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.

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Accepted Answer

> [!TLDR]
> An equation is dimensionally correct if both sides have identical dimensions. Here $\frac{1}{2}mv^2$ and $mgh$ both reduce to $[M L^2 T^{-2}]$, confirming the equation is dimensionally consistent.

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:

| Quantity | Symbol | Dimensions |
|----------|--------|------------|
| Mass | $m$ | $[M]$ |
| Velocity | $v$ | $[L T^{-1}]$ |
| Acceleration | $g$ | $[L T^{-2}]$ |
| Height | $h$ | $[L]$ |

**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:

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

**3. Find the dimensions of the right-hand side**

The right side is $mgh$:

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

**4. Compare both sides**

Left-hand side: $[M L^2 T^{-2}]$ …

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 understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

The Core Idea

The Rules (Memorize These)

1. Non-zero digits are always significant

2. Zeros between non-zero digits are significant

3. Leading zeros are never significant

4. Trailing zeros are significant only if there's a decimal point

5. Exact numbers have infinite sig figs

Why This Matters: Calculations

Multiplication and Division

Addition and Subtraction

Why this formula?

Significant Figures: Why the Rules Work

1. The Fundamental Idea: Uncertainty is the Key

2. Rule for Addition and Subtraction

Why this holds

3. Rule for Multiplication and Division

Why this holds

4. Why These Rules Are Different …

Step-by-step dimensional check

Setting up the exponent table

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

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

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

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

Mistake 4: Forgetting to check both sides separately

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

Quantity	Symbol	Dimensions
Mass	$m$	$[M]$
Velocity	$v$	$[L T^{-1}]$
Acceleration	$g$	$[L T^{-2}]$
Height	$h$	$[L]$

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

Q.Let us consider an equation 12 mv2=mgh\dfrac{1}{2}\,m v^2 = m g h21​mv2=mgh where mmm is the mass of the body, vvv its velocity, ggg is the acceleration due to gravity and hhh is the height. Check whether this equation is dimensionally correct.

Step-by-step dimensional check

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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.

Unlock everything free for 14 days