The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give (a) 2.75 and 2.74 (b) 2.74 and

NCERT Exemplar · Q4

Q. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

(a) 2.75 and 2.74

(b) 2.74 and 2.73

(d) 2.74 and 2.74

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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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Both 2.745 and 2.735 round to 2.74 when the standard round-half-to-even rule (used by NCERT for a dropped digit of exactly 5) is applied — the answer is option (D).

The rounding rule for a dropped digit of exactly 5

When the digit right after the last figure you keep is exactly 5 with nothing beyond it, always rounding up introduces a small upward bias over many numbers. NCERT's Class-11 Physics textbook prescribes the round-half-to-even ("banker's rounding") convention for this exact situation: look at the last retained digit — if it is already even, leave it as is; if it is odd, round it up to make it even.

(This applies only when the dropped digit is exactly 5 with no further non-zero digits after it — an ordinary digit $\geq 5$ followed by more non-zero digits still always rounds up.)

Rounding 2.745 to 3 significant figures

Keeping 3 significant figures means keeping $2.74$ and dropping the trailing $5$ . The last retained digit is $4$ , which is already even, so it stays unchanged:

2.745 \to 2.74

Rounding 2.735 to 3 significant figures …

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Here are the most common mistakes students make when rounding 2.745 and 2.735 to 3 significant figures, along with how to avoid each.

Mistake 1: Forgetting the "Rule for 5" (Rounding to the Nearest Even)

The Error:

Many students think that when the digit to be dropped is exactly 5, you always round up.

They round 2.745 → 2.75 (correct)
But they also round 2.735 → 2.74 (incorrect here — see below)

Why it’s wrong:

The standard convention (especially in Indian exams like JEE, NEET, and CBSE) is:

If the digit to be dropped is exactly 5, round to the nearest even digit.

2.745: The digit before 5 is 4 (even). Dropping the 5 keeps it even → 2.74 (not 2.75).
2.735: The digit before 5 is 3 (odd). Dropping the 5 rounds up to make it even → 2.74.

How to Avoid:

Memorise: "Round to even when it's exactly 5."
Practice with a few examples:
- 3.45 → 3.4 (4 is even)
- 3.35 → 3.4 (3 is odd, so round up to 4)

Mistake 2: Counting Significant Figures Incorrectly

The Error:

Students sometimes think 2.745 has 4 significant figures and try to round to 3 by looking at the wrong digit.

Why it’s wrong:

2.745 has 4 significant figures (2, 7, 4, 5).
To round to 3 significant figures, you keep the first three digits (2, 7, 4) and look at the fourth digit (5) to decide.

How to Avoid:

Always count from the first non-zero digit from the left.
For 2.745: digits are 2, 7, 4, 5 → 4 sig figs.
For 2.735: digits are 2, 7, 3, 5 → 4 sig figs.
The last digit you keep is the 3rd sig fig. The next digit tells you what to do.

Mistake 3: Applying the "Round Up for 5" Rule Blindly

The Error:

Students apply a blanket rule: "If the next digit is 5 or more, round up."

This gives: 2.745 → 2.75 and 2.735 → 2.74.

Why it’s wrong:

This rule works for most cases, but not when the digit to be dropped is exactly 5 with no non-zero digits after it. The "round to even" rule overrides it.

How to Avoid:

When the digit to be dropped is exactly 5 (and nothing after it), use the even-digit rule.
If there are non-zero digits after the 5 (e.g., 2.7451), then you always round up because it's more than halfway.

Mistake 4: Confusing "3 Significant Figures" with "3 Decimal Places"

The Error:

Students think rounding to 3 significant figures means keeping 3 digits after the decimal point.

Why it’s wrong: …

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

KCET 2025Set D-41 markMCQ
$Q. Select the INCORRECT statement/s from the following:$
(a) 22 books have infinite significant figures
(b) In the answer of calculation $2.5 \times 1.25$ has four significant figures
(c) Zero's preceding to first non-zero digit are significant
(d) In the answer of calculation $12.11 + 18.0 + 1.012$ has three significant figures (A) b, c and d only (B) b and c only (C) b and d only (D) a and b only
›Reveal solutionSolution
Evaluate each of the four statements against the significant-figure rules (exact numbers, multiplication rule, leading zeros, addition rule) and collect the false ones.

Statement (a): "22 books have infinite significant figures."

"22 books" is an exact counted number, not a measurement. Counting is not subject to measurement uncertainty — there are precisely $22$ , not $22 \pm 0.5$ . Exact numbers (and defined constants such as $1\ \mathrm{km} = 1000\ \mathrm{m}$ ) are treated as having infinite significant figures, so they never limit the precision of a calculation.

$\Rightarrow$ (a) is CORRECT.

Statement (b): "In the answer of $2.5 \times 1.25$ there are four significant figures."

The multiplication/division rule: the result carries as many significant figures as the factor with the fewest.

