Give the number of significant figures in each of the following a. $1.230 \times 10^4$ b. 0.002030 c. $1.23 \times 10^4$ d. $1.89 \times 10^{-4}$

Give the number of significant figures in each of the following a. 1.230 × 10⁴ b. 0.002030 c. 1.23 × 10⁴ d. 1.

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1.230 \times 10^4

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Concept understanding — Significant Figures

Significant figures are the digits in a measurement that are known reliably, plus the first uncertain one. They are how a number honestly advertises its own precision: writing a length as 2.50 m claims more than 2.5 m, because the trailing zero says the hundredths place was actually measured. Two skills live here — counting the significant figures a number already carries, and not manufacturing precision when you compute with them.

1 — The least count sets the precision. Every instrument can resolve only down to its least count (LC) — the smallest division it can read. A metre scale marked in millimetres has LC = 1 mm; a vernier calliper has LC = 1 MSD − 1 VSD (equivalently 1 MSD / n when n vernier divisions span n−1 main divisions), typically 0.1 mm; a screw gauge / micrometer has LC = pitch / (number of circular-scale divisions), typically 0.01 mm. A measurement is read as main-scale reading + (coinciding division × LC), corrected for any zero error (a non-zero reading when the jaws are closed: a positive zero error is subtracted, a negative one is added).

2 — Counting significant figures. The rules: (i) every non-zero digit is significant; (ii) zeros between non-zero digits are significant (3.05 → 3 s.f.); (iii) leading zeros are never significant — they only fix the decimal point (0.0047 → 2 s.f.); (iv) trailing zeros are significant only if there is a decimal point (4.50 → 3 s.f., but 4500 is ambiguous); (v) scientific notation removes the ambiguity — 4.5 × 10³ shows 2 s.f., 4.50 × 10³ shows 3. A change of unit never changes the count: 5.60 cm and 0.0560 m both have 3 s.f.

3 — Rounding. To round to a required number of significant figures or decimal places: if the first dropped digit is > 5 round up, < 5 round down, and for exactly 5 the common convention rounds up (some texts round to the nearest even digit — state which you use). Rounding to N significant figures and to N decimal places are different operations — don't confuse them.

4 — Arithmetic doesn't create precision. The result of a calculation can be no more precise than its least-precise input. Addition and subtraction: the result keeps the least number of decimal places among the operands (12.3 + 4.56 = 16.9, one decimal). Multiplication and division: the result keeps the least number of significant figures (2.5 × 3.42 = 8.6, two s.f.). The classic trap is applying the wrong rule — using least-significant-figures on a sum, or least-decimals on a product.

5 — Exact numbers carry infinite precision. Counted objects (a set of 20 readings), defined conversion factors (1 inch = 2.54 cm exactly), and pure mathematical constants (the 2 in 2πr, the 4/3 in a sphere's volume) have infinitely many significant figures and never limit a result — only the measured quantities do.

Order of magnitude. Rounding a quantity to the nearest power of ten gives its order of magnitude: write it as a × 10ⁿ with 1 ≤ a < 10, then bump the power up by one when a ≥ √10 ≈ 3.16. So 6.0 × 10⁷ is of order 10⁸, while 2.0 × 10⁷ is of order 10⁷.

How this concept is examined. JEE Main asks these as fast single-correct or numerical items: count the significant figures in a given measurement, find the least count or the corrected reading of a vernier / screw gauge, round to a stated precision, or report the significant figures in the result of an addition or a multiplication. The physics is minimal; the marks reward applying the right rule — decimals for sums, significant figures for products — and keeping exact numbers out of the precision count.

Significant figures are introduced in the NCERT Class 11 Physics Units and Measurement chapter, and 'significant figures rules class 11 physics' or 'significant figures important questions' are consistently searched terms during board exam preparation. This rounding-and-precision framework is also a quick, reliable scoring topic in JEE Main and state CET physics papers.

a. $1.230 \times 10^4$ — coefficient 1.230 has four digits, all significant (scientific notation) → 4.

b. 0.002030 — leading zeros not significant; digits 2, 0, 3, 0 are significant (the internal zero and the trailing zero after the decimal point both count) → 4.

c. $1.23 \times 10^4$ — coefficient 1.23 → 3.

d. $1.89 \times 10^{-4}$ — coefficient 1.89 → 3.

✓Final answer

a. 4 b. 4 c. 3 d. 3

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

Showing the 12 most recent of 13 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. The significant figures of number 20340 are :$
(a) 3
(b) 4
(c) 5
(d) 1
›Reveal solutionSolution
20340 has 4 significant figures: the trailing zero after the last non-zero digit is not counted as significant when there is no decimal point.

Rules for counting significant figures:

All non-zero digits are significant.

Zeros between two non-zero digits are significant.

