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

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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

(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.

KEAM 2025Set pha-2025-0424A4 marksMCQ
$Q. What is the value of the sum of 12.1100 +18.0 +1.012 as per the rule of significant figures? (A) 31.1220 (B) 31.122 (C) 31.12 (D) 31.1 (E) 31$
›Reveal solutionSolution

By the significant-figure rule for addition, $12.1100+18.0+1.012=31.1$ .

Arithmetic sum: $12.1100+18.0+1.012=31.122$ .

For addition/subtraction, the result is rounded to the least number of decimal places among the terms. Here $18.0$ has only one …
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KEAM 2024Set eng-2024-06074 marksMCQ
$1.1\times10^{-2}$
›Reveal solutionSolution

The cube's volume computes to $1.331\times10^{-6}$ , rounded to 2 significant figures $=1.3\times10^{-6}\,m^3$ .

Volume of a cube is $V=a^3$ :

$V=(1.1\times10^{-2})^3=1.331\times10^{-6}\,m^3.$ …
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KEAM 2024Set pha-2024-06104 marksMCQ
$Q. The number of significant figures in 0.0500L is (A) one (B) two (C) three (D) four (E) five$
›Reveal solutionSolution

Leading zeros are not significant; trailing zeros after a decimal are.

In $0.0500\text{ L}$ : the leading zeros (before the 5) are not significant, but the 5 and the two trailing zeros after the decimal point are significant. That gives three …
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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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