Why significant figures exist
Every measurement involves error. A vernier callipers cannot resolve below its least count; a stopwatch cannot beat the reaction time of the person who clicks it. NCERT §1.3 puts the principle as plainly as possible: the reported result of a measurement should be a number that includes all digits that are known reliably plus the first digit that is uncertain. Those reliable digits together with the first uncertain one are the significant figures. When the period of a simple pendulum is reported as $T = 1.62$ s, the digits 1 and 6 are certain while the digit 2 is uncertain — three significant figures in total, no more, no less.
Reporting more digits is not "more accurate" — it implies a precision the apparatus does not deliver. Reporting fewer throws away information the apparatus did deliver. Significant-figure discipline is the contract between experimenter and reader: write 1.62 s and you say the next decimal is unknown; write 1.6 s and you say you cannot resolve to the second decimal at all. Two different physical claims about the same pendulum.
NCERT adds a corollary: changing the unit cannot change the number of significant figures. $2.308$ cm has four sig-figs. In metres, $0.02308$ m — still four. In millimetres, $23.08$ mm — still four. Sig-figs are a property of the measurement, not of the unit. The parent Units and Measurement chapter introduces the idea; here we drill into the counting rules and the arithmetic that follows.
"The reliable digits plus the first uncertain digit are known as significant digits or significant figures."
NCERT Class 11 Physics, §1.3
Counting rules — the six cases NEET tests
NCERT lays out the rules through worked examples; NIOS §1.2.3 lists them as numbered cases. The two sources agree on every rule. The table below collates them in NEET-ready form, with one short illustration for each.
| Rule | Statement | Example | Sig-figs |
|---|---|---|---|
| 1 | All non-zero digits are significant. | 315.58 |
5 |
| 2 | Zeros between two non-zero digits are significant, regardless of decimal location. | 5\,300\,405.003 |
10 |
| 3 | For a number less than 1, zeros to the right of the decimal but to the left of the first non-zero digit are not significant. The zero conventionally placed before the decimal is never significant. | 0.002308 (only digits 2, 3, 0, 8 count) |
4 |
| 4 | Trailing zeros in a number without a decimal point are not significant. | 123 m = 12\,300 cm = 123\,000 mm |
3 |
| 5 | Trailing zeros in a number with a decimal point are significant. | 3.500 or 0.06900 |
4 each |
| 6 | In scientific notation $a \times 10^b$, every digit shown in $a$ is significant — scientific notation removes all ambiguity. | 4.700 × 10³ mm |
4 |
Rule 4 is the one that catches NEET students. Suppose a length is reported as $4.700$ m. The trailing zeros after the decimal are clearly there to convey precision — if they were not, the report would simply read $4.7$ m. Now switch units: $4.700$ m $= 470.0$ cm $= 4700$ mm $= 0.004700$ km. Rule 4 would naively assign $4700$ mm only two significant figures (the zeros being trailing, decimal-less). NCERT notes the contradiction explicitly: a change of units cannot reduce the number of significant figures of an already-measured quantity. The repair is rule 6 — always write the value as $4.700 \times 10^3$ mm and there is no ambiguity left.
A seventh case sits outside the counting rules but matters in arithmetic: exact numbers carry infinite significant figures. The "2" in $s = 2\pi r$, the "10" in "10 balls", the integer count in $T = t/n$ — none of them constrain the precision of the result. NCERT §1.3 records this as rule (6) of the listing; we use it heavily in the rounding section.
Worked: NCERT Exercise 1.10 — counting practice
This is the canonical NCERT drill — twelve numbers spread across the trap cases. We work each by applying the table above, then list a few more to stretch the pattern.
State the number of significant figures in:
(a) $0.007$ m$^2$ — leading zeros are not significant (rule 3). Only the digit 7 counts. 1 sig-fig.
(b) $2.64 \times 10^{24}$ kg — scientific notation; every digit in the mantissa is significant (rule 6). Digits 2, 6, 4. 3 sig-figs.
(c) $0.2370$ g cm$^{-3}$ — the leading 0 before the decimal does not count; the trailing 0 after a decimal does (rules 3 and 5). Digits 2, 3, 7, 0. 4 sig-figs.
(d) $6.320$ J — trailing zero after a decimal is significant (rule 5). Digits 6, 3, 2, 0. 4 sig-figs.
(e) $6.032$ N m$^{-2}$ — the embedded zero between non-zero digits is significant (rule 2). Digits 6, 0, 3, 2. 4 sig-figs.
(f) $0.0006032$ m$^2$ — leading zeros do not count (rule 3); the embedded zero between 6 and 3 does (rule 2). Digits 6, 0, 3, 2. 4 sig-figs.