Trailing zeros in a number with no decimal point are NOT significant — they only fix the position of the decimal point / order of magnitude.

Applying this to 20340:

2 → significant

0 (between 2 and 3) → significant

3 → significant

4 → significant

trailing 0 → not significant (placeholder only)

So the number of significant figures = 4.

✓Final answer
$The correct option is (b) 4.$
CBSE 2026Set sz1 markMCQ
$Q. The number of significant figures in 6638 is:$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution

All non-zero digits in a measured number are significant, so 6638 has 4 significant figures.

The rules for counting significant figures state that all non-zero digits are significant. In the number 6638, the digits are 6, 6, 3, and 8 — none of them is zero, so all four count as significant figures.

✓Final answer
$The correct option is (d) 4.$
CBSE 2026Set ANNUAL1 mark
$Q. Write the answer in one word/one sentence: How many significant digits are there in the number 0.03000?$
›Reveal solutionSolution
0.03000 has 4 significant figures: the '3' and the three trailing zeros after it.

Rules for counting significant figures:

All non-zero digits are significant.

Zeros between non-zero digits are significant.

Leading zeros (before the first non-zero digit) are NOT significant — they only fix the position of the decimal point.

Trailing zeros after a decimal point ARE significant, because they indicate the precision to which the quantity was measured.

In 0.03000:

The zeros before the '3' (i.e., '0.0') are leading zeros — not significant.

The digit '3' is significant.

The three zeros after the '3' are trailing zeros after the decimal point — significant.

So the significant digits are 3, 0, 0, 0 — a total of 4.

✓Final answer
$The number 0.03000 has 4 significant figures.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. The numbers of significant digits in 0.0065 and 2.0065 are respectively$
(a) 4, 5
(b) 2, 3
(c) 4, 3
(d) 2, 5
›Reveal solutionSolution
0.0065 -> 2 significant figures; 2.0065 -> 5 significant figures. Answer (D).

Rules for significant figures:

Leading zeros (zeros only used to fix the decimal point) are NOT significant.

Zeros trapped between non-zero digits ARE significant.

For 0.0065: the two zeros after the decimal are leading zeros, so only 6 and 5 count -> 2 significant figures.

For 2.0065: the digits are 2, 0, 0, 6, 5. Here the zeros lie between non-zero digits, so all five count -> 5 significant figures.

✓Final answer
$(D) 2, 5.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. A cube has a side of length 1.2 x 10^-2 m. Its volume up to correct significant figures is$
(a) 1.7 x 10^-6 m^3
(b) 1.73 x 10^-6 m^3
(c) 1.78 x 10^-6 m^3
(d) 1.732 x 10^-6 m^3
›Reveal solutionSolution

Volume = side^3 = 1.728 x 10^-6 m^3, but since the given length has only 2 significant figures, the answer must be rounded to 2 sig figs: 1.7 x 10^-6 m^3.

Side of cube, a = 1.2 x 10^-2 m. This value has exactly 2 significant figures (the digits 1 and 2).

Volume V = a^3 = (1.2 x 10^-2)^3 = 1.728 x 10^-6 m^3 (raw calculator value).

By the rules of significant figures, the result of a multiplication/power operation cannot have more significant figures than the least precise measurement used in the calculation. Since 'a' has only 2 sig figs, V must be rounded to 2 sig figs: 1.728 -> 1.7.

So V = 1.7 x 10^-6 m^3.

✓Final answer
$(a) 1.7 x 10^-6 m^3.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. State the number of significant figures in 0.06900$
(a) 1
(b) 2
(c) 4
(d) 3
›Reveal solutionSolution
0.06900 has 4 significant figures.

Rules for counting significant figures: all non-zero digits are significant; zeros between non-zero digits are significant; leading zeros (before the first non-zero digit) are NEVER significant — they only locate the decimal point; trailing zeros after the decimal point, once a non-zero digit has appeared, ARE significant (they show measurement precision).

In 0.06900:

The zeros before the 6 (0.0) are leading zeros → not significant.

The digits 6 and 9 are non-zero → significant.

The two zeros after the 9 are trailing zeros after the decimal point → significant.

So the significant digits are 6, 9, 0, 0 → 4 significant figures.

✓Final answer
$The correct option is (c) 4 significant figures.$
CBSE 2025Set hz1 markMCQ
$Q. The number of significant figures in 2.64 x 10^24 Kg is:$
(a) 3
(b) 4
(c) 24
(d) 1
›Reveal solutionSolution

Only the digits actually measured count as significant; a power-of-ten multiplier in scientific notation is never counted. 2.64 x 10^24 kg has 3 significant figures.

When a number is written in scientific notation as N x 10^n, the significant figures are exactly the digits in N (the coefficient), regardless of how large or small the exponent n is. This convention exists precisely so that changing units (which changes only the exponent) never changes how precisely a quantity is known.