$2.5$ has 2 significant figures.

$1.25$ has 3 significant figures.

The raw product is
$2.5 \times 1.25 = 3.125$
but it must be rounded to the smaller count, 2 significant figures:
$\Rightarrow 3.1$
The claim of four significant figures is wrong (it just reports every digit the calculator shows).

$\Rightarrow$ (b) is INCORRECT.

Statement (c): "Zeros preceding the first non-zero digit are significant."

Leading zeros are never significant — they are placeholders that merely fix the decimal point. For example $0.0025$ has only 2 significant figures ( $2$ and $5$ ); writing it as $2.5 \times 10^{-3}$ makes this obvious, since the leading zeros vanish entirely in scientific notation.

$\Rightarrow$ (c) is INCORRECT.

Statement (d): "In the answer of $12.11 + 18.0 + 1.012$ there are three significant figures." …
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COMEDK 2024Set 2024-A1 markMCQ
$1.5 \mathrm{~kg}$
›Reveal solutionSolution
$P = \dfrac{mgh}{t} = 3.465\,\text{W}$ . The least precise datum (mass $1.5\,\text{kg}$ , two significant figures) fixes the precision, so the answer to the appropriate significant figures is $3.5\,\text{W}$ — option (D).

Concept

The motor lifts the load against gravity, so the work done equals the gain in gravitational potential energy, $W = mgh$ , and the power is that work divided by the time, $P = mgh/t$ . The phrase "appropriate significant figures" is the real point of the question: a calculated result can carry no more significant figures than the least precise measurement used.

Solution

Formula: $P = \dfrac{mgh}{t}$ .

Substitute: $P = \dfrac{1.5 \times 9.81 \times 1.128}{4.79}$ .

Evaluate: $1.5 \times 9.81 = 14.715$ ; $14.715 \times 1.128 = 16.59852\,\text{J}$ ; $16.59852 / 4.79 = 3.465\,\text{W}$ . …
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KCET 2023Set D-21 markMCQ
$\sqrt{3}$
›Reveal solutionSolution

Use the BCC body-diagonal contact relation to get the edge $a$ from the radius, then cube it — and convert Å to metres carefully.

1. The BCC radius–edge relation

In a body-centred cubic cell the atoms touch along the body diagonal, whose length is $\sqrt3\,a$ and which contains 4 radii:
$4r = \sqrt{3}\,a \qquad\Longrightarrow\qquad a = \frac{4r}{\sqrt3}$
2. Substitute $r = \sqrt3$ Å
$a = \frac{4 \times \sqrt3}{\sqrt3} = 4\ \text{Å}$
The $\sqrt3$ was chosen precisely so that it cancels — a clean edge length of 4 Å.

3. Volume of the cubic cell
$V = a^3 = (4\ \text{Å})^3 = 64\ \text{Å}^3$
4. Convert to $\mathrm{m^3}$

Since $1\ \text{Å} = 10^{-10}\ \mathrm{m}$ ,
$1\ \text{Å}^3 = (10^{-10})^3 = 10^{-30}\ \mathrm{m^3}$ $V = 64 \times 10^{-30}\ \mathrm{m^3} = \mathbf{6.4 \times 10^{-29}\ \mathrm{m^3}}$
5. Reading the distractors …
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Q.The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

The Core Idea

The Rules (Memorize These)

1. Non-zero digits are always significant

2. Zeros between non-zero digits are significant

3. Leading zeros are never significant

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

5. Exact numbers have infinite sig figs

Why This Matters: Calculations

Multiplication and Division

Addition and Subtraction

Why this formula?

Significant Figures: Why the Rules Work

1. The Fundamental Idea: Uncertainty is the Key

2. Rule for Addition and Subtraction

Why this holds

3. Rule for Multiplication and Division

Why this holds

4. Why These Rules Are Different …

The rounding rule for a dropped digit of exactly 5

Rounding 2.745 to 3 significant figures

Rounding 2.735 to 3 significant figures …

Method: The "Even-Digit Rule" (also called Banker's Rounding or Round-Half-to-Even)

Steps

Step 1: Identify the digit to keep

Step 2: Apply the Even-Digit Rule

Mistake 1: Forgetting the "Rule for 5" (Rounding to the Nearest Even)

Mistake 2: Counting Significant Figures Incorrectly

Mistake 3: Applying the "Round Up for 5" Rule Blindly

Mistake 4: Confusing "3 Significant Figures" with "3 Decimal Places"

Concept

Solution

1. The BCC radius–edge relation

2. Substitute $r = \sqrt3$ Å

3. Volume of the cubic cell

4. Convert to $\mathrm{m^3}$

5. Reading the distractors …

Q.The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give

The rounding rule for a dropped digit of exactly 5

Rounding 2.745 to 3 significant figures

Rounding 2.735 to 3 significant figures …

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