Answer Key-Deducing Relations Among Physical Quantities Using Dimensional Analysis:

 Here is the Answer Key for the NEP-based worksheet on Deducing Relations Among Physical Quantities Using Dimensional Analysis:

✅ Section A: Conceptual Understanding

1. Definitions:
   (a) Dimensional Formula: An expression showing which base quantities and their powers represent a physical quantity. E.g., volume = [M⁰ L³ T⁰].
   (b) Dimensional Equation: An equation equating the physical quantity with its dimensional formula. E.g., [F] = [M L T⁻²].
   (c) Dimensionless Quantity: A physical quantity with no dimensions. E.g., angle (radian), refractive index.

2. Limitations:

   • It cannot determine the dimensionless constants (like 2π, ½).

   • It cannot differentiate between two physical quantities with the same dimensions (e.g., torque and work).

3. An equation is dimensionally consistent if all the terms in it have the same dimensions.

4. Because dimensional analysis only considers powers of base units, and constants like $2\pi$ are pure numbers without dimensions.

✅ Section B: Application-Based Questions

5. Simple Pendulum

   $$T = k \cdot l^x \cdot g^y$$

   Dimensions:
   $[T] = [L]^x [L T^{-2}]^y = L^{x+y} T^{-2y}$
   Equating dimensions:

   $$x + y = 0,\quad -2y = 1 \Rightarrow y = -\tfrac{1}{2},\ x = \tfrac{1}{2}$$ $$\boxed{T = k \sqrt{\frac{l}{g}}}$$

6. Wave Speed

   $$v = k \cdot T^a \cdot \mu^b$$

   Dimensions:
   $[L T^{-1}] = [M L T^{-2}]^a [M L^{-1}]^b$
   $= M^{a+b} L^{a - b} T^{-2a}$
   Equating powers:

   $$a + b = 0,\quad a - b = 1,\quad -2a = -1 \Rightarrow a = \tfrac{1}{2},\ b = -\tfrac{1}{2}$$ $$\boxed{v = k \sqrt{\frac{T}{\mu}}}$$

7. Free Fall Time

   $$t = k \cdot h^x \cdot g^y$$

   Dimensions:
   $[T] = [L]^x [L T^{-2}]^y = L^{x+y} T^{-2y}$

   $$x + y = 0,\quad -2y = 1 \Rightarrow y = -\tfrac{1}{2},\ x = \tfrac{1}{2}$$ $$\boxed{t = k \sqrt{\frac{h}{g}}}$$

✅ Section C: Higher-Order Thinking

8. Angular Momentum $Q = m^x v^y r^z$
   $[Q] = [M L^2 T^{-1}]$,
   $[m]^x [L T^{-1}]^y [L]^z = M^x L^{y+z} T^{-y}$
   Matching exponents:

   $$x = 1,\quad y + z = 2,\quad -y = -1 \Rightarrow y = 1,\ z = 1$$ $$\boxed{Q = k \cdot m \cdot v \cdot r}$$

9. Potential Energy:
   The formula $U = -\frac{G M m}{r}$ contains a negative sign and constant $G$.
   Dimensional analysis can confirm form like:

   $$U \propto \frac{Mm}{r}$$

   But cannot determine the value or sign of the gravitational constant $G$.
   ✅ Yes, it can be verified dimensionally but not derived exactly.

10. Centripetal Force

$$F = k \cdot m^x v^y r^z \Rightarrow [M L T^{-2}] = M^x L^{y+z} T^{-y}$$

Matching:

$$x = 1,\quad y + z = 1,\quad -y = -2 \Rightarrow y = 2,\ z = -1$$ $$\boxed{F = k \cdot \frac{m v^2}{r}}$$

✅ Section D: Competency Activity

Given:

$$\Phi = k \cdot t^a \cdot q^b \cdot l^c$$

Assuming $\Phi$ behaves like electric potential (V).