Stretch numbers: $5300405.003$ has 10 (rule 2 for all interior zeros, rule 5 for the trailing 003). $5000$ written without a decimal has 1 (rule 4). $5000.$ with a trailing decimal point has 4. $5.000 \times 10^3$ also has 4 — the unambiguous form.
Notice the asymmetry: every time a zero is on the wrong side of a decimal point, the rule flips. The mental model that helps is "zeros only count when they are doing precision work" — that is, when removing them would degrade the implied accuracy of the measurement.
Arithmetic with significant figures
Once you can count sig-figs, the next question is how they combine. NCERT §1.3.1 lays down two rules — one for multiplication/division, one for addition/subtraction — and they look similar but are subtly different. Confusing them is the most common NEET error in this topic.
Multiplication and division — track significant figures
Rule: The result keeps as many significant figures as the input with the fewest significant figures.
A mass is measured as $m = 4.237$ g (4 sig-figs) and a volume as $V = 2.51$ cm$^3$ (3 sig-figs). Find the density.
Arithmetic: $\rho = \dfrac{4.237}{2.51} = 1.68804780876\ldots$ g cm$^{-3}$.
The least-precise input has 3 sig-figs (the volume), so the answer must round to 3 sig-figs: $\rho = \mathbf{1.69}$ g cm$^{-3}$.
Reporting $1.688$ g cm$^{-3}$ would falsely claim that the third decimal is meaningful when the volume measurement does not support it. Reporting $1.7$ g cm$^{-3}$ throws away a sig-fig you legitimately have.
Speed of light $c = 3.00 \times 10^8$ m s$^{-1}$ (3 sig-figs). One year $= 3.1557 \times 10^7$ s (5 sig-figs). Compute one light-year.
Arithmetic: $1$ ly $= (3.00 \times 10^8)(3.1557 \times 10^7) = 9.4671 \times 10^{15}$ m.
Limiting input is $c$ at 3 sig-figs. Round: $1$ ly $= \mathbf{9.47 \times 10^{15}}$ m.
A plate has length $3.003$ m (4 sig-figs) and width $2.26$ m (3 sig-figs). Find its area.
Arithmetic: $A = 3.003 \times 2.26 = 6.78678$ m$^2$.
Least-precise input is the width at 3 sig-figs. Round: $A = \mathbf{6.79}$ m$^2$.
Addition and subtraction — track decimal places
Rule: The result keeps as many decimal places as the input with the fewest decimal places. Notice the shift in vocabulary — not sig-figs, but decimal places.
Add the masses $436.32$ g, $227.2$ g and $0.301$ g.
Arithmetic sum: $436.32 + 227.2 + 0.301 = 663.821$ g.
Decimal places: $436.32$ has 2; $227.2$ has 1; $0.301$ has 3. The least is 1 decimal place. Round: $\mathbf{663.8}$ g.
If you had used the multiplication rule by mistake, you would round to 4 sig-figs and report $663.8$ g (coincidentally correct here) — but in other cases the two rules diverge sharply. The next example shows that.
Subtract: $0.307$ m $-$ $0.304$ m.
Arithmetic: $0.307 - 0.304 = 0.003$ m.
Both inputs have 3 decimal places, so the result keeps 3 decimal places: $\mathbf{0.003}$ m $= 3 \times 10^{-3}$ m.
If you had used the multiplication rule (least sig-figs = 3), you would report $3.00 \times 10^{-3}$ m — implying two more decimals of precision than the measurement supports. Subtraction destroys sig-figs. Both inputs had 3 sig-figs; the answer carries only 1. That asymmetry is what NCERT §1.3.3 calls out explicitly.
One more subtraction example that NEET 2020 lifted straight from this rule: $9.99$ m $-$ $0.0099$ m. Decimal places: $9.99$ has 2; $0.0099$ has 4. The minimum is 2, so the result must end at 2 decimal places. Arithmetic gives $9.9801$ m; rounded, $\mathbf{9.98}$ m. The answer keeps 3 sig-figs even though the second number had 2 — sig-fig counts in subtraction can shrink dramatically.
Rounding off — including the round-to-even rule
Once a calculation is done, the result must be rounded to the correct number of sig-figs (or decimal places, per the operation). NCERT §1.3.2 sets out three sub-rules.
- Drop < 5: keep the preceding digit. $1.743$ rounded to 3 sig-figs becomes $1.74$ — the dropped 3 is less than 5.
- Drop > 5: raise the preceding digit by 1. $2.746$ rounded to 3 sig-figs becomes $2.75$ — the dropped 6 is greater than 5.
- Drop is exactly 5 — apply the round-to-even rule. If the preceding digit is even, drop the 5 and leave it. If the preceding digit is odd, drop the 5 and raise the preceding digit by 1.
