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Dimensional Formula

The dimensional formula is a way to express physical quantities in terms of the fundamental quantities: Mass (M), Length (L), and Time (T). It helps verify the correctness of equations, derive relationships between quantities, and convert units systematically.

Understanding Dimensional Formula

1. Definition:
The dimensional formula expresses a physical quantity in terms of the powers (exponents) of the fundamental quantities—mass (M), length (L), and time (T). For example:

Velocity = Distance ÷ Time → Dimensional Formula: $[𝑀^{0} 𝐿^{1} 𝑇^{- 1}]$ .

2. Fundamental Quantities and Their Dimensions:

Physical Quantity	Symbol	Dimension
Mass	M	$[𝑀]$
Length	L	$[𝐿]$
Time	T	$[𝑇]$

3. Derived Quantities and Their Dimensions:
Derived quantities are expressed using combinations of fundamental dimensions.

Example:
- Force = Mass × Acceleration → $[𝐹] = [𝑀] \cdot [𝐿 𝑇^{- 2}] = [𝑀^{1} 𝐿^{1} 𝑇^{- 2}]$ .

Applications of Dimensional Formula

1. Checking the Correctness of Equations:
An equation is dimensionally correct if both sides have the same dimensions.

Example: $𝑣 = 𝑢 + 𝑎 𝑡$
- Left-hand side ( $𝑣$ ): $[𝐿 𝑇^{- 1}]$ .
- Right-hand side ( $𝑢 + 𝑎 𝑡$ ): $[𝐿 𝑇^{- 1}] + [𝐿 𝑇^{- 2}] \cdot [𝑇] = [𝐿 𝑇^{- 1}]$ .
- Both sides match, so the equation is dimensionally correct.

2. Deriving Relationships Between Physical Quantities:
If the relationship between variables is unknown, dimensional analysis can help derive it.

Example: Derive the formula for the period of a pendulum ( $𝑇$ ).
- Assume $𝑇$ depends on length ( $𝑙$ ), acceleration due to gravity ( $𝑔$ ), and mass ( $𝑚$ ).
- $𝑇 \propto 𝑙^{𝑎} 𝑔^{𝑏} 𝑚^{𝑐}$ .
- Substituting dimensions: $[𝑇] = [𝐿]^{𝑎} [𝐿 𝑇^{- 2}]^{𝑏} [𝑀]^{𝑐}$ .
- Solving gives $𝑎 = \frac{1}{2}$ , $𝑏 = - \frac{1}{2}$ , $𝑐 = 0$ , so $𝑇 \propto \sqrt{\frac{𝑙}{𝑔}}$ .

3. Converting Units:
Dimensional formulas help convert units from one system to another.

Example: Convert $1 N$ (Newton) into dyne.
- $1 N = 1 kg \cdot {m/s}^{2}$ .
- In CGS units: $1 dyne = 1 g \cdot {cm/s}^{2}$ .
- Conversion factor: $1 N = 10^{5} dyne$ .

Limitations of Dimensional Analysis

Cannot determine numerical constants (e.g., cannot derive $𝐸 = \frac{1}{2} 𝑚 𝑣^{2}$ ).
Does not apply to logarithmic, exponential, or trigonometric functions.
Assumes all variables are independent, which may not always be true.

Quick Revision Points

Fundamental Quantities: Mass ( $𝑀$ ), Length ( $𝐿$ ), Time ( $𝑇$ ).
Derived Quantities: Expressed as powers of $𝑀$ , $𝐿$ , and $𝑇$ .
Applications: Check correctness of equations, derive relationships, and convert units.
Limitations: Cannot handle constants, non-dimensional functions, or dependent variables.

Previous Year Questions and Answers

Q1: What is the dimensional formula of velocity?
A1: Velocity = Distance ÷ Time → Dimensional Formula:

[𝑀^{0} 𝐿^{1} 𝑇^{- 1}]

Q2: Write the dimensional formula of force.
A2: Force = Mass × Acceleration → Dimensional Formula:

[𝑀^{1} 𝐿^{1} 𝑇^{- 2}]

Q3: Check the dimensional correctness of the equation

𝑣 = 𝑢 + 𝑎 𝑡

.
A3:

Left-hand side ( $𝑣$ ): $[𝐿 𝑇^{- 1}]$ .
Right-hand side ( $𝑢 + 𝑎 𝑡$ ): $[𝐿 𝑇^{- 1}] + [𝐿 𝑇^{- 2}] \cdot [𝑇] = [𝐿 𝑇^{- 1}]$ .
Both sides match, so the equation is dimensionally correct.

Q4: Derive the dimensional formula of energy.
A4: Energy = Force × Distance →

[𝑀^{1} 𝐿^{1} 𝑇^{- 2}] \cdot [𝐿] = [𝑀^{1} 𝐿^{2} 𝑇^{- 2}]

Q5: Why is dimensional analysis not applicable to exponential functions?
A5: Dimensional analysis assumes that physical quantities are expressed as products or ratios of fundamental dimensions, which does not apply to exponential or logarithmic functions.

Expected Questions

Q1: What is the dimensional formula of pressure?
A1: Pressure = Force ÷ Area →

[𝑀^{1} 𝐿^{1} 𝑇^{- 2}] \div [𝐿^{2}] = [𝑀^{1} 𝐿^{- 1} 𝑇^{- 2}]

Q2: Check the dimensional correctness of the equation

𝑠 = 𝑢 𝑡 + \frac{1}{2} 𝑎 𝑡^{2}

.
A2:

Left-hand side ( $𝑠$ ): $[𝐿]$ .
Right-hand side ( $𝑢 𝑡 + \frac{1}{2} 𝑎 𝑡^{2}$ ): $[𝐿 𝑇^{- 1}] \cdot [𝑇] + [𝐿 𝑇^{- 2}] \cdot [𝑇^{2}] = [𝐿] + [𝐿] = [𝐿]$ .
Both sides match, so the equation is dimensionally correct.

Q3: Derive the dimensional formula of power.
A3: Power = Work ÷ Time →

[𝑀^{1} 𝐿^{2} 𝑇^{- 2}] \div [𝑇] = [𝑀^{1} 𝐿^{2} 𝑇^{- 3}]

Q4: What is the dimensional formula of angular velocity?
A4: Angular velocity = Angle ÷ Time →

[𝑀^{0} 𝐿^{0} 𝑇^{- 1}]

Q5: Can dimensional analysis determine the value of constants like

𝐺

(gravitational constant)?
A5: No, dimensional analysis cannot determine numerical constants; it only provides the relationship between physical quantities.

Unit 1.7 – Dimensional Formula