Q.The length, breadth and thickness of a rectangular sheet of metal are 4.234 m, 1.005 m, and 2.01 cm respectively. Give the area and volume of the sheet to correct significant figures.
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🔒 Start your 14-day free trial to unlock the full solution →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.
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 m=100 cm are exact by definition.
When in doubt, write the number in scientific notation. 1.20×103 clearly has 3 sig figs, while 1.2×103 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.
3.14×2.5=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±0.05 (uncertainty in the tenths place)
- B=4.56±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:
Δ(A+B)=(ΔA)2+(ΔB)2≈0.052+0.0052≈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:
12.3×4.56=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±0.05 → relative uncertainty = 12.30.05≈0.00407 (0.407%)
- B=4.56±0.005 → relative uncertainty = 4.560.005≈0.00110 (0.110%)
For multiplication, relative uncertainties add (approximately):
A×BΔ(A×B)≈(AΔA)2+(BΔB)2
Plugging in:
≈0.004072+0.001102≈0.00422 (0.422%)
Now, the absolute uncertainty in the product:
Δ(A×B)≈0.00422×(12.3×4.56)≈0.00422×56.088≈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 …
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 cm=0.0201 m (3 s.f.). Length =4.234 m (4 s.f.), breadth =1.005 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. …
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 8.72 m2 and the volume is 0.0855 m3.
Setting up
The sheet has three given dimensions:
- Length l=4.234 m
- Breadth b=1.005 m
- Thickness t=2.01 cm=0.0201 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×b, two of size b×t, and two of size t×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×b were wanted, the thickness would never have been given at all.
Counting significant figures
- l=4.234 m → 4 significant figures
- b=1.005 m → 4 significant figures
- t=0.0201 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=4.234×1.005=4.25517 m2
- bt=1.005×0.0201=0.0202005 m2
- tl=0.0201×4.234=0.0851034 m2
- Sum: 4.25517+0.0202005+0.0851034=4.3604839 m2
- A=2×4.3604839=8.7209678 m2
- Round to 3 significant figures: A=8.72 m2
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 m → 4 significant figures
- Breadth b=1.005 m → 4 significant figures
- Thickness t=2.01 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 cm=2.01×10−2 m=0.0201 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=4.234×1.005=4.25517 m2
bt=1.005×0.0201=0.0202005 m2
tl=0.0201×4.234=0.0851034 m2
A=2(4.25517+0.0202005+0.0851034)=2×4.3604839=8.7209678 m2
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:
A=8.72 m2
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 4.234×1.005×2.01 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 cm=0.0201 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 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, 4.234×1.005=4.255 m2, and report that as "the area of the sheet."
Why it's wrong:
A thickness is explicitly given — 2.01 cm — which means this is not a flat rectangle but a thin rectangular slab (a cuboid) with six faces. If "area" only meant l×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(4.234×1.005+1.005×0.0201+0.0201×4.234)=2(4.25517+0.0202005+0.0851034)≈8.72 m2 (3 s.f.)
Mistake 6: Reporting area or volume with too many or too few significant figures
The Error:
Giving area as 8.7209678 m2 (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: …
Showing the 12 most recent of 20 on this concept.
- CBSE 2026Set ANNUAL1 markMCQQ.How many significant figures does 0.057 have?(a) 2(b) 4(c) 3(d) 0
›Reveal solutionSolution
Leading zeros used only to fix the decimal point are never significant. 0.057 has exactly 2 significant figures: 5 and 7.
Rule for counting significant figures:
- All non-zero digits are significant.
- Zeros between two non-zero digits are significant.
- Leading zeros (to the left of the first non-zero digit) are NOT significant -- they only locate the decimal point.
- Trailing zeros after a decimal point ARE significant. …
- CBSE 2026Set ANNUAL1 markMCQQ.How many significant figures are there in 0.0052?(a) 4(b) 1(c) 3(d) 2
›Reveal solutionSolution
0.0052 has exactly 2 significant figures (5 and 2); leading zeros are not counted.
Significant figures are the meaningful digits in a measured or calculated quantity - they indicate the precision of a measurement. The rules for counting them: (1) all non-zero digits are significant; (2) zeros between non-zero digits are significant; (3) leading zeros (to the left of the first non-zero digit) are NEVER significant - they only serve to locate the decimal point; (4) trailing zeros after a decimal point ARE significant.
…
- CBSE 2026Set ANNUAL1 markMCQQ.If π = 3.14, then the value of π^2 is:(a) 9.860(b) 9.9(c) 9.86(d) 9.8596
›Reveal solutionSolution
By the rules of significant figures, a result cannot have more significant figures than the least precise value used to calculate it; since pi = 3.14 has 3 significant figures, pi^2 must also be reported to 3 significant figures, i.e., 9.86.
When multiplying or dividing measured quantities, the result should be rounded off to the same number of significant figures as the quantity with the fewest significant figures used in the calculation. This is because a calculation cannot manufacture precision beyond what the original measurement actually had.
