The length, breadth and thickness of a rectangular sheet of metal are $4.234\ \text{m}$, $1.005\ \text{m}$, and $2.01\ \text{cm}$ respectively. Give the area and volume of the sheet to correct significant figures.

The length, breadth and thickness of a rectangular sheet of metal are 4.234m, 1.005m, and 2.01cm respectively.

Exercises · 1.11

4.234\ \text{m}

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Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

Imagine you're measuring the length of a table with a ruler that has marks every millimeter. You see the table edge falls somewhere between 152.3 cm and 152.4 cm. You estimate it as 152.35 cm. But here's the truth: you're certain about 152.3, pretty sure about the 0.05, and guessing about anything beyond that. Significant figures are simply a way to communicate how much of that number you actually know.

The Core Idea

Every measurement has uncertainty. Significant figures (or "sig figs") are the digits in a number that carry meaningful information about its precision. They include all the digits you're sure of, plus one more that you estimate.

Note

A digit is "significant" if removing it would change the precision of the measurement. Zeros can be tricky — they might just be placeholders.

The Rules (Memorize These)

1. Non-zero digits are always significant

$123.45$ has 5 sig figs. Simple.

2. Zeros between non-zero digits are significant

$1002$ has 4 sig figs. The zeros are "sandwiched" — they're part of the measurement.

3. Leading zeros are never significant

$0.00123$ has 3 sig figs. Those zeros just tell you where the decimal point is.

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

$1200$ has 2 sig figs (no decimal — zeros are placeholders)
$1200.$ has 4 sig figs (decimal tells us those zeros were measured)
$1200.0$ has 5 sig figs

5. Exact numbers have infinite sig figs

If you count 5 apples, that's exactly 5 — no uncertainty. Conversion factors like $1 \text{ m} = 100 \text{ cm}$ are exact by definition.

Tip

When in doubt, write the number in scientific notation. $1.20 \times 10^3$ clearly has 3 sig figs, while $1.2 \times 10^3$ has 2.

Why This Matters: Calculations

When you multiply or add measurements, the uncertainty propagates. You can't claim more precision than your least precise measurement.

Multiplication and Division

The result should have the same number of sig figs as the measurement with the fewest sig figs.

$3.14 \times 2.5 = 7.85$ but you report $7.9$ (2 sig figs, because 2.5 has only 2)

Addition and Subtraction

The result should have the same decimal places as the measurement with the fewest decimal places.

$12.11 + 18.0 = 30.11$ but you report $30.1$ (one decimal place, because 18.0 has one) …

Why this formula?

Significant Figures: Why the Rules Work

Let’s start with the core idea: significant figures (sig figs) are a way to honestly report how precise a measurement is. The rules for addition/subtraction and multiplication/division aren’t arbitrary — they come directly from how uncertainty propagates through calculations.

1. The Fundamental Idea: Uncertainty is the Key

Every measurement has an uncertainty (error). When we say a length is 12.3 cm, we mean:

The true value lies somewhere between 12.25 cm and 12.35 cm (assuming ±0.05 cm uncertainty).
The last digit (3) is uncertain; the digits before it (1 and 2) are certain.

Why this matters: When we combine measurements, the uncertainty in the result depends on the uncertainties of the inputs. Sig fig rules are a shortcut for this uncertainty propagation.

2. Rule for Addition and Subtraction

Statement: The result should have the same number of decimal places as the measurement with the fewest decimal places.

Example:

$12.3 + 4.56 = 16.86$ → round to 16.9 (one decimal place, like 12.3)

Why this holds

Consider two measurements:

$A = 12.3 \pm 0.05$ (uncertainty in the tenths place)
$B = 4.56 \pm 0.005$ (uncertainty in the hundredths place)

When we add:

Certain digits: 12.3 has certainty up to the tenths place. 4.56 has certainty up to the hundredths place.
The weaker link: The tenths place of A is uncertain. So in the sum, the hundredths place (from B) is meaningless — because we don’t even know the tenths place of A exactly.

Mathematically, the absolute uncertainty in the sum is:

\Delta(A+B) = \sqrt{(\Delta A)^2 + (\Delta B)^2} \approx \sqrt{0.05^2 + 0.005^2} \approx 0.0502

This uncertainty is ~0.05, which affects the tenths place. So reporting the hundredths place is false precision.

