The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in

NCERT Exemplar · Q5

Q. The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is

(a)

164 \pm 3

^2

(b)

163.62 \pm 2.6

^2

(c)

163.6 \pm 2.6

^2

(d)

163.62 \pm 3

^2

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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 area works out to $163.62\ \text{cm}^2$ , and propagating the least-count error gives $\pm 3\ \text{cm}^2$ ; rounded consistently, the reported result is $164 \pm 3\ \text{cm}^2$ — option (A).

Setting up

Length $l = 16.2\ \text{cm}$ and breadth $b = 10.1\ \text{cm}$ , each measured to the nearest $0.1\ \text{cm}$ — so each has a least count of $0.1\ \text{cm}$ , and we take the absolute error as $\Delta l = \Delta b = 0.1\ \text{cm}$ .

Computing the area

A = l \times b = 16.2 \times 10.1 = 163.62\ \text{cm}^2

Propagating the error

For a product $A = l \times b$ , relative (fractional) errors add:

\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}

\frac{\Delta l}{l} = \frac{0.1}{16.2} \approx 0.00617, \qquad \frac{\Delta b}{b} = \frac{0.1}{10.1} \approx 0.00990

\frac{\Delta A}{A} \approx 0.00617 + 0.00990 = 0.01607

\Delta A \approx 163.62 \times 0.01607 \approx 2.63\ \text{cm}^2

Rounded to one significant figure (errors are conventionally quoted to 1 sig fig): $\Delta A \approx 3\ \text{cm}^2$ .

Reporting the final result …

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Method: Significant Figures in Multiplication with Error Propagation

We use the rule for multiplication with significant figures combined with maximum possible error (or absolute error) propagation.

Step 1: Compute the area (nominal value)

Given:

Length $l = 16.2$ cm (3 significant figures)
Breadth $b = 10.1$ cm (3 significant figures)

Area:

A = l \times b = 16.2 \times 10.1 = 163.62 \ \text{cm}^2

Since both measurements have 3 significant figures, the product should also be reported to 3 significant figures:

A = 164 \ \text{cm}^2 \quad (\text{rounded to 3 s.f.})

Step 2: Find the absolute error in each measurement

Each measurement is given to 1 decimal place, so the least count (smallest division) is $0.1$ cm.

Absolute error in each:

\Delta l = \Delta b = 0.05 \ \text{cm} \quad (\text{half of the least count})

Step 3: Compute the relative error in area

For multiplication:

\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}

Substitute:

\frac{\Delta A}{A} = \frac{0.05}{16.2} + \frac{0.05}{10.1}

Calculate each:

$\frac{0.05}{16.2} \approx 0.003086$
$\frac{0.05}{10.1} \approx 0.004950$

Sum:

\frac{\Delta A}{A} \approx 0.008036

Step 4: Compute absolute error in area

\Delta A = A \times 0.008036 \approx 163.62 \times 0.008036 \approx 1.315

Round to 1 significant figure in the error (standard practice):

\Delta A \approx 1 \ \text{cm}^2 \quad \text{or} \quad \Delta A \approx 1.3 \ \text{cm}^2

But the options show $\pm 3$ or $\pm 2.6$ . Let’s check the maximum possible error method (more common in Indian exams):

Step 5: Maximum possible error (alternative approach)

Using:

\Delta A = (\Delta l) \times b + (\Delta b) \times l

Substitute: …

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Common Mistakes in Significant Figures & Error Calculation (Area of Rectangle)

Students often lose marks on this exact type of problem. Here are the most frequent errors and how to avoid each.

Mistake 1: Reporting too many decimal places in the area

The error:

Students compute $16.2 \times 10.1 = 163.62$ and leave it as is — but the least precise measurement (16.2 or 10.1) has only 3 significant figures. The area must be rounded to 3 significant figures.

Correct approach:

Area = $16.2 \times 10.1 = 163.62$
Round to 3 significant figures → $164$ cm²
This eliminates options (B), (C), and (D) immediately.

Rule: In multiplication/division, the result has the same number of significant figures as the factor with the fewest significant figures.

Mistake 2: Confusing absolute error with relative error

The error:

Students compute $\Delta A = \Delta l + \Delta b$ (adding absolute errors directly) and get $0.1 + 0.1 = 0.2$ , then write $163.62 \pm 0.2$ — which is wrong for multiplication.

Correct approach:

For multiplication, use relative errors:

\frac{\Delta A}{A} = \frac{\Delta l}{l} + \frac{\Delta b}{b}

Here:

$\Delta l = 0.1$ cm (since 16.2 implies uncertainty of $\pm 0.1$ )
$\Delta b = 0.1$ cm
$l = 16.2$ , $b = 10.1$

So:

\frac{\Delta A}{A} = \frac{0.1}{16.2} + \frac{0.1}{10.1} \approx 0.00617 + 0.00990 = 0.01607

Then:

\Delta A = A \times 0.01607 = 163.62 \times 0.01607 \approx 2.63 \approx 3 \text{ cm}^2

Key: For multiplication, relative errors add, not absolute errors.

Mistake 3: Reporting error with too many decimal places

The error:

Writing $\Delta A = 2.63$ or $2.6$ — but the area itself is rounded to 3 significant figures ( $164$ ), so the error must be rounded to 1 significant figure (or match the precision of the area).

Correct approach:

$\Delta A \approx 2.63$ rounds to $3$ cm² (1 significant figure)
Final answer: $164 \pm 3$ cm² → Option (A)

Rule: Error is reported to 1 significant figure (unless the leading digit is 1, in which case 2 sig figs may be used — but here 2.63 → 3 is standard).

Mistake 4: Forgetting to round the area before computing error

The error:

Using $A = 163.62$ to compute $\Delta A$ , then rounding both separately — but the area must be rounded to match the significant figures of the inputs.

Correct approach: …

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Same / Similar Concept — real previous-year questions on the same or a closely similar concept, not this exact question.

Showing the 12 most recent of 20 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. How many significant figures does 0.057 have?$
(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. …
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CBSE 2026Set ANNUAL1 markMCQ
$Q. How many significant figures are there in 0.0052?$
(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.

…

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CBSE 2026Set ANNUAL1 markMCQ
$Q. If π = 3.14, then the value of π^2 is:$
(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: …

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CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The number of significant figures in 2.005 is ______.$
›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.

…

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CBSE 2026Set ANN1 mark
$Q. Find the number of significant figures in the measurement, 0.04597 g.$
›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. …

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CBSE 2025Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005 are:$
(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. …
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CBSE 2024Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005 is:$
(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. …
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CBSE 2024Set ANNUAL1 mark
$Q. Number of significant figures in 0.00300 is ____.$
›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 …

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CBSE 2023Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005:$
(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. …
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CBSE 2023Set ANNUAL1 markMCQ
$Q. Round off the number 19.95 into three significant figures.$
(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).

…

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CBSE 2023Set ANNUAL1 mark
$Q. 5.74 g of a substance occupies 1.2 cm^3. Express its density in proper significant figures.$
›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.

…

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CBSE 2023Set ANNUAL1 markMCQ
$Q. If radius of circle is 2.12 cm, then express its area must be:$
(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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Q.The length and breadth of a rectangular sheet are 16.2 cm and 10.1 cm, respectively. The area of the sheet in appropriate significant figures and error is

Setting up

Computing the area

Propagating the error

Reporting the final result …

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