$$[V] = \frac{W}{q} = \frac{M L^2 T^{-2}}{A T} = M L^2 T^{-3} A^{-1}$$

Dimensions:

$$[M L^2 T^{-3} A^{-1}] = [T]^a [A T]^b [L]^c = A^b T^{a+b} L^c$$

Matching exponents:

$$A^{-1} = A^b \Rightarrow b = -1$$ $$T^{-3} = T^{a + b} \Rightarrow a + (-1) = -3 \Rightarrow a = -2$$ $$L^2 = L^c \Rightarrow c = 2$$ $$\boxed{a = -2,\ b = -1,\ c = 2} \Rightarrow \Phi = k \cdot \frac{l^2}{q \cdot t^2}$$

Dimensional Analysis and Its Applications

 

📘 Dimensional Analysis and Its Applications (1.6)

Dimensional analysis is a powerful tool in physics that deals with the study and application of dimensions of physical quantities. It ensures dimensional consistency in mathematical equations, which means that only quantities with the same dimensions can be added or subtracted.

This method is used to:

1. Verify the correctness of equations by ensuring both sides have the same dimensional formula.

2. Derive relationships among physical quantities when the exact equation is unknown.

3. Convert units from one system to another by treating units algebraically (like cancelling common terms).

4. Check the homogeneity of physical laws without needing specific values.

It is based on the principle of homogeneity, which states that the dimensions of all terms in a physical equation must be the same. Thus, dimensional analysis acts as a preliminary test for the validity of physical equations and helps build a strong foundation in understanding the physical behaviour of systems.

📘 Comprehensive Note on: Deducing Relation Among Physical Quantities Using Dimensional Analysis (Section 1.6.2)

The method of dimensions is a powerful analytical tool in physics that helps deduce the possible mathematical form of a relation among physical quantities. This technique assumes that if a physical quantity depends on certain other quantities, then this dependence can be expressed in the form of a product of powers of those quantities. However, it is important to note that only the form of the relation can be deduced — dimensionless constants (like π, 2, etc.) cannot be determined by this method.

🔶 General Form of Dimensional Analysis:

Suppose a physical quantity $Q$ depends on other physical quantities $A, B, C, \ldots$, then we assume:

$$Q = k \cdot A^x \cdot B^y \cdot C^z \ldots$$

Where:

• $k$ is a dimensionless constant.

• $x, y, z, \ldots$ are unknown exponents to be determined.

• All quantities are replaced by their dimensional formulae, and the powers of each fundamental quantity on both sides are equated to solve for $x, y, z \ldots$

🔷 Example: Time Period of a Simple Pendulum

Let the time period $T$ depend on:

• Length of the pendulum $l$,

• Mass of the bob $m$,

• Acceleration due to gravity $g$

Assume:

$$T = k \cdot l^x \cdot g^y \cdot m^z$$

Write the dimensional formula of each quantity:

• $[T] = [T^1]$

• $[l] = [L]$

• $[g] = [L T^{-2}]$

• $[m] = [M]$

Now substitute:

$$[T] = [L]^x [L T^{-2}]^y [M]^z = [L^{x+y} T^{-2y} M^z]$$

Compare the powers of fundamental quantities on both sides:

$$\begin{aligned} \text{LHS:} & \quad [L^0 M^0 T^1] \\ \text{RHS:} & \quad [L^{x+y} M^z T^{-2y}] \end{aligned}$$

Equating the exponents:

$$\begin{aligned} x + y &= 0 \quad \text{(1)} \\ -2y &= 1 \quad \text{(2)} \\ z &= 0 \quad \text{(3)} \end{aligned}$$

Solving:

From (2): $y = -\frac{1}{2}$
From (1): $x = \frac{1}{2}$
From (3): $z = 0$

So, the relation becomes:

$$T = k \cdot l^{1/2} \cdot g^{-1/2} = k \cdot \sqrt{\frac{l}{g}}$$

The actual equation is:

$$T = 2\pi \sqrt{\frac{l}{g}}$$

Note: Dimensional analysis cannot determine the value of $k$ (in this case, $2\pi$).

