🏠 NEET Home

NEET 2027 · Physics · Class 11 · Ch. 1 Units and Measurement

Significant Figures & Rounding

Counting rules, the two arithmetic rules, the even/odd convention for a dropped 5, and order of magnitude. Almost every NEET paper from 2018 to 2024 carried an item from this section.

PRIORITY 2TOPIC 7 + 9 + 12 OF 1520 QUESTIONS12 NCERT-BASEDNCERT §1.3, 1.3.1, 1.3.2

01Concept in plain language

Significant figures are an honesty system. They let a written number say how carefully it was measured, without any separate statement of error.

Read a length off a metre scale and you can be sure of the centimetres and the millimetres. The fraction of a millimetre you can only estimate. NCERT's definition follows directly: all the digits known reliably, plus the first digit that is uncertain.

So 287.5 cm has four significant figures. The 2, 8 and 7 were read off the scale; the 5 was estimated. Writing 287.53 would be a lie, because that instrument cannot resolve hundredths of a millimetre. Writing 287 would be a waste, because you did estimate that last digit.

Exactly one uncertain digit

This is the discipline of the whole system. Not two, not none. That is why a calculator answer of 1.68804780876 g cm⁻³ is, in NCERT's own word, absurd — it claims eleven reliable digits from data that supports three.

Changing the unit changes nothing

2.308 cm, 0.02308 m, 23.08 mm and 23080 μm are the same measurement written four ways. All four have four significant figures. It follows that any zero which appears or disappears purely because you changed the unit cannot be significant — it is a placeholder. That single observation generates every counting rule in the next section.

Scientific notation removes every ambiguity

Write the measurement as a × 10b with 1 ≤ a < 10, and count the digits in a. Leading zeros disappear, trailing zeros become unambiguous, and the power of ten never enters the count.

4700 mm could be two, three or four figures. 4.700 × 10³ mm is definitively four. NCERT recommends scientific notation for exactly this reason.

02What a significant figure is

SIGNIFICANT FIGURES = ALL RELIABLE DIGITS + THE FIRST UNCERTAIN ONE 28.5 28.7 28.9 edge of the object falls here 2 8 7 . 5 cm read directly — certain estimated 287.5 cm has four significant figures. Writing 287.53 would be dishonest: the scale cannot resolve that far. Exactly one uncertain digit is allowed — never two.
Fig. 1 — A reading of 287.5 cm. Three digits come straight off the scale; the fourth is estimated. That estimated digit is significant, and it is the last one that may be.
FOUR NUMBERS, FOUR RULES — THE WHOLE COUNTING SYSTEM0.0071leading zeros never count2.3084zeros between non-zeros always count123003trailing zeros, no decimal point → not counted0.069004trailing zeros after a decimal → countedScientific notation removes every ambiguity: 1.23 × 10⁴ is unambiguously three figures, 1.2300 × 10⁴ is five.
Fig. 2 — Four numbers, four rules. Leading zeros never count, sandwiched zeros always count, and trailing zeros count only when a decimal point is present.
ROUNDING WHEN THE DROPPED DIGIT IS EXACTLY 5 2.745 → 2.74 preceding digit 4 is even — simply drop the 5 2.735 → 2.74 preceding digit 3 is odd — raise it by one The ordinary cases are unchanged: 2.746 → 2.75 (more than 5, round up) and 1.743 → 1.74 (less than 5, leave alone). The even/odd convention exists so that rounding errors cancel over many calculations instead of always pushing upward. In multi-step work, carry one extra digit through the intermediate steps and round only at the end.
Fig. 3 — The convention when the dropped digit is exactly 5. Both of NCERT's worked examples land on 2.74, one from above and one from below.

03The counting rules

rules 4 and 5 differ only by the presence of a decimal point
RuleStatementExampleCount
1All non-zero digits are significant2343
2Zeros between two non-zero digits are always significant, wherever the decimal point is6.032 · 20034 · 4
3For a number less than 1, zeros to the right of the decimal point but to the left of the first non-zero digit are not significant0.0023084
4Trailing zeros in a number without a decimal point are not significant12300 · 5003 · 1
5Trailing zeros in a number with a decimal point are significant3.500 · 0.069004 · 4
6In scientific notation, every digit of the mantissa is significant; the power of ten is irrelevant4.700 × 10³4
7Exact numbers — counted integers and pure numerical factors — have infinitely many significant figuresthe 2 in r = d/2 · the n in T = t/n∞

