EDUNES ONLINE EDUCATION: Dimensional Formulae and Dimensional Equations

Edunes Online Education

Section 1.5 – Dimensional Formulae and Dimensional Equations from Class 11 Physics (Units and Measurements)

📘 1.5 – Dimensional Formulae and Dimensional Equations

🧠 What is a Dimensional Formula?

A dimensional formula is an expression that shows how and which of the base physical quantities (Mass $[M]$ , Length $[L]$ , Time $[T]$ , etc.) are used to represent a given physical quantity.

\text{Dimensional Formula} = [M^a L^b T^c A^d K^e \text{mol}^f cd^g]

Here,

$a, b, c, d, e, f, g$ are integers (positive, negative, or zero)
Each represents the power of the respective base quantity

📌 Examples of Dimensional Formulae

Physical Quantity	Dimensional Formula
Volume $V$	$[M^0 L^3 T^0]$
Speed $v$	$[M^0 L T^{-1}]$
Acceleration $a$	$[M^0 L T^{-2}]$
Density $ρ$	$[M^1 L^{-3} T^0]$
Force $F$	$[M^1 L^1 T^{-2}]$
Energy $E$	$[M^1 L^2 T^{-2}]$

🧮 What is a Dimensional Equation?

A dimensional equation is formed when a physical quantity is equated to its dimensional formula.

For example:

Volume: $[V] = [M^0 L^3 T^0]$
Velocity: $[v] = [M^0 L T^{-1}]$
Force: $[F] = [M L T^{-2}]$
Density: $[ρ] = [M L^{-3} T^0]$

🧩 Deriving Dimensional Equations from Known Relations

You can derive the dimensional equation of a quantity from the physical law or formula it follows.

📌 Example 1: Force

We know:

F = m \cdot a

Where:

Mass $m = [M]$
Acceleration $a = [L T^{-2}]$

\Rightarrow [F] = [M] \cdot [L T^{-2}] = [M L T^{-2}]

So the dimensional equation of force is:

[F] = [M L T^{-2}]

📌 Example 2: Density

We know:

ρ = \frac{m}{V}

Where:

Mass $m = [M]$
Volume $V = [L^3]$

\Rightarrow [ρ] = \frac{[M]}{[L^3]} = [M L^{-3}]

So the dimensional equation of density is:

[ρ] = [M L^{-3} T^0]

✅ Why Dimensional Equations Are Useful

🔎 Check dimensional consistency of physical equations
🧩 Derive new relations between physical quantities
🌐 Convert units from one system to another
🧪 Help identify hidden physical relations in complex formulas

⚠️ Limitations

Dimensional analysis cannot determine constants (like $\frac{1}{2}, 2\pi$ , etc.)
It cannot distinguish between scalar and vector quantities
It fails if a physical quantity is expressed as a sum/difference of different dimensional terms

🧾 Summary Table

Quantity	Formula	Dimensional Equation
Volume $V$	$L \times B \times H$	$[V] = [M^0 L^3 T^0]$
Velocity $v$	$\frac{d}{t}$	$[v] = [M^0 L T^{-1}]$
Force $F$	$m \cdot a$	$[F] = [M L T^{-2}]$
Pressure $P$	$\frac{F}{A}$	$[P] = [M L^{-1} T^{-2}]$
Energy $E$	$F \cdot d$	$[E] = [M L^2 T^{-2}]$
Density $ρ$	$\frac{m}{V}$	$[ρ] = [M L^{-3} T^0]$

🔚 Conclusion

Dimensional formulae and dimensional equations form the backbone of unit analysis and physical reasoning in physics. They allow us to:

Express physical quantities independent of any unit system
Test equations for correctness
Develop insights into physical laws

They act like a "grammar" in the language of physics, helping ensure that all expressions and equations are physically and mathematically valid.

Based on the principles of NEP 2020, which emphasizes conceptual clarity, critical thinking, and real-world application over rote memorization, here is a high-level worksheet designed to challenge your understanding of Dimensional Formulae and Dimensional Equations.

🚀 Higher-Order Thinking Skills (HOTS) Worksheet

Topic: Units and Measurements | Level: Grade 11 Physics (Advanced)

Part A: Conceptual Depth & Critical Reasoning

Answer the following by diving deeper into the "Why" and "How."

The Constant Conundrum: Dimensional analysis is a powerful tool, yet it cannot determine dimensionless constants like $2\pi$ or $1/2$. Explain why these constants are "invisible" to dimensional formulas and suggest one method a physicist might use to find their actual values.
The Homogeneity Challenge: An ambitious student proposes the equation $v^2 = u^2 + \frac{5}{2}as$ for a new type of motion.
- Show that this equation is dimensionally consistent.
- Critically analyze whether dimensional consistency guarantees the physical correctness of this equation.
Limitations of Addition: Why is it physically meaningless to add a quantity with dimensions $[L]$ to a quantity with dimensions $[M]$? Relate your answer to the Principle of Homogeneity.

Part B: Advanced Derivations

Use the method of dimensions to explore relationships between physical variables.

Planetary Harmonics: The time period ($T$) of a planet orbiting a star depends on the Radius of the orbit ($R$), the Mass of the star ($M$), and the Universal Gravitational Constant ($G$). Given that the dimensional formula for $G$ is $[M^{-1} L^3 T^{-2}]$, derive the relationship between $T, R, M,$ and $G$.
Fluid Dynamics: The terminal velocity ($v$) of a spherical ball falling through a viscous liquid depends on the radius ($r$) of the ball, the coefficient of viscosity ($\eta$) of the liquid, and the effective force ($F$). Derive a functional relationship between these variables.
- Note: Dimensions of $\eta$ are $[M L^{-1} T^{-1}]$.

