Class 11 > Unit # 01: Physical Quantities & Measurements > Homogeneity of Equations Using Dimension

Verification of Physics Equations Using Dimensional Analysis

One of the primary applications of dimensions is checking the correctness of physical equations. According to the Principle of Homogeneity of Dimensions, an equation is dimensionally correct if the dimensions of all the terms on both sides (LHS and RHS) are exactly identical. Below are five common examples illustrating this verification process.

Example 1: First Equation of Motion ($v_f = v_i + at$)

Where: $v_f$ = final velocity, $v_i$ = initial velocity, $a$ = acceleration, $t$ = time

Left Hand Side (LHS):
Dimension of $v_f = \text{Velocity} = $ [LT⁻¹]

Right Hand Side (RHS):
Dimension of $v_i = $ [LT⁻¹]
Dimension of $at = \text{[LT⁻²]} \times \text{[T]} = $ [LT⁻¹]

Since both terms on the RHS possess identical dimensions to the LHS, the terms can be added algebraically:

$$\text{LHS} = \text{RHS} = [LT^{-1}]$$

✓ Verified: The equation is dimensionally correct.

Example 2: Second Equation of Motion ($S = v_i t + \frac{1}{2}at^2$)

Where: $S$ = distance, $v_i$ = initial velocity, $a$ = acceleration, $t$ = time

Left Hand Side (LHS):
Dimension of $S = \text{Length} = $ [L]

Right Hand Side (RHS):
Dimension of $v_i t = \text{[LT⁻¹]} \times \text{[T]} = $ [L]
Dimension of $\frac{1}{2}at^2 = \text{[LT⁻²]} \times \text{[T²]} = $ [L]  (Note: Constant numbers like $\frac{1}{2}$ are dimensionless)

$$\text{LHS} = \text{RHS} = [L]$$

✓ Verified: The equation is dimensionally correct.

Example 3: Einstein's Mass-Energy Equivalence ($E = mc^2$)

Where: $E$ = energy, $m$ = mass, $c$ = speed of light

Left Hand Side (LHS):
Dimension of $E = \text{Energy (or Work)} = $ [ML²T⁻²]

Right Hand Side (RHS):
Dimension of $m = $ [M]
Dimension of $c^2 = (\text{[LT⁻¹]})^2 = $ [L²T⁻²]
Combining them: $mc^2 = $ [ML²T⁻²]

$$\text{LHS} = \text{RHS} = [ML^2T^{-2}]$$

✓ Verified: The equation is dimensionally correct.

Example 4: Kinetic Energy Formula ($K.E. = \frac{1}{2}mv^2$)

Where: $K.E.$ = kinetic energy, $m$ = mass, $v$ = velocity

Left Hand Side (LHS):
Dimension of $K.E. = $ [ML²T⁻²]

Right Hand Side (RHS):
Dimension of $m = $ [M]
Dimension of $v^2 = (\text{[LT⁻¹]})^2 = $ [L²T⁻²]
Combining them: $\frac{1}{2}mv^2 = $ [ML²T⁻²]  (Constant $\frac{1}{2}$ is dimensionless)

$$\text{LHS} = \text{RHS} = [ML^2T^{-2}]$$

✓ Verified: The equation is dimensionally correct.

Example 5: Time Period of a Simple Pendulum ($T = 2\pi\sqrt{\frac{l}{g}}$)

Where: $T$ = time period, $l$ = length of pendulum, $g$ = acceleration due to gravity

Left Hand Side (LHS):
Dimension of $T = \text{Time} = $ [T]

Right Hand Side (RHS):
The constant $2\pi$ is dimensionless. Let's analyze the radical term $\sqrt{\frac{l}{g}}$:
Dimension of length $l = $ [L]
Dimension of gravitational acceleration $g = $ [LT⁻²]

Substituting into the ratio expression:

$$\text{RHS} = \sqrt{\frac{[L]}{[LT^{-2}]}} = \sqrt{\frac{1}{[T^{-2}]}} = \sqrt{[T^2]} = [T]$$ $$\text{LHS} = \text{RHS} = [T]$$

✓ Verified: The equation is dimensionally correct.