✅ Key Points to Remember:

• The method assumes product-type dependency (no addition/subtraction).

• It helps deduce possible relationships between variables.

• Dimensionless constants like $2, \pi, e$, etc., cannot be determined.

• It is limited to dimensional consistency, not actual physical derivation.

• It cannot distinguish between different quantities having the same dimensions (e.g., torque and energy).

📌 Applications:

• Verifying the correctness of derived equations.

• Estimating relations where experimental data or derivations are unavailable.

• Developing initial hypotheses for further theoretical or experimental verification.

NEP-based (National Education Policy, India) Worksheet on “Deducing Relations Among Physical Quantities Using Dimensional Analysis” designed for Class 11 Physics under experiential, competency-based learning. It includes conceptual understanding, application, and analysis-level questions, aligned with NEP’s core goals.

📝 Worksheet: Dimensional Analysis & Deducing Relations

Subject: Physics | Grade: 11 | Topic: Dimensional Analysis
Learning Outcome: Learners will deduce relationships among physical quantities using the method of dimensions and understand its applications and limitations.

✨ Section A: Conceptual Understanding (Recall & Comprehension)

1. Define:
   (a) Dimensional formula
   (b) Dimensional equation
   (c) Dimensionless quantity

2. State two limitations of the method of dimensions in deducing physical relations.

3. What does it mean for an equation to be dimensionally consistent?

4. Why can’t we find the value of constants like $2\pi$, $e$, or numerical coefficients using dimensional analysis?

✍️ Section B: Application-Based Questions

5. The time period $T$ of a simple pendulum depends on its length $l$, and acceleration due to gravity $g$. Use dimensional analysis to derive the formula for $T$.
   (Hint: $T = k \cdot l^x \cdot g^y$)

6. The speed $v$ of a wave on a string depends on the tension $T$ in the string and the mass per unit length $\mu$. Use dimensional analysis to find the relation between them.
   (Hint: $T \rightarrow [M L T^{-2}],\ \mu \rightarrow [M L^{-1}]$)

7. Derive the expression for the time $t$ taken by a body to fall freely from a height $h$ under gravity $g$, using dimensional analysis.

🧠 Section C: Higher-Order Thinking (Analysis & Reasoning)

8. A quantity $Q$ depends on mass $m$, velocity $v$, and radius $r$ such that:

   $$Q = k \cdot m^x \cdot v^y \cdot r^z$$

   If $Q$ represents angular momentum, find the values of $x, y, z$ using dimensional analysis.

9. Can dimensional analysis be used to derive the formula for gravitational potential energy $U = -\frac{G M m}{r}$? Justify your answer.

10. Suppose the centripetal force $F$ acting on a body moving in a circle of radius $r$ with speed $v$ depends on mass $m$, speed $v$, and radius $r$. Use dimensional analysis to derive the form of the equation.

🎯 Section D: Competency-Based Skill Practice (Activity)

🔶 Activity:
Imagine you are a scientist who just discovered a new physical quantity called “flarence” (symbol $\Phi$) which is believed to depend on time $t$, charge $q$, and length $l$. Your task is to assume a formula and use dimensional analysis to find the relationship.

$$\Phi = k \cdot t^a \cdot q^b \cdot l^c$$

• Find the dimensional formula of $\Phi$, assuming it behaves like electric potential.

• Determine the values of $a, b, c$.

🔚 Submission:

• Write your answers clearly with all steps.

• Highlight dimensional formulas used.

• Discuss the result of your analysis in one sentence.

Dimensional Consistency of Equations

CBSE Class 11 Physics: Chapter 1 – Units and Measurements (Section 1.6.1)

📘 1.6.1 – Dimensional Consistency of Equations

⚙️ Principle of Dimensional Homogeneity

The principle of homogeneity states:

An equation is dimensionally consistent (or homogeneous) if the dimensions of each term on both sides of the equation are the same.