The third sub-rule is the one most students mis-remember. NCERT puts it as: "if the preceding digit is even, the insignificant digit is simply dropped and, if it is odd, the preceding digit is raised by 1." So $2.745$ rounded to 3 sig-figs is $2.74$ (preceding digit 4 is even — drop the 5). And $2.735$ rounded to 3 sig-figs is also $2.74$ (preceding digit 3 is odd — raise to 4). This is sometimes called banker's rounding; over many calculations it eliminates the systematic upward bias that "always round 5 up" would introduce.
One operational note NCERT adds: in multi-step calculations, retain one extra digit in intermediate steps and round only the final answer. NCERT illustrates this with $1/9.58$. The reciprocal calculated to 3 sig-figs is $0.104$; the reciprocal of $0.104$ to 3 sig-figs is $9.62$ — and we have lost the original. But if we keep $1/9.58 = 0.1044$ in the intermediate step, taking the reciprocal of that to 3 sig-figs returns $9.58$. The extra digit absorbs the rounding wobble.
Worked: NCERT Exercise 1.11 — rectangular sheet
This is the canonical NEET-style problem that fuses sig-fig counting with the arithmetic rules. Read the units carefully — one of the dimensions is given in centimetres and must be converted.
The length, breadth and thickness of a rectangular sheet of metal are $4.234$ m, $1.005$ m, and $2.01$ cm respectively. Find the area and the volume of the sheet to the correct number of significant figures.
Step 1 — convert to consistent units. Thickness $t = 2.01$ cm $= 2.01 \times 10^{-2}$ m $= 0.0201$ m. The conversion does not change sig-figs (still 3).
Step 2 — count sig-figs. $\ell = 4.234$ m (4), $b = 1.005$ m (4), $t = 0.0201$ m (3).
Step 3 — surface area of the two large faces. $A_\text{face} = 2\ell b = 2 \times 4.234 \times 1.005 = 8.510340$ m$^2$. Multiplication; least sig-figs in $\ell, b$ is 4; round to 4 sig-figs: $A_\text{face} = \mathbf{8.510}$ m$^2$.
Step 4 — total surface area (all six faces). $A = 2(\ell b + b t + \ell t) = 2(4.234 \times 1.005 + 1.005 \times 0.0201 + 4.234 \times 0.0201)$. Arithmetic gives $A = 2(4.25517 + 0.0202005 + 0.0851034) = 2 \times 4.3604739 = 8.7209478$ m$^2$. The thickness limits us to 3 sig-figs: $A = \mathbf{8.72}$ m$^2$. (If the problem intends only the area of the rectangular face $\ell \times b$, that single face has area $\ell b = 4.255$ m$^2$ to 4 sig-figs.)
Step 5 — volume. $V = \ell \times b \times t = 4.234 \times 1.005 \times 0.0201$ m$^3$. Arithmetic: $V = 0.085524\ldots$ m$^3$. Multiplication; least sig-figs in the inputs is 3 (thickness); round: $V = \mathbf{8.55 \times 10^{-2}}$ m$^3$ $= 0.0855$ m$^3$.
The lesson is mechanical but important: the input with the fewest sig-figs governs the whole answer. The 4 sig-figs in the length and breadth do not buy you any extra precision once the thickness, known to only 3 sig-figs, enters the product.
Order of magnitude — squeezing 60 powers of ten into one number
Physical quantities span ranges so vast that decimal notation becomes useless. The radius of a proton is about $10^{-15}$ m; the radius of the observable universe is about $10^{26}$ m. That is 41 orders of magnitude in a single quantity. The vocabulary that handles such ranges cleanly is order of magnitude.
NCERT §1.3 gives the formal recipe. Write any positive number in scientific notation as
$$N = a \times 10^b, \qquad 1 \le a < 10.$$Round $a$ to $1$ if $a \le 5$, or to $10$ if $5 < a \le 10$ (in which case the power increments by 1). The exponent that remains is the order of magnitude of $N$. When an estimate alone is wanted, the quantity is "of the order of $10^b$".