04Formula sheet

Arithmetic with significant figures
Multiplication and division
answer keeps the least number of significant figures
4.237 g ÷ 2.51 cm³ = 1.69 g cm⁻³, three figures
Addition and subtraction
answer keeps the least number of decimal places
436.32 + 227.2 + 0.301 = 663.8 g, one decimal place
Powers and roots
same as multiplication — least number of significant figures
(7.203)³ = 373.7 m³
Exact factors
no limit imposed
½, 2π, 6 in 6a², counted integers
Intermediate steps
carry one extra digit, round only at the end
prevents rounding errors from accumulating
Rounding and order of magnitude
Dropped digit > 5
raise the preceding digit
2.746 → 2.75
Dropped digit < 5
leave the preceding digit unchanged
1.743 → 1.74
Dropped digit exactly 5, preceding even
drop it
2.745 → 2.74
Dropped digit exactly 5, preceding odd
raise the preceding digit
2.735 → 2.74
Scientific notation
a × 10b, 1 ≤ a < 10
Order of magnitude
b if a ≤ 5 ; b + 1 if 5 < a ≤ 10
the cut is at 5, per NCERT

05The two arithmetic rules

Everything in this section reduces to knowing which of two rules applies. Get the operation right and the rule follows.

Multiplication and division: count significant figures

The result cannot be more precise than the weakest input. NCERT's example: 4.237 g (four figures) divided by 2.51 cm³ (three figures) gives a density of 1.69 g cm⁻³ — three figures, matching the weaker input.

Addition and subtraction: count decimal places

Here what matters is where each number's uncertainty sits, not how many digits it has. Adding 436.32 g, 227.2 g and 0.301 g gives 663.821 g arithmetically, but 227.2 g is known only to one decimal place, so the answer is 663.8 g.

NCERT states the warning explicitly

Do not use the multiplication rule for addition. In the example above, counting significant figures would give 664 g, which does not convey the precision properly. And for the subtraction 0.307 m − 0.304 m = 0.003 m, the multiplication rule would suggest writing 3.00 × 10⁻³ m, which claims three figures where only one survives.

Carry an extra digit through intermediate steps

NCERT gives a striking illustration. The reciprocal of 9.58, rounded to three figures, is 0.104. Take the reciprocal of 0.104 to three figures and you get 9.62 — not 9.58. But keep 1/9.58 = 0.1044 through the intermediate step and the original value returns exactly. Round once, at the end.

06Rounding and order of magnitude

The three rounding cases

Two of them are ordinary: a dropped digit greater than 5 rounds the preceding digit up, and a dropped digit less than 5 leaves it alone. The third case — a dropped digit of exactly 5 — is where NCERT departs from school habit.

When the digit to be dropped is exactly 5, look at the digit before it. Even → drop the 5. Odd → raise by one. So 2.745 becomes 2.74 and 2.735 also becomes 2.74. The convention exists so that rounding errors cancel over many calculations rather than always pushing the answer upward.

Order of magnitude

Write the number as a × 10b with a between 1 and 10. Then round a to 1 if it is at most 5, or to 10 if it is more than 5. The power of ten that survives is the order of magnitude.

NCERT's worked pair: the diameter of the Earth is 1.28 × 10⁷ m, so its order of magnitude is 7. The diameter of a hydrogen atom is 1.06 × 10⁻¹⁰ m, order −10. The Earth is therefore 17 orders of magnitude larger than a hydrogen atom.

Where the cut sits

The mantissa threshold is 5, and it is easy to get wrong. 4.9 × 10³ has order 3; 5.1 × 10³ has order 4. Options in NEET are frequently built one order apart to catch exactly this.

07Exceptions and traps

Significant figures are not decimal places

0.0855 has three significant figures and four decimal places. Multiplication counts one; addition counts the other. Choosing the wrong one is the single largest source of error in this topic.

Exact numbers never limit the answer

The 6 in 6a², the ½ in ½mv², the 2 in r = d/2, the number of oscillations you counted — all have infinitely many significant figures. Only measured quantities constrain the result.

A change of units cannot change the count

NCERT states this as an important remark and then builds several rules on it. If a conversion appears to change the number of significant figures, the extra digits are placeholders.

The exactly-5 rule overrides school rounding

2.745 → 2.74, not 2.75. But note the rule applies only when the dropped digit is exactly 5 with nothing after it. 2.7451 → 2.75, because the dropped part now exceeds 5.

Subtraction can destroy significant figures

20.17 − 20.15 = 0.02 g. Two four-figure inputs give a one-figure answer. This is the significant-figure counterpart of the error explosion seen with differences of nearly equal quantities.