Mathematically:

$$\text{If } A = B + C + D \quad \Rightarrow \quad [A] = [B] = [C] = [D]$$

✅ Only like dimensions can be added, subtracted, or equated.

📏 Why Check Dimensional Consistency?

• It helps verify correctness of derived equations.

• It ensures all terms are physically compatible.

• It provides a preliminary check before experimental validation.

• It is independent of unit systems, saving effort in unit conversions.

❗ Limitation: A dimensionally correct equation may not always be physically correct, but a dimensionally incorrect one is certainly wrong.

✅ Example: Kinematic Equation

$$x = x_0 + v_0 t + \frac{1}{2} a t^2$$

Where:

• $x$: displacement

• $x_0$: initial position

• $v_0$: initial velocity

• $a$: acceleration

• $t$: time

Check Dimensions:

• $[x] = [x_0] = \text{Length} = [L]$

• $[v_0 t] = [L T^{-1}] \cdot [T] = [L]$

• $\left[\frac{1}{2} a t^2 \right] = [L T^{-2}] \cdot [T^2] = [L]$

✅ All terms have dimension $[L]$ → dimensionally consistent.

🎯 Dimensional Check for:

$$\frac{1}{2} m v^2 = m g h$$

Where:

• $m$: mass $[M]$

• $v$: velocity $[L T^{-1}]$

• $g$: acceleration $[L T^{-2}]$

• $h$: height $[L]$

🔍 Left Side:

$$\left[\frac{1}{2} m v^2\right] = [M] \cdot [L T^{-1}]^2 = [M] \cdot [L^2 T^{-2}] = [M L^2 T^{-2}]$$

🔍 Right Side:

$$[m g h] = [M] \cdot [L T^{-2}] \cdot [L] = [M L^2 T^{-2}]$$

✅ Both sides = $[M L^2 T^{-2}]$ → dimensionally correct equation

📌 This is the equation of conservation of mechanical energy for a body under gravity.

⚠️ Special Note on Functions

• Arguments of functions like sin, cos, tan, log, exp must be dimensionless.

• For example:

  • $\sin(\theta)$, where $\theta$ is angle in radians (a pure number: $[1]$)

  • $\exp(x)$, $\log(x)$ must have $x$ as dimensionless

🔁 Difference Between Units and Dimensions

• Units: Based on a chosen standard (like meter, second)

• Dimensions: Inherent nature of a quantity (like Length $[L]$, Mass $[M]$)

Testing dimensions is easier and more general than testing units.

✅ Key Takeaways

• An equation is dimensionally consistent if all its terms share the same dimensions.

• A dimensionally inconsistent equation is always wrong.

• Dimensionally consistent ≠ always physically valid.

• Functions like sin, log, exp only accept dimensionless inputs.

• Helps derive or verify formulas without performing actual experiments.

Worksheet: Dimensional Formulae and Dimensional Equations

📝 Worksheet: Dimensional Formulae and Dimensional Equations

✍️ Section A: Fill in the Blanks

1. The dimensional formula of force is ________________.

2. The dimensional formula of energy is ________________.

3. The dimensional formula of pressure is ________________.

4. The dimensional formula of power is ________________.

5. The dimensional formula of momentum is ________________.

📘 Section B: Match the Following

Physical Quantity	Dimensional Formula
A. Work/Energy	i. $[M^1 L^1 T^{-2}]$
B. Acceleration	ii. $[M^1 L^2 T^{-3}]$
C. Power	iii. $[M^1 L^2 T^{-2}]$
D. Force	iv. $[M^0 L T^{-2}]$
E. Velocity	v. $[M^0 L T^{-1}]$

🧠 Section C: Conceptual Questions

1. What is meant by the dimensional formula of a physical quantity?

2. How can you check whether a physical equation is dimensionally correct?

3. Why can’t dimensional analysis help in finding constants like $\frac{1}{2}$, $\pi$, or 2?

4. Can two different physical quantities have the same dimensional formula? Give an example.

5. Why is dimensional analysis called a “preliminary test” for checking the correctness of equations?

🔎 Section D: Derivation Questions

Find the dimensional formula of the following quantities by analyzing their base physical relation.

