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.

Concept: Significant Figures Calculation — Because thickness is given along with length and breadth, this is a thin rectangular slab, so its "area" means the total surface area of all six faces, $A = 2(lb+bt+tl)$, and its volume is $V = lbt$. Both must be rounded to the least number of significant figures among the three measurements.

Step 1: Convert to the same unit.

Thickness $= 2.01\ \text{cm} = 0.0201\ \text{m}$ (3 s.f.). Length $= 4.234\ \text{m}$ (4 s.f.), breadth $= 1.005\ \text{m}$ (4 s.f.). The least is 3 s.f. (from thickness), so both final answers are limited to 3 significant figures.

Step 2: Compute total surface area. …

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

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

Volume …

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

$t$ 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 $(l,b)$, $(b,t)$, $(t,l)$:

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:

Step 4: Calculate the volume …

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:

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:

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: …