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.