Q.Let us consider an equation $\dfrac{1}{2}\,m v^2 = m g h$ where $m$ is the mass of the body, $v$ its velocity, $g$ is the acceleration due to gravity and $h$ is the height. Check whether this equation is dimensionally correct.

Concept: Dimensional homogeneity – every term in a physically valid equation must have identical dimensions.

For the equation $\frac{1}{2} m v^2 = m g h$ to be dimensionally correct, both sides must reduce to the same dimensional formula.

Left side: $\frac{1}{2} m v^2$ has dimensions $[M][LT^{-1}]^2 = [ML^2T^{-2}]$, which is energy (kinetic energy).

Right side: $m g h$ has dimensions $[M][LT^{-2}][L] = [ML^2T^{-2}]$, which is also energy (potential energy). …

The principle of dimensional homogeneity states that any physically meaningful equation must have the same dimensions on both sides. This is a necessary (though not sufficient) condition for correctness. The equation $\frac{1}{2}mv^2 = mgh$ represents the conservation of mechanical energy: kinetic energy equals the potential energy lost. If the dimensions don't match, the equation cannot possibly be correct.

Dimensional analysis strips away numerical constants and focuses purely on the fundamental quantities: mass $[M]$, length $[L]$, and time $[T]$.

Step-by-step dimensional check

1. Identify the dimensions of each physical quantity

We need the dimensions of mass, velocity, acceleration due to gravity, and height:

2. Find the dimensions of the left-hand side

The left side is $\frac{1}{2}mv^2$. The factor $\frac{1}{2}$ is dimensionless (pure number), so we ignore it:

3. Find the dimensions of the right-hand side

The right side is $mgh$:

4. Compare both sides

Left-hand side: $[M L^2 T^{-2}]$ …

Setting up the exponent table

1. Write each base quantity's dimension as an $(M, L, T)$ exponent triple:

   Quantity	Dimension	Triple $(M,L,T)$
   $m$	$[M]$	$(1, 0, 0)$
   $v$	$[LT^{-1}]$	$(0, 1, -1)$
   $g$	$[LT^{-2}]$	$(0, 1, -2)$
   $h$	$[L]$	$(0, 1, 0)$

2. Sum the triples for the left side, $\frac{1}{2}mv^2$. The factor $\frac{1}{2}$ is a pure number, so it contributes $(0,0,0)$ and is dropped. $v^2$ doubles $v$'s triple:

3. Sum the triples for the right side, $mgh$ — each factor's triple is added once, since it's a simple product: $m:(1,0,0) + g:(0,1,-2) + h:(0,1,0) \;\;\Rightarrow\;\; (1,2,-2)$ …

Common Mistakes in Dimensional Analysis of $\frac{1}{2} m v^2 = m g h$

Mistake 1: Ignoring the constant $\frac{1}{2}$

What students do wrong:

Students often think that because $\frac{1}{2}$ is a pure number, it can be ignored in dimensional analysis. That part is actually correct — but some students remove it incorrectly or get confused about whether it affects dimensions.

Correct approach:

Constants like $\frac{1}{2}$, $\pi$, $e$ are dimensionless. They do not affect the dimensional formula. So you can safely ignore them.

✓ Rule: Only check the dimensions of the physical quantities, not the numerical coefficients.

Mistake 2: Writing wrong dimensions for velocity ($v$)

What students do wrong:

Some write $[v] = [L T^{-1}]$ correctly, but then square it incorrectly — e.g., writing $[v^2] = [L^2 T^{-2}]$ as $[L T^{-2}]$ or $[L^2 T^{-1}]$.

Correct approach:

• $v$ = distance / time → $[v] = [L T^{-1}]$
• $v^2$ → $[v^2] = [L^2 T^{-2}]$

✓ Tip: Square the entire dimensional formula, not just the unit.

Mistake 3: Confusing $g$ (acceleration due to gravity) with $g$ (gram)

What students do wrong:

In the equation, $g$ is acceleration due to gravity, but some students treat it as mass (gram) and write $[M]$.

Correct approach:

• $g$ = acceleration = velocity / time → $[g] = [L T^{-2}]$
• Never confuse the symbol $g$ with the unit gram ($g$). In dimensional analysis, symbols represent physical quantities, not units.

✓ Tip: Always identify the physical quantity first: "g is acceleration due to gravity."

Mistake 4: Forgetting to check both sides separately

What students do wrong:

Students try to "cancel" $m$ from both sides without first writing dimensions. This leads to errors — especially if they cancel $m$ but forget that $m$ appears on both sides with the same dimension.

Correct approach:

• Left-hand side (LHS): $\frac{1}{2} m v^2$

  • $[m] = [M]$, $[v^2] = [L^2 T^{-2}]$
  • So $[LHS] = [M][L^2 T^{-2}] = [M L^2 T^{-2}]$

• Right-hand side (RHS): $m g h$

  • $[m] = [M]$, $[g] = [L T^{-2}]$, $[h] = [L]$
  • So $[RHS] = [M][L T^{-2}][L] = [M L^2 T^{-2}]$

✓ Result: Both sides have dimensions $[M L^2 T^{-2}]$ → dimensionally correct.

Mistake 5: Thinking "dimensionally correct" means "numerically correct"

What students do wrong: …