How many significant figures are present in the following? (i) 0.0025 (ii) 208 (iii) 5005 (iv) 126,000 (v) 500

Exercises · 1.19

Q. How many significant figures are present in the following?

(i) 0.0025

(ii) 208

(iii) 5005

(iv) 126,000

(v) 500.0

(vi) 2.0034

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Concept understanding — Unit Conversion

Unit Conversion: Why 1 Metre and 100 Centimetres Are the Same Thing

Imagine you're measuring the length of your desk. You pull out a ruler marked in centimetres and find it's 120 cm long. Your friend, using a metre stick, says it's 1.2 m. You're both right — you've just used different units to describe the same physical length.

That's the core idea: unit conversion is the process of changing how you express a quantity without changing the quantity itself.

The Intuition: Same Quantity, Different Labels

Think of a pizza. Whether you call it "one pizza" or "8 slices," the amount of pizza hasn't changed. You've just used a different unit (pizza vs. slice) to describe it.

Similarly, 1 metre and 100 centimetres are the same length — just like 1 pizza and 8 slices are the same amount. The number changes (1 becomes 100, or 1 becomes 8), but the actual thing being measured stays identical.

Note

This is the most important idea to hold onto: conversion changes the number, not the quantity. If you ever feel like the quantity has changed, you've made a mistake.

The Precise Statement

Unit conversion is the multiplication of a quantity by a conversion factor — a fraction equal to 1 — that cancels the old unit and introduces the new one.

A conversion factor looks like this:

\frac{\text{new unit}}{\text{old unit}} = 1

For length: $\frac{1 \text{ m}}{100 \text{ cm}} = 1$ and $\frac{100 \text{ cm}}{1 \text{ m}} = 1$ .

Why are these fractions equal to 1? Because 1 metre is 100 centimetres. The numerator and denominator describe the same physical length, so their ratio is exactly 1.

How to Convert: The Only Rule You Need

Multiply by a conversion factor that cancels the unit you have and leaves the unit you want.

Let's convert 120 cm to metres:

Start with what you have: $120 \text{ cm}$
Choose the conversion factor that has "cm" in the denominator (to cancel it) and "m" in the numerator: $\frac{1 \text{ m}}{100 \text{ cm}}$
Multiply:

120 \text{ cm} \times \frac{1 \text{ m}}{100 \text{ cm}} = \frac{120}{100} \text{ m} = 1.2 \text{ m}

The "cm" units cancel just like numbers do: $\frac{\text{cm}}{\text{cm}} = 1$ .

Tip

Always write the units explicitly. If the units don't cancel correctly, you've used the wrong conversion factor. This catches 90% of conversion mistakes.

The Reverse: Metres to Centimetres

Now convert 1.2 m to cm. This time, you want "cm" to remain and "m" to cancel. Use $\frac{100 \text{ cm}}{1 \text{ m}}$ :

1.2 \text{ m} \times \frac{100 \text{ cm}}{1 \text{ m}} = 1.2 \times 100 \text{ cm} = 120 \text{ cm}

Notice: when going from a larger unit (m) to a smaller unit (cm), the number gets larger (1.2 → 120). When going from smaller to larger, the number gets smaller (120 → 1.2). This is a useful sanity check.

Common Conversion Factors You'll Use

Quantity	Relationship	Conversion Factors
Length	1 m = 100 cm	$\frac{1 \text{ m}}{100 \text{ cm}}$ , $\frac{100 \text{ cm}}{1 \text{ m}}$
Mass	1 kg = 1000 g	$\frac{1 \text{ kg}}{1000 \text{ g}}$ , $\frac{1000 \text{ g}}{1 \text{ kg}}$
Time	1 h = 60 min	$\frac{1 \text{ h}}{60 \text{ min}}$ , $\frac{60 \text{ min}}{1 \text{ h}}$
Speed	1 km/h = $\frac{1000}{3600}$ m/s	$\frac{1000 \text{ m}}{1 \text{ km}} \times \frac{1 \text{ h}}{3600 \text{ s}}$

Why this formula?

Dimensional Analysis: Why the Key Principles Hold

Dimensional Analysis is a powerful tool in physics and engineering that lets us check the consistency of equations, derive relationships, and convert units. But why does it work? Let's build the reasoning from the ground up.

1. The Core Idea: Physical Quantities Have Dimensions

Every physical quantity (like length, time, mass) can be expressed in terms of fundamental dimensions. The most common set in mechanics is:

L = Length
M = Mass
T = Time

For example:

Speed has dimensions $[L T^{-1}]$
Force has dimensions $[M L T^{-2}]$
Energy has dimensions $[M L^2 T^{-2}]$

Why this matters: Two quantities can only be meaningfully compared or equated if they have the same dimensions. You cannot add apples to oranges — and you cannot add length to time.

2. The Principle of Dimensional Homogeneity

The key formula that underpins everything is:

Every valid physical equation must be dimensionally homogeneous.