Order of magnitude is not the exponent you see

6.7 × 10⁴ has exponent 4 but order of magnitude 5, because the mantissa exceeds 5 and rounds up to 10.

08Numbers to remember

these exact worked values recur in NEET items almost verbatim
Worked value from NCERTResult
Cube of side 7.203 m — surface area311.3 m² (4 sf)
Cube of side 7.203 m — volume373.7 m³ (4 sf)
5.74 g in 1.2 cm³4.8 g cm⁻³ (2 sf)
4.237 g in 2.51 cm³1.69 g cm⁻³ (3 sf)
436.32 + 227.2 + 0.301 g663.8 g (1 decimal place)
0.307 m − 0.304 m0.003 m = 3 × 10⁻³ m (1 sf)
Sheet 4.234 × 1.005 × 2.01 cm — area8.72 m² (3 sf)
Sheet — volume0.0855 m³ (3 sf)
Box 2.30 kg + 20.15 g + 20.17 g2.34 kg (2 decimal places)
20.17 g − 20.15 g0.02 g (2 decimal places, 1 sf)
2.745 → 3 sf2.74 (preceding digit even)
2.735 → 3 sf2.74 (preceding digit odd)
Speed of light rounded for computation3 × 10⁸ m s⁻¹ from 2.99792458 × 10⁸
Diameter of the Earth1.28 × 10⁷ m, order of magnitude 7
Diameter of a hydrogen atom1.06 × 10⁻¹⁰ m, order of magnitude −10
Earth relative to hydrogen atom17 orders of magnitude larger
Value of π for exam use3.14 or 3.142, as the data warrants

09Scientists to remember

lightly examined; one reading is enough
NameWhat to attach to the name
Simon StevinPopularised decimal fractions in Europe (1585), which is what makes place-value precision expressible at all.
John NapierIntroduced the decimal point in its modern role, and logarithms — the tool that made multi-digit computation practical.
Carl Friedrich GaussRound-half-to-even, the convention behind the exactly-5 rule, is sometimes called Gaussian or banker's rounding; it is designed so rounding errors cancel.
Lord KelvinThe measurement dictum quoted at the head of this chapter: knowledge that cannot be expressed in numbers is of a meagre kind.

10Twenty worked questions

About the labels

Items tagged PYQ here are NCERT worked examples and chapter-end exercises, which NTA has used almost verbatim in several papers. Every count and every rounded value on this page was checked by an independent script.

Q01PYQ · NCERT Ex. 1.10(a)Counting

The number of significant figures in 0.007 m² is:

  1. 1
  2. 2
  3. 3
  4. 4
Show full solution
Given
0.007 m².
Asked
Number of significant figures.
Concept
Zeros to the left of the first non-zero digit are only placeholders. They fix the size of the number, not its precision, so they are never counted.
Formula
leading zeros → not significant
Baby steps
  1. The digits present are 0, 0, 0 and 7.
  2. All three zeros lie to the left of the first non-zero digit.
  3. They are placeholders, so none is significant.
  4. Only the 7 counts: 1 significant figure.
  5. Confirm with scientific notation: 0.007 = 7 × 10⁻³ — a single digit.
Answer
(a) 1
Why not others
(a)Correct.
(b)This counts one of the leading zeros.
(c)This counts two of them.
(d)This counts all four digits.
Shortcut
Rewrite the number in scientific notation and count the digits in the mantissa. Leading zeros vanish automatically, which removes the whole difficulty.
Where it goes wrong
The zero written before the decimal point in numbers like 0.1250 is a typographic convention only. NCERT states explicitly that it is never significant.
Q02PYQ · NCERT Ex. 1.10(b)Counting

The number of significant figures in 2.64 × 10²⁴ kg is:

  1. 24
  2. 26
  3. 2
  4. 3
Show full solution
Given
2.64 × 10²⁴ kg.
Asked
Number of significant figures.
Concept
In scientific notation the power of ten is irrelevant to precision. Only the digits of the mantissa are counted.
Formula
a × 10b → count the digits in a only
Baby steps
  1. The mantissa is 2.64.
  2. It has three digits, all non-zero.
  3. The exponent 24 says nothing about how carefully the mass was measured.
  4. So the answer is 3.
Answer
(d) 3
Why not others
(a)The exponent is not a count of significant figures.
(b)Neither the exponent nor the sum of digits is relevant.
(c)This drops one digit of the mantissa.
(d)Correct — this is the mass of the Earth to three figures.
Shortcut
NCERT says it directly: the power of ten is irrelevant to the determination of significant figures, and every digit written in the mantissa is significant.
Where it goes wrong
Larger exponents feel like more precision. They are not. 2.64 × 10²⁴ and 2.64 × 10⁻²⁴ are equally precise: three figures each.
Q03PYQ · NCERT Ex. 1.10(c)Counting