NCERT's two model examples make this concrete. The diameter of the Earth is $1.28 \times 10^7$ m. Since $1.28 \le 5$, $a$ rounds to 1, and the order of magnitude is 7. The diameter of a hydrogen atom is $1.06 \times 10^{-10}$ m. Again $1.06 \le 5$, so the order of magnitude is $-10$. The Earth is therefore $7 - (-10) = 17$ orders of magnitude larger than a hydrogen atom — a factor of $10^{17}$, which is unimaginable as a decimal but a single integer in this language.
| Quantity | Value | Scientific form | Order of magnitude |
|---|---|---|---|
| Radius of a proton | ~ $1$ fm | $1 \times 10^{-15}$ m | $-15$ |
| Radius of a hydrogen atom | $0.53$ Å | $5.3 \times 10^{-11}$ m | $-10$ (since $5.3 > 5$, $a$ rounds to 10) |
| Wavelength of green light | $550$ nm | $5.5 \times 10^{-7}$ m | $-6$ |
| Thickness of a sheet of paper | ~ $0.1$ mm | $1 \times 10^{-4}$ m | $-4$ |
| Height of a human being | ~ $1.7$ m | $1.7 \times 10^{0}$ m | $0$ |
| Diameter of the Earth | $1.28 \times 10^7$ m | $1.28 \times 10^7$ m | $7$ (NCERT) |
| Earth–Sun distance (1 AU) | $1.50 \times 10^{11}$ m | $1.50 \times 10^{11}$ m | $11$ |
| Radius of the Milky Way | ~ $1 \times 10^{21}$ m | $1 \times 10^{21}$ m | $21$ (NIOS Table 1.4) |
| Radius of the observable universe | ~ $1 \times 10^{26}$ m | $1 \times 10^{26}$ m | $26$ (NIOS Table 1.4) |
The hydrogen-atom row demonstrates the rounding subtlety. $5.3 \times 10^{-11}$ m has $a = 5.3 > 5$, so $a$ rounds to $10$ and the expression becomes $10 \times 10^{-11} = 10^{-10}$ m. The order of magnitude is therefore $-10$, not $-11$. Forgetting to apply the $a > 5 \Rightarrow$ round-up step is the most common slip on this kind of question.
Find the order of magnitude of the mass of the Earth, $M_\text{E} = 5.972 \times 10^{24}$ kg.
Here $a = 5.972$. Since $a > 5$, round $a$ up to $10$, increasing the exponent by 1. The order of magnitude is $\mathbf{25}$, not 24.
Find the order of magnitude of Planck's constant, $h = 6.626 \times 10^{-34}$ J s.
$a = 6.626 > 5$, so round to $10$ and the exponent shifts. Order of magnitude: $\mathbf{-33}$. Sanity check — $h$ is enormous on quantum scales but vanishingly small in everyday units, and an exponent of $-33$ captures exactly that.
Why order of magnitude matters — Fermi-style estimation
Order of magnitude is more than a notational convenience. It is the working tool of plausibility checks. If a calculation gives a star a temperature of $10^9$ K when surfaces of ordinary stars are at $10^3$–$10^4$ K, you have a sign error somewhere. If a question asks for the speed of a falling apple and you get a number whose order of magnitude is $10^5$ m s$^{-1}$, you have made a slip — terminal velocity for an apple is order $10^1$ m s$^{-1}$. NEET examiners often build distractors at the wrong order of magnitude precisely to test whether the student is paying attention to size.
The other use is Fermi estimation — back-of-envelope answers to physically meaningful questions using only orders of magnitude. How many atoms in a human body? Mass ~ $10^2$ kg; mass per atom ~ $10^{-26}$ kg; ratio ~ $10^{28}$ atoms. The correct figure (depending on exact composition) is about $7 \times 10^{27}$ — same order of magnitude. The technique sacrifices precision for instant plausibility, which is exactly the trade-off a NEET aspirant should make under time pressure: does my answer have the right power of ten before I argue about the leading digit?
Order of magnitude also tells you when problems are easy and when they are hard. The molar volume of an ideal gas at STP is $22.4$ L $= 2.24 \times 10^{-2}$ m$^3$ — order $-2$. The atomic volume of a mole of hydrogen atoms (worked out in the SI-units page) is about $3 \times 10^{-7}$ m$^3$ — order $-7$. The ratio is $10^5$, which is exactly the NCERT Exercise 1.15 punchline: "most of a gas is empty space".
What this subtopic locked in
- Definition. Sig-figs = reliable digits + first uncertain digit. Reflects instrument precision; invariant under unit change.
- Six counting rules. Non-zeros count; sandwich zeros count; leading zeros never count; trailing zeros count only with a decimal point; scientific notation makes every digit count.
- Arithmetic. Multiplication/division → round to the input with fewest sig-figs. Addition/subtraction → round to the input with fewest decimal places.
- Rounding. Drop < 5 keeps the preceding digit; drop > 5 raises it. Drop exactly 5 → round-to-even (NCERT §1.3.2). So $2.745 \to 2.74$, $2.735 \to 2.74$.
- Intermediate steps. Carry one extra digit through, round only at the end.
- Order of magnitude. Write $N = a \times 10^b$; round $a$ to 1 if $a \le 5$, to 10 if $a > 5$. The resulting power is the order of magnitude. Earth (order 7) is 17 orders of magnitude larger than a hydrogen atom (order $-10$).