Physics
Units and Measurements
Significant Figures

Units and Measurements · JEE & NEET Physics

Significant Figures: notes and previous year questions

Which digits of a measurement can be trusted, how to count them, how to round, and how many to keep in a calculation.

3 JEE Main questions (2019–2024)
3 NEET questions (2022–2026)

Significant Figures in short

Significant figures are the sure digits of a measurement plus the first estimated digit.
More significant figures means a more precise measurement.
Non-zero digits, and zeros between them, always count.
Leading zeros never count; they only place the decimal point.

1What significant figures are

The significant figures of a measurement are all the digits we are sure of, plus the first digit we have to estimate. They show how precise a measurement is.

2Counting rules

Rule	Examples
1. Non-zero digits always count	123 → 3, 56.78 → 4, 9 → 1
2. Zeros between non-zero digits always count	1002 → 4, 50.03 → 4, 2.004 → 4
3. Leading zeros never count: they only place the decimal point	0.0025 → 2, 0.5 → 1
4. End zeros count only when there is a decimal point	2.500 → 4, 250.0 → 4, 100. → 3
No decimal point: end zeros are unclear	2500 → 2, 3 or 4

Key ideaQuick reference

Number	Significant figures	Why
123	3	all non-zero
0.00123	3	leading zeros do not count
1.0023	5	zeros between digits count
1.200	4	end zeros after a decimal point count
1200	2, 3 or 4	unclear: use scientific notation
1200.	4	the decimal point makes the end zeros count
$1.2 \times 10^3$	2	clear from scientific notation
0.0500	3	leading zeros do not; end zeros do
0.00340	3	3, 4 and the end zero
3.1400	5	end zeros after a decimal point count

3Scientific notation and exact numbers

In scientific notation, a number is written $a \times 10^n$ with $1 \le a < 10$ , and every digit of $a$ is significant. This removes all doubt about end zeros:

$4.5 \times 10^3$ has 2 significant figures;
$4.50 \times 10^3$ has 3;
$4.500 \times 10^3$ has 4.

Leading zeros disappear too: $0.0025 = 2.5 \times 10^{-3}$ plainly has 2 significant figures, and $0.00340 = 3.40 \times 10^{-3}$ has 3.

Exact numbers have unlimited significant figures, so they never limit an answer. They come from counting (30 students, 5 apples), from definitions (1 m = 100 cm, 1 hour = 60 min), and from pure factors like the 2 in $d = 2r$ .

Constants such as $\pi = 3.14159\ldots$ are not measured, but their decimals never end. Use more digits of $\pi$ than your data has, and it will never limit the answer.

4Rounding

To round to $n$ significant figures, look at the first digit you drop:

Below 5: leave the last kept digit as it is.
Above 5, or a 5 followed by other non-zero digits: round the last kept digit up.
Just a 5 (followed by nothing, or only zeros): make the last kept digit even. If it is odd, round up; if it is even, leave it.

5Adding and subtracting

When adding or subtracting, round the answer to the fewest decimal places among the numbers. A sum cannot be known further after the point than its least-known part.

6Multiplying and dividing

When multiplying or dividing, give the answer the same number of significant figures as the number with the fewest.

Powers and roots keep the significant figures of the original number: $(2.5)^2 = 6.25 \to 6.2$ , $(1.5)^3 = 3.375 \to 3.4$ , $\sqrt{16.0} = 4.00$ .

In longer calculations, keep at least one extra digit in every middle step and round only once, at the end.

Summary

Key ideas

Significant figures are the sure digits of a measurement plus the first estimated digit.
More significant figures means a more precise measurement.
Non-zero digits, and zeros between them, always count.
Leading zeros never count; they only place the decimal point.
End zeros count only when there is a decimal point; without one they are unclear.
In scientific notation, every digit shown is significant.
Exact numbers (counted or defined) never limit the answer; use enough digits of π.
When rounding, a dropped digit below 5 keeps, above 5 rounds up, and a lone 5 goes to the even digit.
Adding and subtracting: keep the fewest decimal places.
Multiplying and dividing: keep the fewest significant figures.
Powers and roots keep the significant figures of the original number.
Keep extra digits in middle steps and round only once, at the end.