1. Kinetic Energy: $KE = \frac{1}{2}mv^2$

2. Pressure: $P = \frac{F}{A}$

3. Momentum: $p = mv$

4. Gravitational Potential Energy: $U = mgh$

5. Surface Tension: $T = \frac{F}{l}$

🔍 Section E: Dimensional Consistency Check

Check whether the following equations are dimensionally correct.

1. $v = u + at$

2. $s = ut + \frac{1}{2}at^2$

3. $E = mc^3$

4. $T = 2\pi \sqrt{\frac{l}{g}}$ (Time period of a simple pendulum)

5. $F = ma + v$

📐 Section F: Assertion and Reasoning (Write TRUE/FALSE)

1. Assertion: The dimensional formula of work is the same as that of energy.
   Reason: Both involve force applied over a distance.

2. Assertion: Dimensional equations are unique for all physical quantities.
   Reason: Each physical quantity is derived from different base units.

✅ Answer Key

✍️ Section A: Fill in the Blanks

1. $[M^1 L^1 T^{-2}]$

2. $[M^1 L^2 T^{-2}]$

3. $[M^1 L^{-1} T^{-2}]$

4. $[M^1 L^2 T^{-3}]$

5. $[M^1 L^1 T^{-1}]$

📘 Section B: Match the Following

Physical Quantity	Dimensional Formula
A. Work/Energy	iii. $[M^1 L^2 T^{-2}]$
B. Acceleration	iv. $[M^0 L T^{-2}]$
C. Power	ii. $[M^1 L^2 T^{-3}]$
D. Force	i. $[M^1 L^1 T^{-2}]$
E. Velocity	v. $[M^0 L T^{-1}]$

🧠 Section C: Conceptual Questions

1. Dimensional formula expresses a physical quantity in terms of the base quantities (M, L, T, etc.) with their respective powers.

2. By checking whether both sides of the equation have the same dimensional formula.

3. Because dimensional analysis only handles dimensions, not pure numbers or constants without dimension.

4. Yes. Torque and work both have the same dimensional formula: $[M L^2 T^{-2}]$.

5. Because it only verifies dimensional homogeneity, not correctness of constants or functional dependence.

🔎 Section D: Derivation Questions

1. $KE = \frac{1}{2}mv^2$:
   $[M] \cdot [L T^{-1}]^2 = [M L^2 T^{-2}]$

2. $P = \frac{F}{A}$:
   $[M L T^{-2}]/[L^2] = [M L^{-1} T^{-2}]$

3. $p = mv$:
   $[M] \cdot [L T^{-1}] = [M L T^{-1}]$

4. $U = mgh$:
   $[M] \cdot [L T^{-2}] \cdot [L] = [M L^2 T^{-2}]$

5. $T = \frac{F}{l}$:
   $[M L T^{-2}]/[L] = [M T^{-2}]$

🔍 Section E: Dimensional Consistency Check

1. $v = u + at$
   ✅ Correct (All terms have $[L T^{-1}]$)

2. $s = ut + \frac{1}{2}at^2$
   ✅ Correct ($[L]$ on all terms)

3. $E = mc^3$
   ❌ Incorrect
   LHS: $[M L^2 T^{-2}]$
   RHS: $[M] \cdot [L T^{-1}]^3 = [M L^3 T^{-3}]$

4. $T = 2\pi \sqrt{\frac{l}{g}}$
   ✅ Correct
   $[L]/[L T^{-2}] = [T^2] \Rightarrow \sqrt{T^2} = [T]$

5. $F = ma + v$
   ❌ Incorrect
   LHS: $[M L T^{-2}]$, RHS: sum of force and velocity → dimensionally incompatible

📐 Section F: Assertion and Reasoning

1. TRUE – Both involve application of force over distance and have the same dimensional formula.

2. FALSE – Different quantities can share the same dimensional formula (e.g., torque and work).

Dimensional Formulae and Dimensional Equations

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.

