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Applications of Dimensional Analysis

Dimensional analysis is a powerful tool in physics and engineering. It helps verify equations, derive relationships, and convert units systematically. Below are the key applications of dimensional analysis explained in detail.

1. Checking the Correctness of Equations

Dimensional analysis ensures that both sides of an equation have the same dimensions. If they do not match, the equation is incorrect.

Example:
Check if

v^{2} = u^{2} + 2 a s

is dimensionally correct.

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

2. Deriving Relationships Between Physical Quantities

When the exact relationship between variables is unknown, dimensional analysis can help derive it by assuming proportionality.

Example: Derive the formula for the time period (

T

) of a simple pendulum.

Assume $T$ depends on length ( $l$ ), acceleration due to gravity ( $g$ ), and mass ( $m$ ).
Let $T \propto l^{a} g^{b} m^{c}$ .
Substituting dimensions:
- $[T] = [L]^{a} [L T^{- 2}]^{b} [M]^{c}$ .
- Simplify: $[T] = [L^{a + b} T^{- 2 b} M^{c}]$ .
Equating powers:
- For $M$ : $c = 0$ (mass does not affect $T$ ).
- For $T$ : $- 2 b = 1 \Rightarrow b = - \frac{1}{2}$ .
- For $L$ : $a + b = 0 \Rightarrow a = \frac{1}{2}$ .
Final formula: $T \propto \sqrt{\frac{l}{g}}$ .

3. Converting Units Between Systems

Dimensional analysis helps convert units from one system (e.g., SI) to another (e.g., CGS).

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 factors:
- $1 kg = 10^{3} g$ .
- $1 m = 10^{2} cm$ .
Substitute:
- $1 N = (10^{3} g) \cdot (10^{2} cm) / (s^{2}) = 10^{5} dyne$ .

4. Determining the Nature of Physical Quantities

Dimensional analysis helps identify whether a quantity is fundamental or derived.

Example:

Work = Force × Distance → $[M^{1} L^{2} T^{- 2}]$ .
This shows work is a derived quantity.

Quick Revision Points

Checking Equations: Ensure both sides have the same dimensions.
Deriving Relationships: Assume proportionality and solve for exponents.
Unit Conversion: Use conversion factors based on dimensions.
Nature of Quantities: Identify fundamental vs. derived quantities.

Previous Year Questions and Answers

Q1: Use dimensional analysis to check if

E = m c^{2}

is dimensionally correct.
A1:

Left-hand side ( $E$ ): Energy → $[M^{1} L^{2} T^{- 2}]$ .
Right-hand side ( $m c^{2}$ ): Mass × Velocity² → $[M] \cdot [L T^{- 1}]^{2} = [M^{1} L^{2} T^{- 2}]$ .
Both sides match, so the equation is dimensionally correct.

Q2: Derive the dimensional formula for surface tension.
A2: Surface tension = Force ÷ Length →

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

Q3: Convert

1 Joule

into ergs.
A3:

$1 Joule = 1 kg \cdot m^{2} / s^{2}$ .
In CGS units: $1 erg = 1 g \cdot {cm}^{2} / s^{2}$ .
Conversion factors:
- $1 kg = 10^{3} g$ .
- $1 m = 10^{2} cm$ .
Substitute:
- $1 Joule = (10^{3} g) \cdot (10^{2} cm)^{2} / (s^{2}) = 10^{7} ergs$ .

Q4: Derive the relationship between time period (

T

) and length (

l

) for a pendulum.
A4:

Assume $T \propto l^{a} g^{b}$ .
Substituting dimensions: $[T] = [L]^{a} [L T^{- 2}]^{b}$ .
Simplify: $[T] = [L^{a + b} T^{- 2 b}]$ .
Equating powers:
- For $T$ : $- 2 b = 1 \Rightarrow b = - \frac{1}{2}$ .
- For $L$ : $a + b = 0 \Rightarrow a = \frac{1}{2}$ .
Final formula: $T \propto \sqrt{\frac{l}{g}}$ .

Expected Questions

Q1: Use dimensional analysis to check if

F = m a

is dimensionally correct.
A1:

Left-hand side ( $F$ ): Force → $[M^{1} L^{1} T^{- 2}]$ .
Right-hand side ( $m a$ ): Mass × Acceleration → $[M] \cdot [L T^{- 2}] = [M^{1} L^{1} T^{- 2}]$ .
Both sides match, so the equation is dimensionally correct.

Q2: Derive the dimensional formula for pressure.
A2: Pressure = Force ÷ Area →

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

Q3: Convert

1 Watt

into

erg/s

.
A3:

$1 Watt = 1 Joule/s = 1 kg \cdot m^{2} / s^{3}$ .
In CGS units: $1 erg/s = 1 g \cdot {cm}^{2} / s^{3}$ .
Conversion factors:
- $1 kg = 10^{3} g$ .
- $1 m = 10^{2} cm$ .
Substitute:
- $1 Watt = (10^{3} g) \cdot (10^{2} cm)^{2} / (s^{3}) = 10^{7} erg/s$ .

Q4: Derive the relationship between velocity (

v

), radius (

r

), and angular velocity (

ω

).
A4:

Assume $v \propto r^{a} ω^{b}$ .
Substituting dimensions: $[L T^{- 1}] = [L]^{a} [T^{- 1}]^{b}$ .
Simplify: $[L T^{- 1}] = [L^{a} T^{- b}]$ .
Equating powers:
- For $L$ : $a = 1$ .
- For $T$ : $- b = - 1 \Rightarrow b = 1$ .
Final formula: $v \propto r ω$ .

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

Unit 1.8 – Applications of Dimensional Analysis