Here, pi is given as 3.14, which has 3 significant figures.
pi^2 = 3.14 x 3.14 = 9.8596 (this is the raw arithmetic result, but it has 5 significant figures, more precision than 3.14 actually supports)
Rounding 9.8596 to 3 significant figures: …
- CBSE 2026Set ANNUAL1 markQ.Fill in the blank: The number of significant figures in 2.005 is ______.
›Reveal solutionSolution
2.005 has 4 significant figures because zeros sandwiched between non-zero digits always count as significant.
Rules for significant figures: (1) all non-zero digits are significant; (2) zeros between two non-zero digits (captive/sandwiched zeros) are significant; (3) leading zeros are not significant; (4) trailing zeros after a decimal point are significant.
…
- CBSE 2026Set ANN1 markQ.Find the number of significant figures in the measurement, 0.04597 g.
›Reveal solutionSolution
Leading zeros are not significant; the digits 4, 5, 9, 7 count, giving 4 significant figures.
Rule: zeros to the left of the first non-zero digit (leading zeros) are NOT significant; they only locate the decimal point. All non-zero digits are significant. …
- CBSE 2025Set ANNUAL1 markMCQQ.Number of significant figures in 2.005 are:(a) 1(b) 2(c) 3(d) 4
›Reveal solutionSolution
Every digit in 2.005 is significant because zeros sandwiched between non-zero digits always count.
Rules for counting significant figures:
- All non-zero digits are significant.
- Zeros between two non-zero digits are significant ("captive zeros").
- Leading zeros (before the first non-zero digit) are NOT significant.
- Trailing zeros after a decimal point ARE significant.
In 2.005: the digits are 2, 0, 0, 5.
- '2' is a non-zero digit — significant. …
- CBSE 2024Set ANNUAL1 markMCQQ.Number of significant figures in 2.005 is:(a) Two(b) Three(c) Four(d) Infinite
›Reveal solutionSolution
All four digits in 2.005 are significant because the zeros sit between non-zero digits.
Rules for counting significant figures:
- All non-zero digits are significant: the digits 2 and 5 count. …
- CBSE 2024Set ANNUAL1 markQ.Number of significant figures in 0.00300 is ____.
›Reveal solutionSolution
0.00300 has 3 significant figures: the leading zeros used only to fix the decimal point are not counted, but trailing zeros after a decimal point ARE significant.
Rules for significant figures: leading zeros (0.00...) before the first non-zero digit are never significant -- they only locate the decimal point. Zeros appearing after the decimal point AND after a non-zero digit ARE significant, since they indicate …
- CBSE 2023Set ANNUAL1 markMCQQ.Number of significant figures in 2.005:(a) 2(b) 3(c) 4(d) Infinite
›Reveal solutionSolution
Captive zeros (zeros sandwiched between non-zero digits) always count as significant, so 2.005 has 4 significant figures.
Rules used:
- All non-zero digits (2, 5) are significant.
- Zeros between two non-zero digits (captive zeros) are significant. …
- CBSE 2023Set ANNUAL1 markMCQQ.Round off the number 19.95 into three significant figures.(a) 20.1(b) 19.9(c) 19.5(d) 20.0
›Reveal solutionSolution
Rounding 19.95 to three significant figures using the 'round half to even' convention gives 20.0.
19.95 has four significant figures (1, 9, 9, 5). To round to three significant figures, we must decide the fate of the third significant digit (the second 9), based on the digit being dropped (a 5, with nothing after it).
Rule for rounding off when the digit to be dropped is exactly 5: if the preceding digit is even, it is left unchanged; if the preceding digit is odd, it is increased by 1 (this is the 'round half to even' convention used for significant figures, and it avoids a systematic upward bias).
…
- CBSE 2023Set ANNUAL1 markQ.5.74 g of a substance occupies 1.2 cm^3. Express its density in proper significant figures.
›Reveal solutionSolution
Density = 5.74/1.2 = 4.7833... g/cm^3, rounded to 2 significant figures = 4.8 g/cm^3.
Rule for significant figures in division/multiplication: the result must be reported with the same number of significant figures as the input quantity that has the FEWEST significant figures. This keeps the calculated result from claiming more precision than the actual measurements support.
Step 1: Count significant figures in each given quantity.
Mass = 5.74 g has 3 significant figures.
Volume = 1.2 cm^3 has 2 significant figures.
Step 2: Compute the raw quotient.
Density = mass/volume = 5.74 / 1.2 = 4.7833... g/cm^3.
…
- CBSE 2023Set ANNUAL1 markMCQQ.If radius of circle is 2.12 cm, then express its area must be:(a) 14 cm^2(b) 14.1 cm^2(c) 14.11 cm^2(d) 14.1124 cm^2
›Reveal solutionSolution
Rounded to the correct significant figures, the area is 14.1 cm^2.
Area A = πr^2 = 3.14159 × (2.12)^2 = 3.14159 × 4.4944 = 14.1197 cm^2.
…
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