The number of significant figures in 0.2370 g cm⁻³ is:

  1. 2
  2. 3
  3. 4
  4. 5
Show full solution
Given
0.2370 g cm⁻³.
Asked
Number of significant figures.
Concept
The leading zero does not count, but the trailing zero does, because the number carries a decimal point. A trailing zero after a decimal is a deliberate statement of precision.
Formula
leading zeros no · trailing zeros after a decimal yes
Baby steps
  1. Leading 0: placeholder, not counted.
  2. Digits 2, 3, 7: all significant.
  3. Trailing 0 after the decimal point: significant, because it would be pointless to write otherwise.
  4. Total: 4 significant figures.
Answer
(c) 4
Why not others
(a)This counts only 2 and 3.
(b)This drops the trailing zero.
(c)Correct.
(d)This counts the leading zero as well.
Shortcut
A trailing zero after a decimal point is never accidental. Writing 0.2370 instead of 0.237 is the experimenter's way of saying the fourth digit was actually measured.
Where it goes wrong
NCERT's own example: 3.500 and 0.06900 each have four significant figures. Count from the first non-zero digit to the last written digit.
Q04PYQ · NCERT Ex. 1.10(d)Counting

The number of significant figures in 6.320 J is:

  1. 3
  2. 4
  3. 2
  4. 5
Show full solution
Given
6.320 J.
Asked
Number of significant figures.
Concept
Every digit here is either non-zero or a trailing zero after a decimal point, so every digit counts.
Formula
all digits from the first non-zero to the last written one
Baby steps
  1. 6, 3 and 2 are non-zero and therefore significant.
  2. The final 0 follows a decimal point, so it is significant too.
  3. There are no leading zeros to discard.
  4. Total: 4.
Answer
(b) 4
Why not others
(a)This drops the trailing zero.
(b)Correct.
(c)This counts only two digits.
(d)There are only four digits in the number.
Shortcut
Once past the first non-zero digit, count everything you see, including zeros, provided there is a decimal point in the number.
Where it goes wrong
Contrast with 6320 J, which has no decimal point and therefore only three significant figures. The decimal point is doing real work.
Q05PYQ · NCERT Ex. 1.10(e)Counting

The number of significant figures in 6.032 N m⁻² is:

  1. 4
  2. 3
  3. 2
  4. 5
Show full solution
Given
6.032 N m⁻².
Asked
Number of significant figures.
Concept
A zero trapped between two non-zero digits is always significant, wherever the decimal point sits.
Formula
sandwiched zeros → always significant
Baby steps
  1. Digits: 6, 0, 3, 2.
  2. The 0 lies between 6 and 3, so it is a sandwiched zero.
  3. Sandwiched zeros are always significant, regardless of the decimal point.
  4. Total: 4.
Answer
(a) 4
Why not others
(a)Correct.
(b)This discards the sandwiched zero, which is never allowed.
(c)This counts only two digits.
(d)There are only four digits present.
Shortcut
Three zero rules and no fourth: leading zeros never count, sandwiched zeros always count, trailing zeros count only when there is a decimal point.
Where it goes wrong
Students who have just learned that 'zeros do not count' over-apply the rule. It applies only to leading zeros.
Q06PYQ · NCERT Ex. 1.10(f)Counting

The number of significant figures in 0.0006032 m² is:

  1. 7
  2. 3
  3. 5
  4. 4
Show full solution
Given
0.0006032 m².
Asked
Number of significant figures.
Concept
Combine two rules: discard the leading zeros, keep the sandwiched one.
Formula
0.0006032 → count from the 6
Baby steps
  1. Four leading zeros: all placeholders, none counted.
  2. Remaining digits: 6, 0, 3, 2.
  3. The 0 is sandwiched between 6 and 3, so it counts.
  4. Total: 4. In scientific notation, 6.032 × 10⁻⁴.
Answer
(d) 4
Why not others
(a)This counts every digit written, including the placeholders.
(b)This discards the sandwiched zero as well.
(c)This counts one leading zero.
(d)Correct — identical to 6.032, since a change of unit or scale cannot change the count.
Shortcut
This number and 6.032 from the previous question have the same count, which illustrates NCERT's key remark: a change of units never changes the number of significant figures.
Where it goes wrong
Counting the four leading zeros. They only tell you where the decimal point is.
Q07Counting

A length is written as 2.308 cm. Written as 23080 μm, the number of significant figures is:

  1. 5
  2. 3
  3. 4
  4. 2
Show full solution
Given
2.308 cm = 0.02308 m = 23.08 mm = 23080 μm.
Asked
Number of significant figures in the μm form.
Concept
A change of unit cannot change how carefully something was measured, so the count must stay at four. The trailing zero in 23080 is a placeholder created by the unit change, not a measured digit.
Formula
trailing zero with no decimal point → not significant
Baby steps
  1. 2.308 cm has four significant figures: 2, 3, 0, 8.
  2. Converting to micrometres shifts the decimal point but measures nothing new.
  3. In 23080 the final zero has no decimal point after it, so it is not significant.
  4. The count remains 4.
Answer
(c) 4
Why not others
(a)This counts the placeholder zero.
(b)This drops the sandwiched zero.
(c)Correct — exactly the example NCERT uses to introduce the trailing-zero rule.
(d)Far too few.
Shortcut
If a conversion appears to change the count, the extra zeros must be placeholders. Scientific notation settles it: 2.308 × 10⁴ μm, four figures, unambiguous.
Where it goes wrong
This ambiguity is the entire reason NCERT recommends reporting every measurement in scientific notation. In 4700 mm you cannot tell whether the zeros were measured; in 4.700 × 10³ mm you can.
Q08PYQ · NCERT Example 1.1Arithmetic

Each side of a cube is measured to be 7.203 m. Its total surface area, to the appropriate number of significant figures, is:

  1. 311.299254 m²
  2. 311.3 m²
  3. 311 m²
  4. 311.30 m²
Show full solution
Given
Side = 7.203 m (4 significant figures).
Asked
Total surface area, correctly rounded.
Concept
For multiplication, the answer carries as many significant figures as the least precise input. Here there is only one input, with four figures, so the answer has four.
Formula
A = 6a²
Baby steps
  1. A = 6 × (7.203)² = 6 × 51.883209.
  2. = 311.299254 m².
  3. The input has 4 significant figures; the 6 is exact.
  4. Round to 4 figures: 311.3 m².
Answer
(b) 311.3 m²
Why not others
(a)Nine figures cannot be justified by a four-figure measurement.
(b)Correct.
(c)Three figures — this discards precision that was actually measured.
(d)Five figures — one more than the data supports.
Shortcut
Count the significant figures of the input before you start calculating. You then know exactly where to stop, and can avoid writing out digits you will discard.
Where it goes wrong
The 6 in 6a² is an exact counting number with infinitely many significant figures. It never limits the answer.
Q09PYQ · NCERT Example 1.1Arithmetic

For the same cube of side 7.203 m, the volume to the appropriate number of significant figures is:

  1. 373.7 m³
  2. 373.714754 m³
  3. 374 m³
  4. 373.71 m³
Show full solution
Given
Side = 7.203 m (4 significant figures).
Asked
Volume, correctly rounded.
Concept
Cubing does not change the significant-figure rule. The answer still carries four figures, because the single input carried four.
Formula
V = a³
Baby steps
  1. V = (7.203)³ = 373.714754 m³.
  2. Input has 4 significant figures.
  3. Round to 4: 373.7 m³.
  4. Note: the fifth digit is 1, less than 5, so the fourth digit is left unchanged.
Answer
(a) 373.7 m³
Why not others
(a)Correct.
(b)Nine figures, unjustified by the data.
(c)Three figures, discarding measured precision.
(d)Five figures, one too many.
Shortcut
Multiplication, division and powers all follow the same rule: match the least number of significant figures among the inputs.
Where it goes wrong
Do not confuse this with the addition rule, which counts decimal places rather than significant figures. Which operation you are performing decides which rule applies.
Q10PYQ · NCERT Example 1.2Arithmetic

5.74 g of a substance occupies 1.2 cm³. Its density, expressed with the correct number of significant figures, is:

  1. 4.783 g cm⁻³
  2. 4.78 g cm⁻³
  3. 5 g cm⁻³
  4. 4.8 g cm⁻³
Show full solution
Given
Mass = 5.74 g (3 sf); volume = 1.2 cm³ (2 sf).
Asked
Density with correct significant figures.
Concept
In a division, the weakest input sets the limit. The volume has only two significant figures, so the answer can have only two.
Formula
ρ = m / V
Baby steps
  1. ρ = 5.74 / 1.2 = 4.7833... g cm⁻³.
  2. Mass has 3 significant figures; volume has 2.
  3. The smaller count is 2.
  4. Round to 2 figures: 4.8 g cm⁻³.
Answer
(d) 4.8 g cm⁻³
Why not others
(a)Four figures — the volume cannot support this.
(b)Three figures — still one too many.
(c)One figure — this discards precision the data does support.
(d)Correct.
Shortcut
Find the input with the fewest significant figures before calculating. That number is your answer's ceiling, and knowing it in advance saves time.
Where it goes wrong
A poorly measured quantity drags the whole result down. Improving the mass measurement here would achieve nothing until the volume is measured better.
Q11PYQ · NCERT Ex. 1.11Arithmetic