Here N = 2.64, which has three digits: 2, 6 and 4. All three are significant (the leading digit is non-zero, and both trailing digits are explicitly written, so they are taken as reliably known). The factor 10^24 is only a scale factor and contributes nothing to the count.

✓Final answer
$The correct option is (a) 3 significant figures.$
CBSE 2025Set ANNUAL1 markMCQ
$6.0023\, \text{g cm}^{-3}$
(a) 5
(b) 3
(c) 1
(d) 4
›Reveal solutionSolution

In $6.0023\ \text{g cm}^{-3}$ , every digit — 6, 0, 0, 2, 3 — is significant because zeros lying between two non-zero digits are always counted as significant figures.

Counting the digits of $6.0023$ : 6 (non-zero, significant), 0 (between 6 and 0, significant), 0 (between 6 and 2, significant), 2 (non-zero, significant), 3 (non-zero, significant).

So the number of significant figures $=5$ .

✓Final answer
$(a) 5$
CBSE 2025Set sz1 markMCQ
$Q. The numbers 2.745 and 2.735 on rounding off to 3 significant figures will give: (A) 2.75 and 2.74 (B) 2.74 and 2.73 (C) 2.75 and 2.73 (D) 2.74 and 2.74$
›Reveal solutionSolution
Using the round-half-to-even rule, both 2.745 and 2.735 round to 2.74 to three significant figures.

Rounding to 3 significant figures here means keeping two decimal places. The dropped digit is 5 in both numbers, so the special rule applies: raise the preceding digit by 1 if it is odd, leave it unchanged if it is even.

2.745 -> preceding digit is 4 (even) -> left unchanged -> 2.74

2.735 -> preceding digit is 3 (odd) -> raised by 1 -> 2.74

Hence both round to 2.74.

✓Final answer
$The correct option is (D) 2.74 and 2.74.$
CBSE 2024Set sz1 markMCQ
$Q. The number of significant numbers in 23.023 is: (A) 2 (B) 3 (C) 4 (D) 5$
›Reveal solutionSolution

23.023 has 5 significant figures — every digit counts because the internal zero is sandwiched between non-zero digits.

Rules for counting significant figures: (1) all non-zero digits are significant, (2) zeros between two non-zero digits are significant, (3) leading zeros are never significant, (4) trailing zeros after a decimal point are significant.

In 23.023: the digits are 2, 3, 0, 2, 3. The zero is sandwiched between the 3 and the 2, so by rule (2) it is significant. All five digits count.

✓Final answer
$The correct option is (D) 5 significant figures.$
CBSE 2022Set ANNUAL1 markMCQ
$Q. The number of significant figures in 0.007 m2 is$
(a) 1
(b) 3
(c) 4
(d) 7
›Reveal solutionSolution

$0.007\ \text{m}^2$ has only $1$ significant figure — the digit $7$ .

Significant figures are the digits in a measurement that carry real information about its precision. The rule for zeros is: zeros that appear before the first non-zero digit (leading zeros) are never significant — they only locate the decimal point and depend on the choice of unit, not on measurement precision.

Writing $0.007\ \text{m}^2 = 7\times10^{-3}\ \text{m}^2$ makes this explicit: in scientific notation only the digits in the coefficient ( $7$ ) count, and the power of ten carries no significance information. The three zeros in $0.007$ (one before the decimal point and two after it, before the $7$ ) are all leading zeros, so none of them is significant.

Hence $0.007\ \text{m}^2$ has exactly $1$ significant figure.

✓Final answer
$The correct option is (a) 1.$
CBSE 2019Set hz1 mark
$Q. What is the number of significant digits in 0.005 m^2 ?$
›Reveal solutionSolution
Leading zeros are never significant, so 0.005 m^2 has exactly 1 significant figure.

Rules for significant figures:

All non-zero digits are significant.

Zeros between two non-zero digits are significant.

Leading zeros (to the left of the first non-zero digit) are NOT significant - they only locate the decimal point.

Trailing zeros after a decimal point ARE significant.

In 0.005, the digits before the '5' (the zero before the decimal and the two zeros after it) are all leading zeros, so none of them count. Only the digit 5 is significant.

This becomes clearer if the number is written in scientific notation: 0.005 = 5 x 10^-3, where only the coefficient '5' carries significant-figure information.

✓Final answer
$0.005 m^2 has only 1 significant digit (the digit 5).$

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Q.Give the number of significant figures in each of the following a. 1.230×1041.230 \times 10^41.230×104 b. 0.002030 c. 1.23×1041.23 \times 10^41.23×104 d. 1.89×10−41.89 \times 10^{-4}1.89×10−4

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Q.Give the number of significant figures in each of the following a. $1.230 \times 10^4$ b. 0.002030 c. $1.23 \times 10^4$ d. $1.89 \times 10^{-4}$

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