]

Example 7:

Find the dimensions of potential energy. Potential energy = mgh → [M] × [L T⁻²] × [L] = [M L² T⁻²]

JEE / NEET Level Examples

Problem 1

Find the dimensions of the gravitational constant G. Using Newton’s law: F = G (m₁ m₂ / r²)
G = F r² / (m₁ m₂)
Dimensions = [M L T⁻²] × [L²] / [M²] = [M⁻¹ L³ T⁻²], showing its inverse relation with mass.

Problem 2

Which of the following is dimensionless? (a) Refractive index
(b) Force
(c) Power
(d) Momentum
Answer: (a) Refractive index

Problem 3

Find dimensions of surface tension. Surface tension = Force / Length = [M L T⁻²] / [L] = [M T⁻²]

Problem 4

If energy E = 1/2 mv², check its dimensions. Right side = [M] × [L² T⁻²] = [M L² T⁻²]
Matches work → dimensionally correct.

Problem 5

A physical quantity X is given as X = (a² b) / √c.
Dimensions of a = [L], b = [M], c = [T].
Find dimensions of X. Solution: X = [L² × M] / [T¹/²] = [M L² T⁻¹/²]

Problem 6

Find the dimensions of angular momentum. L = mvr → [M] × [L T⁻¹] × [L] = [M L² T⁻¹]

Problem 7

Find the dimensions

Problem 7

Find the dimensions of the coefficient of viscosity. Viscosity η = (shearing stress) / (rate of strain). Shearing stress = Force/Area = [M L T⁻²] / [L²] = [M L⁻¹ T⁻²] Rate of strain = velocity gradient = [T⁻¹] Thus, η = [M L⁻¹ T⁻²] / [T⁻¹] = [M L⁻¹ T⁻¹]

Problem 8

Find the dimensions of universal gas constant R. Using PV = nRT: R = PV / nT Pressure = [M L⁻¹ T⁻²], Volume = [L³] Thus R = [M L⁻¹ T⁻²] × [L³] = [M L² T⁻²] Since per mole per Kelvin → multiply with Θ⁻¹ N⁻¹ Final: [M L² T⁻² Θ⁻¹ N⁻¹]

Problem 9

A quantity Y is defined as Y = (F² / (m r³)). Find its dimensions. Force = [M L T⁻²] Mass = [M] Distance = [L] Thus Y = [M² L² T⁻⁴] / [M L³] = [M L⁻¹ T⁻⁴]

Problem 10

Find the dimensions of permittivity ε₀. Using Coulomb’s law: F = (1 / (4π ε₀)) (q₁ q₂ / r²) ε₀ = q² / (F r²) Charge q = [I T] Thus ε₀ = [I² T²] / ([M L T⁻²] × [L²]) = [M⁻¹ L⁻³ T⁴ I²]

Problem 11

A force F acting on a body is given by F = k v² r, where k is a constant. Find the dimensions of k. F = [M L T⁻²] v² = [L² T⁻²] r = [L] Thus k = F / (v² r) = [M L T⁻²] / ([L² T⁻²] × [L]) = [M L⁻²]

Problem 12

Find the dimensions of Boltzmann constant k. Using KE = (3/2) kT → k = KE / T = [M L² T⁻²] / [Θ] = [M L² T⁻² Θ⁻¹]

Problem 13

If centripetal force F = m v² / r, show that both sides match dimensionally. RHS = [M] × [L² T⁻²] / [L] = [M L T⁻²] Matches force → dimensionally valid.

Problem 14

Find the dimensions of magnetic field B. Force on charge q: F = q v B Thus B = F / (q v) = [M L T⁻²] / ([I T] [L T⁻¹]) = [M T⁻² I⁻¹]

Problem 15

Find the dimensions of Planck’s constant (from E = hν). h = E / ν = [M L² T⁻²] / [T⁻¹] = [M L² T⁻¹]

FAQs

Q1. What are dimensions of a physical quantity?

Dimensions show how a physical quantity depends on the fundamental quantities like mass (M), length (L), time (T), and others. They are written in the form [Mᵃ Lᵇ Tᶜ …].

Q2. How are dimensions different from units?

Dimensions indicate the nature of a physical quantity in terms of base quantities, while units are the standard magnitudes used to measure that quantity (such as metre, second, kilogram).

Q3. Can two different physical quantities have the same dimensions?

Yes. Work, energy, and torque all have the same dimensions [M L² T⁻²], even though they represent different physical concepts.

Q4. What is a dimensionless quantity?

A dimensionless quantity has no dependence on fundamental units. It is often a ratio of similar quantities, like refractive index, strain, or relative density.

Q5. How do dimensions help in checking the correctness of formulas?

By ensuring that all terms in an equation have identical dimensions. If dimensions do not match, the equation is definitely incorrect.

Q6. Can dimensional analysis give exact numerical constants in a formula?

No. Dimensional analysis can reveal the form of a relation but not numerical constants like 1/2, 2, or π.

Q7. Why must arguments of trigonometric or exponential functions be dimensionless?

Because such functions are defined mathematically only for pure numbers. If the argument carries dimensions, the expression has no physical or mathematical meaning.

Q8. Is dimensional analysis useful in competitive exams?

Yes. It helps eliminate wrong options quickly, check formulas during exams, and solve problems involving unit conversion or proportionality.

Conclusion

Dimensions of physical quantities provide a powerful framework for understanding the structure of physical laws, verifying equations, deriving relations, and solving numerical problems with confidence. The deeper students engage with dimensional analysis, the more intuitive problem solving becomes across mechanics, electricity, waves, and modern physics.

At Deeksha Vedantu, we emphasize conceptual clarity, repeated practice, and exam-oriented thinking so that learners not only understand dimensions theoretically but also apply them smoothly in JEE, NEET, KCET, and board exams.