A rectangular sheet of metal has length 4.234 m, breadth 1.005 m and thickness 2.01 cm. Its total surface area, to the correct number of significant figures, is:

  1. 8.7209 m²
  2. 8.721 m²
  3. 8.72 m²
  4. 8.7 m²
Show full solution
Given
l = 4.234 m (4 sf), b = 1.005 m (4 sf), t = 2.01 cm = 0.0201 m (3 sf).
Asked
Total surface area, correctly rounded.
Concept
Convert everything to a single unit first, then apply the multiplication rule using the weakest input, which here is the three-figure thickness.
Formula
A = 2(lb + bt + tl)
Baby steps
  1. t = 2.01 cm = 0.0201 m.
  2. lb = 4.234 × 1.005 = 4.25517 m².
  3. bt = 1.005 × 0.0201 = 0.0202005 m²; tl = 0.0201 × 4.234 = 0.0851034 m².
  4. A = 2(4.25517 + 0.0202005 + 0.0851034) = 8.7209478 m².
  5. Least significant-figure count is 3 (from 2.01), so A = 8.72 m².
Answer
(c) 8.72 m²
Why not others
(a)Five figures, unsupported by the 3-figure thickness.
(b)Four figures, still one too many.
(c)Correct — this is the NCERT answer.
(d)Two figures, discarding measured precision.
Shortcut
Scan the data for the weakest measurement before computing. Here 2.01 cm has three figures and everything else has four, so the answer will have three no matter what.
Where it goes wrong
Forgetting to convert 2.01 cm into metres. Mixing units inside a single formula is a far more expensive error than mis-rounding.
Q12PYQ · NCERT Ex. 1.11Arithmetic

For the same sheet (4.234 m × 1.005 m × 2.01 cm), the volume to the correct number of significant figures is:

  1. 0.08552892 m³
  2. 0.0855 m³
  3. 0.086 m³
  4. 0.08553 m³
Show full solution
Given
l = 4.234 m, b = 1.005 m, t = 0.0201 m.
Asked
Volume, correctly rounded.
Concept
Same rule, same weakest input. Note that the leading zeros in the answer are placeholders and are not counted among the three significant figures.
Formula
V = l × b × t
Baby steps
  1. V = 4.234 × 1.005 × 0.0201.
  2. = 4.25517 × 0.0201 = 0.08552892 m³.
  3. Least count of significant figures is 3.
  4. V = 0.0855 m³ — the digits 8, 5, 5 are the three significant ones.
Answer
(b) 0.0855 m³
Why not others
(a)Seven figures, unjustified.
(b)Correct — the NCERT answer.
(c)Two significant figures, one too few.
(d)Four significant figures, one too many.
Shortcut
When the answer is less than 1, count significant figures starting from the first non-zero digit. The zeros after the decimal point but before the 8 do not count.
Where it goes wrong
Reading 0.0855 as five significant figures because it has five characters after the decimal point. Significant figures and decimal places are different things.
Q13PYQ · NCERT Ex. 1.12(a)Addition rule

The mass of a box measured by a grocer's balance is 2.30 kg. Two gold pieces of masses 20.15 g and 20.17 g are added to it. The total mass of the box, to correct significant figures, is:

  1. 2.34 kg
  2. 2.34032 kg
  3. 2.3 kg
  4. 2.340 kg
Show full solution
Given
Box 2.30 kg; gold pieces 20.15 g and 20.17 g.
Asked
Total mass with correct significant figures.
Concept
For addition the rule counts decimal places, not significant figures. Convert to a common unit first, then match the input with the fewest decimal places.
Formula
in addition, keep the least number of decimal places
Baby steps
  1. Convert: 20.15 g = 0.02015 kg, 20.17 g = 0.02017 kg.
  2. Sum = 2.30 + 0.02015 + 0.02017 = 2.34032 kg.
  3. The box mass, 2.30 kg, is known to only 2 decimal places.
  4. Round to 2 decimal places: 2.34 kg.
Answer
(a) 2.34 kg
Why not others
(a)Correct.
(b)Five decimal places — the balance cannot support this.
(c)One decimal place — the box mass was known to two.
(d)Three decimal places, one too many.
Shortcut
Addition and subtraction → decimal places. Multiplication and division → significant figures. Decide which operation you have before choosing the rule.
Where it goes wrong
Applying the significant-figure rule here would give 2.3 kg, because 2.30 has three figures and 20.15 has four... and that reasoning is simply the wrong rule. NCERT warns about this explicitly.
Q14PYQ · NCERT Ex. 1.12(b)Subtraction rule