Part C: Real-World Application & Data Analysis

NEP focuses on "Learning by Doing" and cross-curricular links.

The "Physics Grammar" Audit: Look at the following table derived from the Edunes Summary Table. Identify the error in the Pressure entry and provide the corrected derivation.

Quantity	Formula	Dimensional Equation
Pressure ($P$)	$F / A$	$[M L^{1} T^{-2}]$

Unit System Architect: Imagine you are creating a new system of units where Force ($F$), Length ($L$), and Time ($T$) are the fundamental base quantities instead of Mass, Length, and Time.
- What would be the dimensional formula for Mass in your new "FLT" system?
- What would be the dimensional formula for Energy?

Part D: Self-Reflection & Evaluation

The "Gap" Analysis: Based on your study of Dimensional Analysis Limitations, list three specific scenarios in a laboratory setting where relying only on dimensional analysis would lead to an incorrect conclusion.

Pro-Tip for NEP Success: Don't just solve for the variables. Try to visualize how changing the "Dimensions" of a problem changes our understanding of the universe!

Here is the step-by-step solution key for the Higher-Order Thinking Skills Worksheet. These solutions emphasize the logic behind the "Physics Grammar" of dimensions.

🔑 Solution Key: Advanced Dimensional Analysis

Part A: Conceptual Depth

1. The Constant Conundrum

Reasoning: Dimensions represent the nature of a physical quantity (length, mass, time). Pure numbers or ratios (like $2\pi$ or $1/2$) do not have a physical nature—they are scaling factors. Since dimensional analysis only tracks base units, these constants are lost.
Method to find them: To find the exact value of such constants, one must use experimental data or rigorous calculus-based derivations.

2. The Homogeneity Challenge

Consistency Check: * LHS ($v^2$): $[L T^{-1}]^2 = [L^2 T^{-2}]$
- RHS Term 1 ($u^2$): $[L T^{-1}]^2 = [L^2 T^{-2}]$
- RHS Term 2 ($as$): $[L T^{-2}] \cdot [L] = [L^2 T^{-2}]$
- Since all terms have the same dimensions, the equation is consistent.
Critical Analysis: No, consistency does not guarantee correctness. The numerical coefficient $5/2$ is incorrect (the real kinematic equation uses $2$). Dimensional analysis ensures the units match, but not the accuracy of the numbers.

3. Limitations of Addition

Reasoning: You cannot add "5 kilograms" to "2 meters." Physical addition represents the scaling of the same property. The Principle of Homogeneity states that only quantities with identical dimensional formulas can be added or subtracted to maintain a logically sound physical law.

Part B: Advanced Derivations

4. Planetary Harmonics ($T \propto R^a M^b G^c$)

Set up the equation: $[T] = [L]^a [M]^b [M^{-1} L^3 T^{-2}]^c$
Combine powers: $[M^0 L^0 T^1] = [M^{b-c} L^{a+3c} T^{-2c}]$
Solve for powers:
1. $-2c = 1 \Rightarrow c = -1/2$
2. $b - c = 0 \Rightarrow b = -1/2$
3. $a + 3c = 0 \Rightarrow a + 3(-1/2) = 0 \Rightarrow a = 3/2$
Result: $T \propto \sqrt{\frac{R^3}{GM}}$ (This is Kepler’s Third Law).

5. Fluid Dynamics ($v \propto r^a \eta^b F^c$)

Dimensional Equation: $[L T^{-1}] = [L]^a [M L^{-1} T^{-1}]^b [M L T^{-2}]^c$
Combine powers: $[M^0 L^1 T^{-1}] = [M^{b+c} L^{a-b+c} T^{-b-2c}]$
Solve for powers:
1. $b + c = 0 \Rightarrow b = -c$
2. Substitute in $T$: $-(-c) - 2c = -1 \Rightarrow c - 2c = -1 \Rightarrow c = 1$
3. Therefore, $b = -1$.
4. Substitute in $L$: $a - (-1) + 1 = 1 \Rightarrow a + 2 = 1 \Rightarrow a = -1$
Result: $v \propto \dfrac{F}{r\eta}$ (Related to Stokes' Law).

Part C: Real-World Application

6. The "Physics Grammar" Audit

Error: The table listed Pressure as $[M L^1 T^{-2}]$.
Correct Derivation: * $P = \text{Force} / \text{Area}$
- $[P] = [M L T^{-2}] / [L^2]$
- Correct Result: $[M L^{-1} T^{-2}]$

7. Unit System Architect (FLT System)

Mass in FLT: * Since $F = M \cdot A \Rightarrow M = F / A$
- $A = [L T^{-2}]$
- Dimensional Formula: $[F L^{-1} T^2]$
Energy in FLT:
- Energy = Force $\cdot$ Distance
- Dimensional Formula: $[F L]$

Part D: Self-Reflection

8. The "Gap" Analysis

Scenario 1: Distinguishing between Kinetic Energy ($1/2 mv^2$) and Potential Energy ($mgh$)—they look identical dimensionally.
Scenario 2: Solving equations involving Trigonometric or Logarithmic functions (arguments must be dimensionless).
Scenario 3: Determining if a quantity is a vector or scalar (e.g., Work and Torque have the same dimensions $[M L^2 T^{-2}]$, but one is a scalar and the other a vector).