Every equation

Significant figures: $\text{sure digits} + \text{first estimated digit}$
Leading zeros: $0.0025 = 2.5 \times 10^{-3}\ \text{(2 figures)}$
End zeros with a point: $2.500\ \text{(4 figures)}$
Scientific notation: $4.50 \times 10^3\ \text{(3 figures)}$
Exact factor: $d = 2 \times 5.00 = 10.0\ \text{cm}$
Rounding, lone 5: $12.45 \to 12.4,\ 12.55 \to 12.6$
Adding: $12.3 + 0.45 + 0.006 = 12.756 \to 12.8$
Subtracting: $123.456 - 12.1 = 111.356 \to 111.4$
Multiplying: $2.5 \times 3.42 = 8.55 \to 8.6$
Dividing: $12.5 \div 0.25 = 5.0 \times 10^1$
Power: $(1.5)^3 = 3.375 \to 3.4$
Root: $\sqrt{16.0} = 4.00$

Previous year questions with solutions

Real JEE and NEET questions on significant figures. Try each one before you open the solution.

Q1NEET 2026One correct option

Each side of a metallic cube of mass 5.580 kg is measured to the 9.0 cm . Keeping the significant figures in view, the density of the material of the cube can be best expressed as $X\times {10}^{3} \mathrm{kg}m^{-3}$ where the value of $X$ is:

A7.654
B7.6
C7.65
D7.7

Show answer and solution

Answer: Option D

Density is mass divided by volume, so put the side into metres first: $9.0\ \mathrm{cm} = 0.090\ \mathrm{m}$ , and the volume is $0.090^3 = 7.29 \times 10^{-4}\ \mathrm{m^3}$ , which gives $\dfrac{5.580}{7.29 \times 10^{-4}} = 7.654 \times 10^{3}\ \mathrm{kg\,m^{-3}}$ . The side carries only two significant figures, so the density keeps two, and $7.654$ rounds to $7.7$ . Round once at the end — trimming the volume to $7.3 \times 10^{-4}$ on the way would have given $7.6$ and lost the mark.

Q2NEET 2023One correct option

The diameter of a spherical bob, when measured with vernier callipers yielded the following values : $3.33 \mathrm{cm},3.32 \mathrm{cm},3.34 \mathrm{cm},3.33 \mathrm{cm}$ and $3.32 \mathrm{cm}$ . The mean diameter to appropriate significant figures is :

A $3.328 \mathrm{cm}$
B $3.3 \mathrm{cm}$
C $3.33 \mathrm{cm}$
D $3.32 \mathrm{cm}$

Show answer and solution

Answer: Option C

Add the five readings and divide: $\dfrac{3.33 + 3.32 + 3.34 + 3.33 + 3.32}{5} = \dfrac{16.64}{5} = 3.328\ \mathrm{cm}$ . The $5$ is an exact count of readings, so it limits nothing, and the only limit is the readings themselves, each measured to three significant figures. Quote the mean to three as well, which rounds $3.328$ to $3.33\ \mathrm{cm}$ .

Q3NEET 2022One correct option

The area of a rectangular field (in m²) of length 55.3 m and breadth 25 m after rounding off the value for correct significant digits is

A138 $\times$ 10¹
B1382
C1382.5
D14 $\times$ 10²

Show answer and solution

Answer: Option D

Area is length times breadth: $55.3 \times 25 = 1382.5\ \mathrm{m^2}$ . The breadth is given as $25\ \mathrm{m}$ , only two significant figures, and a product carries no more figures than its least precise input, so the area is allowed two. Rounding $1382.5$ to two figures keeps the $1$ and the $3$ , and the $8$ behind them pushes the $3$ up: $1400\ \mathrm{m^2}$ , which option D writes as $14 \times 10^{2}$ .

Practice questions, easy to hard

Three questions from the significant figures practice ladder: one easy, one medium, one hard.

Q4One correct option

Round $9.97$ to two significant figures.

This one is awkward on purpose. The two figures you keep are $9$ and $9$ , and pushing that second $9$ up has to carry, the way $99$ becomes $100$ .

A $9.9$
B $9.0$
C $11$
D $10$

Show answer and solution

Answer: Option D

A $9$ has nowhere to go on its own, so pushing it up turns $9.9$ into $10$ . Written plainly the answer is $10$ , and written as $1.0 \times 10^{1}$ it also shows the two figures you are claiming.

Q5One correct option

Here are the two rules side by side, on the same pair of numbers, $12.5$ and $0.24$ .

The sum comes to $12.74$ and the product comes to exactly $3$ . Which line writes both of them properly?

ASum $12.74$ , product $3.00$
BSum $12.7$ , product $3.0$
CSum $12.7$ , product $3.00$
DSum $12.74$ , product $3.0$

Show answer and solution

Answer: Option B

For the sum you count decimal places: $12.5$ has one, so the sum gets one, $12.7$ . For the product you count significant figures: $0.24$ has two, so the product gets two, written $3.0$ . Same two numbers, different rule, different answer.