Dimensions of Physical Quantities

Dimensions of Physical Quantities from Section 1.4 – CBSE Class 11 Physics (Units and Measurements)

📘 1.4 – Dimensions of Physical Quantities

🔍 What are Dimensions?

The dimension of a physical quantity refers to the power (or exponent) to which a base quantity must be raised to represent that physical quantity.

For any derived quantity, we express it in terms of fundamental dimensions like:

$$\text{[Length]} = [L], \quad \text{[Mass]} = [M], \quad \text{[Time]} = [T], \quad \text{[Electric Current]} = [A], \\ \text{[Temperature]} = [K], \quad \text{[Luminous Intensity]} = [cd], \quad \text{[Amount of Substance]} = [mol]$$

These seven are the fundamental or base dimensions in physics.

🔢 Representing Dimensions

The notation [Q] refers to the dimensions of a physical quantity Q.

For example:

• If $Q = \text{Force}$, then $[Q] = [M L T^{-2}]$

• If $Q = \text{Speed}$, then $[Q] = [L T^{-1}]$

📐 Examples in Mechanics

In mechanics, all quantities can be expressed using only [M], [L], [T].

📦 Example 1: Volume

Volume = Length × Breadth × Height
Each is of dimension $[L]$, so:

$$[\text{Volume}] = [L] \cdot [L] \cdot [L] = [L^3]$$

It has:

• 3 dimensions in length

• 0 in mass $[M^0]$

• 0 in time $[T^0]$

🧲 Example 2: Force

$$\text{Force} = \text{Mass} \times \text{Acceleration}$$ $$[\text{Mass}] = [M], \quad [\text{Acceleration}] = \frac{[L]}{[T^2]} = [L T^{-2}]$$ $$[\text{Force}] = [M] \cdot [L T^{-2}] = [M L T^{-2}]$$

Force has:

• 1 dimension in mass

• 1 in length

• –2 in time

🚗 Example 3: Velocity

Velocity = Displacement / Time

$$[\text{Velocity}] = \frac{[L]}{[T]} = [L T^{-1}]$$

Same applies to:

• Initial velocity

• Final velocity

• Average velocity

• Speed

Even though magnitudes vary, their dimensions remain the same.

✅ Important Characteristics of Dimensional Representation

• Only the type or nature of quantity is described.

• It does not involve magnitude or unit.

• Different quantities with same dimensional formula are called dimensionally similar.

• It helps:

  • Check dimensional consistency of equations

  • Derive relations between quantities

  • Convert units between systems

🧮 Zero Dimensions in a Base Quantity

A quantity can have zero power in a particular base quantity, meaning it doesn't depend on it.

For example:

• Volume: $[M^0 L^3 T^0]$

• Speed: $[M^0 L^1 T^{-1}]$

🚫 Not All Physical Properties Are Dimensioned

• Quantities like refractive index, strain, angle (in radians) are dimensionless.

• Still physical, but have no associated base dimension.

🧠 Summary Table

Physical Quantity	Dimensional Formula
Speed/Velocity	$[L T^{-1}]$
Acceleration	$[L T^{-2}]$
Force	$[M L T^{-2}]$
Work/Energy	$[M L^2 T^{-2}]$
Power	$[M L^2 T^{-3}]$
Pressure	$[M L^{-1} T^{-2}]$
Density	$[M L^{-3}]$
Momentum	$[M L T^{-1}]$

🧾 Conclusion

Understanding dimensions is essential for:

• Analyzing physical relationships

• Verifying equations

• Deriving new formulas

• Performing conversions across unit systems

By using dimensional analysis, we gain powerful insight into the structure of physical laws, without the need for detailed experiments at every step.