For the same two gold pieces of masses 20.15 g and 20.17 g, the difference in their masses, to correct significant figures, is:

  1. 0.020 g
  2. 0.0200 g
  3. 0.02000 g
  4. 0.02 g
Show full solution
Given
20.15 g and 20.17 g, each known to 2 decimal places.
Asked
The difference, correctly expressed.
Concept
Subtraction follows the decimal-place rule as well. Both inputs have two decimal places, so the answer keeps two decimal places — which here leaves only one significant figure.
Formula
in subtraction, keep the least number of decimal places
Baby steps
  1. 20.17 − 20.15 = 0.02 g.
  2. Both inputs are known to 2 decimal places.
  3. The answer therefore carries 2 decimal places: 0.02 g.
  4. Note that four-figure inputs have produced a one-figure answer.
Answer
(d) 0.02 g
Why not others
(a)Three decimal places, more than the data allows.
(b)Four decimal places.
(c)Five decimal places.
(d)Correct.
Shortcut
Subtraction is the only operation that can reduce the number of significant figures. Two four-figure masses have produced a single-figure difference.
Where it goes wrong
This is the significant-figure face of the error result seen in the errors file: subtracting nearly equal numbers destroys precision. NCERT's other example is 0.307 − 0.304 = 0.003 m.
Q15Rounding

The number 2.745 rounded off to three significant figures is:

  1. 2.75
  2. 2.7
  3. 2.74
  4. 2.8
Show full solution
Given
2.745, to be rounded to three significant figures.
Asked
The rounded value.
Concept
When the digit to be dropped is exactly 5, NCERT applies the even/odd convention: if the preceding digit is even, drop the 5; if odd, raise the preceding digit by one.
Formula
dropped digit exactly 5 → preceding even, drop; preceding odd, raise
Baby steps
  1. Keeping three figures means dropping the final 5.
  2. The preceding digit is 4, which is even.
  3. So the 5 is simply dropped and the 4 is left unchanged.
  4. Result: 2.74.
Answer
(c) 2.74
Why not others
(a)This rounds up, which the convention forbids for an even preceding digit.
(b)This keeps only two significant figures.
(c)Correct.
(d)This rounds to two figures and rounds up as well.
Shortcut
The convention exists so rounding errors cancel over many calculations instead of always pushing the result upward. Half the time you drop, half the time you raise.
Where it goes wrong
Ordinary school rounding ('5 always rounds up') gives 2.75 and the wrong option. NCERT explicitly overrides that habit for the exactly-5 case.
Q16Rounding

The number 2.735 rounded off to three significant figures is:

  1. 2.73
  2. 2.74
  3. 2.7
  4. 2.8
Show full solution
Given
2.735, to be rounded to three significant figures.
Asked
The rounded value.
Concept
Same convention, opposite case: the preceding digit is now odd.
Formula
dropped digit exactly 5, preceding digit odd → raise by one
Baby steps
  1. Dropping the final 5 leaves three figures.
  2. The preceding digit is 3, which is odd.
  3. So it is raised by one, from 3 to 4.
  4. Result: 2.74.
Answer
(b) 2.74
Why not others
(a)This drops the 5, which applies only when the preceding digit is even.
(b)Correct — and note that 2.745 and 2.735 both round to 2.74, which is the point of the convention.
(c)Two significant figures.
(d)Two figures, rounded up.
Shortcut
Both of NCERT's worked examples land on 2.74, one from above and one from below. If you remember that coincidence you have remembered both halves of the rule.
Where it goes wrong
The convention applies only when the dropped digit is exactly 5 with nothing after it. For 2.7451 the digit dropped exceeds 5, so the ordinary rule applies and the answer is 2.75.
Q17Arithmetic

The mass of an object is 4.237 g and its volume is 2.51 cm³. Its density to the correct number of significant figures is:

  1. 1.69 g cm⁻³
  2. 1.688 g cm⁻³
  3. 1.7 g cm⁻³
  4. 1.68804780876 g cm⁻³
Show full solution
Given
m = 4.237 g (4 sf); V = 2.51 cm³ (3 sf).
Asked
Density, correctly rounded.
Concept
The division rule again: the answer carries the smaller of the two significant-figure counts.
Formula
ρ = m / V
Baby steps
  1. ρ = 4.237 / 2.51 = 1.68804780876 g cm⁻³.
  2. Mass 4 sf, volume 3 sf → answer 3 sf.
  3. The fourth digit is 8, greater than 5, so the third digit rounds up from 8 to 9.
  4. ρ = 1.69 g cm⁻³.
Answer
(a) 1.69 g cm⁻³
Why not others
(a)Correct — the NCERT worked example in §1.3.1.
(b)Four figures, one too many.
(c)Two figures, one too few.
(d)Eleven figures, which NCERT calls absurd and irrelevant.
Shortcut
A calculator will always give you eleven digits. Deciding in advance how many you are allowed to keep is part of the physics, not an afterthought.
Where it goes wrong
Rounding 1.688 to 1.68 rather than 1.69. The digit after the third figure is 8, so the third figure must go up.
Q18Comparison

Which of the following numbers has the greatest number of significant figures?

  1. 500
  2. 0.00500
  3. 5.00 × 10³
  4. 5.000
Show full solution
Given
Four numbers.
Asked
Which has the most significant figures.
Concept
Apply all three zero rules and compare. Every option is designed to test a different one.
Formula
count from the first non-zero digit to the last significant one
Baby steps
  1. 500: trailing zeros with no decimal point → 1 significant figure.
  2. 0.00500: leading zeros discarded, trailing zeros after a decimal kept → 3.
  3. 5.00 × 10³: mantissa has three digits → 3.
  4. 5.000: all four digits significant → 4, the greatest.
Answer
(d) 5.000
Why not others
(a)Only 1 — the zeros are placeholders.
(b)3 significant figures.
(c)3 significant figures.
(d)Correct, with 4.
Shortcut
Note that 500 and 5.000 look similar and differ by a factor of four in precision. The decimal point carries the entire message.
Where it goes wrong
Option (a) is the reason NCERT recommends scientific notation. If the experimenter genuinely measured three figures, they should have written 5.00 × 10².
Q19Order of magnitude

The order of magnitude of 2.34 × 10⁻⁵ is:

  1. −6
  2. −4
  3. −5
  4. −3
Show full solution
Given
2.34 × 10⁻⁵.
Asked
Its order of magnitude.
Concept
Write the number as a × 10b with 1 ≤ a < 10. Round a to 1 if it is at most 5, or to 10 if it exceeds 5. The resulting power of ten is the order of magnitude.
Formula
a ≤ 5 → order = b ; a > 5 → order = b + 1
Baby steps
  1. Here a = 2.34 and b = −5.
  2. Since 2.34 ≤ 5, round a down to 1.
  3. The number is then approximately 1 × 10⁻⁵ = 10⁻⁵.
  4. Order of magnitude = −5.
Answer
(c) −5
Why not others
(a)This would require a to round down past a full power of ten.
(b)This applies the b + 1 rule, which is for a > 5.
(c)Correct.
(d)Two orders out.
Shortcut
The cut is at 5, not at 5.5 or at 1. NCERT sets it explicitly: round to 1 for a ≤ 5, and to 10 for 5 < a ≤ 10.
Where it goes wrong
Order of magnitude is not simply 'the exponent you see'. It depends on the mantissa, which is exactly what the next question tests.
Q20Order of magnitude

The order of magnitude of 6.7 × 10⁴ is:

  1. 4
  2. 5
  3. 6
  4. 3
Show full solution
Given
6.7 × 10⁴.
Asked
Its order of magnitude.
Concept
Here the mantissa exceeds 5, so it rounds up to 10 and pushes the exponent up by one.
Formula
a > 5 → order = b + 1
Baby steps
  1. a = 6.7, b = 4.
  2. Since 6.7 > 5, round a up to 10.
  3. The number becomes 10 × 10⁴ = 10⁵.
  4. Order of magnitude = 5.
Answer
(b) 5
Why not others
(a)This ignores the rounding of the mantissa.
(b)Correct.
(c)Two orders out.
(d)One order too low.
Shortcut
NCERT's own examples: the diameter of the Earth, 1.28 × 10⁷ m, has order 7; the diameter of a hydrogen atom, 1.06 × 10⁻¹⁰ m, has order −10. So the Earth is 17 orders of magnitude larger.
Where it goes wrong
Reading the exponent straight off. For 6.7 × 10⁴ the exponent is 4 but the order of magnitude is 5, and NEET builds distractors on precisely that gap.