This means: the dimensions on the left-hand side must equal the dimensions on the right-hand side.

Why must this hold?

Consider an equation like:

v = u + at

Left side: $[v] = L T^{-1}$
Right side: $[u] = L T^{-1}$ , $[at] = (L T^{-2})(T) = L T^{-1}$

Both sides have dimensions $L T^{-1}$ . If they didn't match, the equation would be physically meaningless — you'd be comparing quantities that cannot be equal in any real experiment.

Reasoning: Physical laws describe relationships between measurable quantities. If the dimensions don't match, the equation cannot represent a real physical relationship, because the numerical value would depend on the arbitrary choice of units.

3. The Buckingham Pi Theorem: Why We Can Derive Relationships

This is the deeper mathematical reason. The Buckingham Pi Theorem states:

If a physical problem involves $n$ variables and $k$ fundamental dimensions, then it can be reduced to $n-k$ independent dimensionless groups (called $\pi$ groups).

Why does this work?

Imagine you have a relationship:

f(Q_1, Q_2, \dots, Q_n) = 0

where each $Q_i$ has dimensions. Because the equation must be dimensionally homogeneous, we can rearrange it into a function of dimensionless products only:

F(\pi_1, \pi_2, \dots, \pi_{n-k}) = 0

The reasoning: Dimensions act as constraints. Each fundamental dimension (M, L, T) gives one constraint. So if you have $n$ variables and $k$ constraints, you only have $n-k$ independent dimensionless combinations.

Example: For a simple pendulum, the period $T$ depends on length $L$ , mass $m$ , and gravity $g$ . That's 4 variables with 3 dimensions (M, L, T). So $4-3 = 1$ dimensionless group: $\pi = \frac{T^2 g}{L}$ . This tells us $T \propto \sqrt{L/g}$ without solving any differential equation.

4. Why We Can Convert Units Using Dimensional Analysis

The conversion factor formula:

\text{Value in new unit} = \text{Value in old unit} \times \left(\frac{\text{old unit}}{\text{new unit}}\right)^{\text{dimension exponent}}

Why this works: …

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Significant figures count all certain digits plus the first uncertain digit. The answers are: (i) 2,

(ii) 3,

(iii) 4,

(iv) 3,

(v) 4,

(vi) 5.

The Core Idea: What Are Significant Figures?

Significant figures (or significant digits) are the digits in a number that carry meaningful information about its precision. They include all digits that are known with certainty, plus one digit that is estimated (the first uncertain digit). The rules for counting them are designed to distinguish between digits that are truly measured and digits that are merely placeholders.

The trickiest part is handling zeros. Zeros can be significant or not, depending on where they appear. The key is to ask: Is this zero actually measured, or is it just holding a decimal place?

Step-by-Step Counting

Let’s go through each number one by one.

1. (i) 0.0025

Leading zeros (zeros to the left of the first non-zero digit) are never significant. They only locate the decimal point. Here, the first non-zero digit is 2. The zeros before it are placeholders. The digits 2 and 5 are both significant.

Tip

A quick trick: write the number in scientific notation. $0.0025 = 2.5 \times 10^{-3}$ . The coefficient $2.5$ has two digits — that’s the number of significant figures.

So, 2 significant figures.

2. (ii) 208

All non-zero digits are always significant. The zero here is between two non-zero digits (2 and 8). Such “captive zeros” are always significant because they are part of the measured value — you wouldn’t write 208 if you only knew it was roughly 200.

So, 3 significant figures.

3. (iii) 5005

Again, the two zeros are captive between 5 and 5. They are significant. All four digits count.

So, 4 significant figures.

4. (iv) 126,000

This is a classic trap. The number has no decimal point. The trailing zeros (zeros at the end of a whole number) are ambiguous — they might be significant or just placeholders. By convention, without a decimal point, trailing zeros are not considered significant. Only the non-zero digits (1, 2, 6) count.

Watch out

A common mistake is to count all zeros in a number like 126,000. Without a decimal, you cannot assume those zeros were measured. If the measurement was precise to the nearest thousand, the zeros are just placeholders. If it was precise to the nearest unit, the number would be written as 126,000. (with a decimal point) to show that all zeros are significant.

So, 3 significant figures.

5. (v) 500.0 …

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Understanding Significant Figures

1. Concept First — The Idea Being Tested

Scientific Notation and significant figures are tools to express the precision of a measurement — not just its value. The core idea is:

Significant figures are the digits in a number that carry meaningful information about its precision.

Why does this matter?

In science and exams, a number like 500 could mean "exactly 500" or "roughly 500" depending on how it's written. Significant figures remove that ambiguity. The rules tell us which zeros count and which are just placeholders.

Intuition:

Leading zeros (like in 0.0025) are never significant — they only locate the decimal point.
Trailing zeros after a decimal point (like in 500.0) are always significant — they show the measurement was precise to that decimal place.
Trailing zeros without a decimal point (like in 126,000) are ambiguous — they may or may not be significant.
All non-zero digits are always significant.
Zeros between non-zero digits (like in 5005) are always significant.