Key takeaway: The result’s last significant digit is in the same decimal place as the least precise measurement’s last digit.

3. Rule for Multiplication and Division

Statement: The result should have the same number of significant figures as the measurement with the fewest significant figures.

Example:

$12.3 \times 4.56 = 56.088$ → round to 56.1 (three sig figs, like both inputs)

Why this holds

Let’s use relative uncertainty (percentage error):

$A = 12.3 \pm 0.05$ → relative uncertainty = $\frac{0.05}{12.3} \approx 0.00407$ (0.407%)
$B = 4.56 \pm 0.005$ → relative uncertainty = $\frac{0.005}{4.56} \approx 0.00110$ (0.110%)

For multiplication, relative uncertainties add (approximately):

\frac{\Delta(A \times B)}{A \times B} \approx \sqrt{\left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2}

Plugging in:

\approx \sqrt{0.00407^2 + 0.00110^2} \approx 0.00422 \text{ (0.422\%)}

Now, the absolute uncertainty in the product:

\Delta(A \times B) \approx 0.00422 \times (12.3 \times 4.56) \approx 0.00422 \times 56.088 \approx 0.237

This uncertainty (~0.2) affects the tenths place of the result. So the result 56.088 has uncertainty in the first decimal — meaning only three digits (5, 6, and the uncertain 1) are meaningful. That’s three sig figs, matching the input with fewer sig figs (both have three here).

Key takeaway: The number of sig figs in the result is limited by the least precise measurement’s number of sig figs, because relative uncertainty is dominated by the measurement with the largest relative error.

4. Why These Rules Are Different …

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The sheet is a thin rectangular slab, so its "area" means the total surface area of all six faces — length, breadth, and thickness all contribute. Using $A = 2(lb + bt + tl)$ and $V = lbt$ , and rounding each to the significant figures set by the least precise measurement (thickness, with 3 significant figures), the total surface area is $\boxed{8.72\ \text{m}^2}$ and the volume is $\boxed{0.0855\ \text{m}^3}$ .

Setting up

The sheet has three given dimensions:

Length $l = 4.234\ \text{m}$
Breadth $b = 1.005\ \text{m}$
Thickness $t = 2.01\ \text{cm} = 0.0201\ \text{m}$

Because a thickness is given, this is not a flat two-dimensional rectangle — it is a thin rectangular slab (a cuboid) with six faces: two of size $l\times b$ , two of size $b\times t$ , and two of size $t\times l$ . "The area of the sheet" therefore means the total surface area of the slab, not just the area of its largest face. If only $l \times b$ were wanted, the thickness would never have been given at all.

Counting significant figures

$l = 4.234\ \text{m}$ → 4 significant figures
$b = 1.005\ \text{m}$ → 4 significant figures
$t = 0.0201\ \text{m}$ → 3 significant figures (leading zeros don't count; 2, 0, 1 do)

The least precise measurement is the thickness, with 3 significant figures. Since thickness enters both the area and volume calculations, both final answers are limited to 3 significant figures.

Total surface area

A = 2(lb + bt + tl)

$lb = 4.234 \times 1.005 = 4.25517\ \text{m}^2$
$bt = 1.005 \times 0.0201 = 0.0202005\ \text{m}^2$
$tl = 0.0201 \times 4.234 = 0.0851034\ \text{m}^2$
Sum: $4.25517 + 0.0202005 + 0.0851034 = 4.3604839\ \text{m}^2$
$A = 2 \times 4.3604839 = 8.7209678\ \text{m}^2$
Round to 3 significant figures: $A = 8.72\ \text{m}^2$

Volume …

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Method: Total Surface Area of a Thin Slab + Significant Figures

Method Name: Because length, breadth, and thickness are all given, the sheet is treated as a thin rectangular slab (a cuboid), not a flat 2-D rectangle. Its "area" means the total surface area of all six faces, $A = 2(lb+bt+tl)$ , and its volume is $V = lbt$ . Both results are then rounded using the Rule of Least Precise Measurement: a product carries only as many significant figures as its least precise factor.

Step 1: Identify significant figures in each given value

Length $l = 4.234\ \text{m}$ → 4 significant figures
Breadth $b = 1.005\ \text{m}$ → 4 significant figures
Thickness $t = 2.01\ \text{cm}$ → 3 significant figures

⚠️ Important: Thickness is in cm, while length and breadth are in m. Convert to the same unit before calculating.

Step 2: Convert thickness to metres

t = 2.01\ \text{cm} = 2.01 \times 10^{-2}\ \text{m} = 0.0201\ \text{m}

$t$ still has 3 significant figures.

Step 3: Calculate the total surface area

Because thickness is given, the sheet is a slab with six faces — two of each pair $(l,b)$ , $(b,t)$ , $(t,l)$ :

A = 2(lb + bt + tl)

lb = 4.234 \times 1.005 = 4.25517\ \text{m}^2

bt = 1.005 \times 0.0201 = 0.0202005\ \text{m}^2

tl = 0.0201 \times 4.234 = 0.0851034\ \text{m}^2

A = 2(4.25517 + 0.0202005 + 0.0851034) = 2 \times 4.3604839 = 8.7209678\ \text{m}^2

Apply the significant figure rule: the least number of significant figures among $l$ , $b$ , $t$ is 3 (from $t$ ), so round $A$ to 3 significant figures:

\boxed{A = 8.72\ \text{m}^2}

Step 4: Calculate the volume …

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Here are the most common mistakes students make on this classic significant figures problem, along with the reasoning to avoid each.

Mistake 1: Forgetting to convert units before adding

The Error:

Students directly multiply $4.234 \times 1.005 \times 2.01$ without noticing that thickness is in cm while length and breadth are in m. This gives a wildly wrong volume.

How to Avoid:

Always check units first. Write them down beside each value.
Convert everything to the same unit before any calculation.
Here: $2.01\ \text{cm} = 0.0201\ \text{m}$ .

Mistake 2: Using the wrong rule for multiplication/division

The Error:

Students apply the addition/subtraction rule (look at decimal places) to multiplication.

How to Avoid:

Multiplication/Division: Round to the least number of significant figures, not decimal places.

Mistake 3: Counting significant figures incorrectly in the thickness

The Error:

Thinking $2.01\ \text{cm}$ has 2 significant figures (because of the leading digit '2') or 4 significant figures (because of the trailing '01').

How to Avoid:

Captive zeros between non-zero digits are significant. So $2.01$ has 3 significant figures (2, 0, 1).

Mistake 4: Rounding intermediate results too early

The Error:

Rounding an intermediate product before finishing the calculation, which accumulates rounding error.

How to Avoid:

Do the full calculation first with all digits, and round only the final answer.

Mistake 5: Computing only length × breadth and calling it "the area"

The Error:

Students multiply just the length and breadth, $4.234 \times 1.005 = 4.255\ \text{m}^2$ , and report that as "the area of the sheet."

Why it's wrong:

A thickness is explicitly given — $2.01\ \text{cm}$ — which means this is not a flat rectangle but a thin rectangular slab (a cuboid) with six faces. If "area" only meant $l \times b$ , the thickness would be completely irrelevant to the area calculation, and the question would never have given it. "The area of the sheet" here means the total surface area of the slab:

A = 2(lb + bt + tl)

How to Avoid:

Whenever a thickness (or any third dimension) is given alongside length and breadth for a physical "sheet" or "slab," compute the total surface area, not just one face.
Compute all three face-pair products ( $lb$ , $bt$ , $tl$ ), sum them, and double the sum:

A = 2(4.234\times1.005 + 1.005\times0.0201 + 0.0201\times4.234) = 2(4.25517+0.0202005+0.0851034) \approx 8.72\ \text{m}^2\ (\text{3 s.f.})

Mistake 6: Reporting area or volume with too many or too few significant figures

The Error:

Giving area as $8.7209678\ \text{m}^2$ (all raw digits) or keeping 4 significant figures because length and breadth each have 4.

Why it happens:

Not identifying which measurement has the least significant figures.

How to avoid:

Identify the limiting factor: …

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Q.The length, breadth and thickness of a rectangular sheet of metal are 4.234 m4.234\ \text{m}4.234 m, 1.005 m1.005\ \text{m}1.005 m, and 2.01 cm2.01\ \text{cm}2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.

Setting up

Counting significant figures

Total surface area

Volume …

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Q.The length, breadth and thickness of a rectangular sheet of metal are $4.234\ \text{m}$ , $1.005\ \text{m}$ , and $2.01\ \text{cm}$ respectively. Give the area and volume of the sheet to correct significant figures.

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