2. Step-by-Step — With Reasoning

We'll apply these rules to each number.

(i) 0.0025

Digits: 0, 0, 0, 2, 5
Leading zeros (the first three zeros) are not significant — they only position the decimal.
Non-zero digits 2 and 5 are significant.
Count: 2 significant figures.

Reasoning: If we write in scientific notation: $0.0025 = 2.5 \times 10^{-3}$ . The coefficient 2.5 has two digits — that's the precision.

(ii) 208

Digits: 2, 0, 8
Non-zero digits 2 and 8 are significant.
Zero between non-zero digits (the 0) is significant — it's part of the measured value.
Count: 3 significant figures.

Reasoning: The zero is "sandwiched" — removing it would change the number to 28, which is different. So it counts.

(iii) 5005

Digits: 5, 0, 0, 5
Non-zero digits (first and last) are significant.
Zeros between non-zero digits (both zeros) are significant.
Count: 4 significant figures.

Reasoning: Same logic as 208 — the zeros are trapped between 5s, so they are part of the measurement.

(iv) 126,000

Digits: 1, 2, 6, 0, 0, 0
Non-zero digits 1, 2, 6 are significant.
Trailing zeros (the three zeros at the end) — no decimal point is shown.
- Without a decimal point, trailing zeros are ambiguous. By standard convention, they are not considered significant unless specified otherwise (e.g., by scientific notation).
Count: 3 significant figures.

Reasoning: 126,000 could mean $1.26 \times 10^5$ (3 sig figs) or $1.26000 \times 10^5$ (6 sig figs). In the absence of a decimal point, we assume the simpler case: only the non-zero digits count.

(v) 500.0

Digits: 5, 0, 0, 0
Non-zero digit 5 is significant.
Zeros after the decimal point — all three zeros are significant because the decimal point tells us the measurement was precise to the tenths place.
Count: 4 significant figures.

Reasoning: Writing 500.0 instead of 500 is a deliberate choice — it says "I measured this to the nearest 0.1 unit." Every digit after the decimal is part of that precision.

(vi) 2.0034

Digits: 2, 0, 0, 3, 4 …

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Common Mistakes in Scientific Notation (and How to Avoid Them)

Scientific notation is a compact way to write very large or very small numbers as:

a \times 10^n

where $1 \leq a < 10$ and $n$ is an integer.

Here are the most frequent errors students make, with the exact examples you gave.

Mistake 1: Misplacing the Decimal Point (Wrong $a$ )

Example with 0.0048:

✗ Wrong: $0.48 \times 10^{-2}$ (here $a = 0.48$ , which is less than 1)
✓ Correct: $4.8 \times 10^{-3}$

Why it happens: Students stop too early — they move the decimal but don't check that $a$ is between 1 and 10.

How to avoid: After writing $a \times 10^n$ , always check: is $1 \leq a < 10$ ? If $a$ is less than 1 or greater than or equal to 10, you're not done.

Mistake 2: Wrong Sign of the Exponent

Example with 0.0048:

✗ Wrong: $4.8 \times 10^{3}$ (positive exponent for a small number)
✓ Correct: $4.8 \times 10^{-3}$

Why it happens: Confusing "number of places moved" with "direction." Moving the decimal to the right (for numbers < 1) gives a negative exponent.

How to avoid: Use this rule:

Small number (less than 1) → negative exponent
Large number (greater than 10) → positive exponent

Mistake 3: Counting Trailing Zeros Incorrectly

Example with 234,000:

✗ Wrong: $2.34 \times 10^{5}$ (counted 5 places, but it's actually 5)
✗ Wrong: $2.34 \times 10^{4}$ (counted only 4 places)
✓ Correct: $2.34 \times 10^{5}$

Why it happens: The comma in 234,000 confuses the count. The decimal is after the last zero: 234,000. → move to between 2 and 3 → that's 5 places left.

How to avoid: Write the number without commas first: 234000. Then count the jumps from the original decimal position to the new one.

Mistake 4: Forgetting That Trailing Zeros After a Decimal Matter

Example with 500.0:

✗ Wrong: $5 \times 10^{2}$ (loses the precision of the trailing zero)
✓ Correct: $5.000 \times 10^{2}$ (or $5.0 \times 10^{2}$ )

Why it happens: Students think "500.0 is just 500" — but in scientific notation, the digits after the decimal show the precision of the measurement.

How to avoid: Keep all significant digits from the original number. If the original has 500.0 (4 significant figures), your $a$ must have 4 digits: $5.000$ .

Mistake 5: Forgetting That 8008 Already Has a Decimal

Example with 8008:

✗ Wrong: $8.008 \times 10^{4}$ (moved 4 places instead of 3)
✓ Correct: $8.008 \times 10^{3}$ …

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Q.How many significant figures are present in the following?

The Core Idea: What Are Significant Figures?

Step-by-Step Counting

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