Q.Two students performed the same experiment separately and each one of them recorded two readings of mass which are given below. Correct reading of mass is 3.0 g. On the basis of given data, mark the correct option out of the following statements. Student A — Readings:
Student B — Readings:
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.
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 m=100 cm are exact by definition.
When in doubt, write the number in scientific notation. 1.20×103 clearly has 3 sig figs, while 1.2×103 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.
3.14×2.5=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)
These two rules are different! For multiplication, count sig figs. For addition, count decimal places. Mixing them up is the most common mistake.
A Concrete Example
You measure a rectangular field:
- Length: 152.3 m (4 sig figs)
- Width: 45.0 m (3 sig figs)
Area = 152.3×45.0=6853.5 m²
But your width measurement only has 3 sig figs, so you report 6.85×103 m² (or 6850 m², but that's ambiguous — use scientific notation).
The Big Picture
Significant figures aren't about being pedantic. They're about honesty in science. When you write 3.0 instead of 3, you're telling the reader: "I measured this to the tenths place, and it was exactly 3.0 — not 2.9, not 3.1." That's valuable information.
Final rule of thumb: Your answer cannot be more precise than your least precise measurement. Sig figs enforce that.
Queries such as "significant figures rules class 11 physics" and "significant figures calculation examples" are common around exam season, reflecting how central this topic is to the Units and Measurements chapter of the NCERT/CBSE Class 11 Physics curriculum. It's also a frequent source of numerical-based questions in JEE Main and NEET.
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±0.05 (uncertainty in the tenths place)
- B=4.56±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:
Δ(A+B)=(ΔA)2+(ΔB)2≈0.052+0.0052≈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:
12.3×4.56=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±0.05 → relative uncertainty = 12.30.05≈0.00407 (0.407%)
- B=4.56±0.005 → relative uncertainty = 4.560.005≈0.00110 (0.110%)
For multiplication, relative uncertainties add (approximately):
A×BΔ(A×B)≈(AΔA)2+(BΔB)2
Plugging in:
≈0.004072+0.001102≈0.00422 (0.422%)
Now, the absolute uncertainty in the product:
Δ(A×B)≈0.00422×(12.3×4.56)≈0.00422×56.088≈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
| Operation | Rule | Why different? |
|---|---|---|
| + / − | Decimal places | Uncertainty is absolute — it depends on the position of the last digit |
| × / ÷ | Sig figs | Uncertainty is relative — it depends on the fraction of the value |
Example to see the difference:
- 1000+0.001=1000 (decimal places rule: 1000 has 0 decimal places, so result is 1000)
- 1000×0.001=1 (sig figs rule: 1000 has 4 sig figs? Actually ambiguous — but if 1000 has 1 sig fig, result is 1×100)
5. The Deeper Reason: It’s All About Honest Reporting
The rules exist because:
- Measurements have inherent uncertainty — no measurement is exact.
- Calculations propagate uncertainty — the result cannot be more precise than the least precise input.
- Sig figs are a practical shortcut — they avoid doing full error propagation for every calculation, while still giving a reasonable estimate of precision.
Bottom line: The rules aren’t arbitrary — they’re derived from the mathematics of uncertainty. When you round to the correct number of sig figs, you’re saying: “This is how precisely I actually know the answer, given the precision of my measurements.”
Quick Exam Tip
- Addition/Subtraction: Look at decimal places — the weakest link is the one with fewest decimals.
- Multiplication/Division: Look at sig figs — the weakest link is the one with fewest sig figs.
- Mixed operations: Follow order of operations, applying the appropriate rule at each step.
The key idea here is distinguishing between accuracy and precision in experimental measurements.
- Accuracy refers to how close a measurement (or the average of multiple measurements) is to the true or accepted value.
- Precision refers to how close multiple measurements are to each other, indicating the reproducibility of the measurements.
Let's analyze each student's data against the correct reading of 3.0 g:
-
Student A:
- Readings: 3.01 g, 2.99 g
- Average reading: 23.01+2.99=26.00=3.00 g. This average is exactly the correct reading, so Student A's results are accurate.
- The readings (3.01 g and 2.99 g) are very close to each other (difference of 0.02 g), indicating high reproducibility. So, Student A's results are precise.
-
Student B:
- Readings: 3.05 g, 2.95 g
- Average reading: 23.05+2.95=26.00=3.00 g. This average is exactly the correct reading, so Student B's results are accurate.
- The readings (3.05 g and 2.95 g) are more spread out (difference of 0.10 g) compared to Student A's readings. While the average is accurate, the individual readings are not as close to each other as Student A's, making them less precise.
Comparing the options, Student A's results are both accurate and precise.
The correct statement is that the results of student A are both precise and accurate.
Accuracy refers to how close measurements are to the true value, while precision refers to how close repeated measurements are to each other. Student A's results are both accurate and precise, while Student B's results are accurate but not precise. Therefore, the correct option is (ii).
When evaluating experimental data, two key terms are used: accuracy and precision. Understanding the distinction between them is crucial.
Accuracy describes how close a measured value (or the average of several measured values) is to the true or accepted value. A measurement is accurate if it has a small systematic error.
Precision describes how close repeated measurements are to each other. It reflects the reproducibility of the measurements. A set of measurements is precise if they are tightly clustered, regardless of whether they are close to the true value. Precision is related to random error.
Think of it like target practice:
- If all your shots are tightly grouped around the bullseye, you are both accurate and precise.
- If all your shots are tightly grouped but far from the bullseye, you are precise but not accurate.
- If your shots are scattered all over the target, but their average position is the bullseye, you are accurate but not precise.
- If your shots are scattered and far from the bullseye, you are neither accurate nor precise.
Let's apply these concepts to the given data.
-
Identify the True Value
The problem states that the correct reading of mass is 3.0 g. This is our true value.
-
Analyze Student A's Readings
Student A's readings are 3.01 g and 2.99 g.
-
Assess Accuracy for Student A:
To assess accuracy, we compare the average of the readings to the true value.
Average for Student A =23.01 g+2.99 g=26.00 g=3.00 g.
The average reading (3.00 g) is exactly equal to the true value (3.0 g). This indicates that Student A's results are accurate.
-
Assess Precision for Student A:
To assess precision, we look at how close the individual readings are to each other.
The readings are 3.01 g and 2.99 g. The difference between them is 3.01−2.99=0.02 g. This is a very small spread, meaning the readings are very close to each other. This indicates that Student A's results are precise.
-
Conclusion for Student A: Student A's results are both accurate and precise.
-
-
Analyze Student B's Readings
Student B's readings are 3.05 g and 2.95 g.
-
Assess Accuracy for Student B:
Average for Student B =23.05 g+2.95 g=26.00 g=3.00 g.
The average reading (3.00 g) is exactly equal to the true value (3.0 g). This indicates that Student B's results are accurate.
-
Assess Precision for Student B:
The readings are 3.05 g and 2.95 g. The difference between them is 3.05−2.95=0.10 g. This spread (0.10 g) is significantly larger than Student A's spread (0.02 g). The readings are not very close to each other. This indicates that Student B's results are not precise.
-
Conclusion for Student B: Student B's results are accurate but not precise.
-
-
Evaluate the Options
Based on our analysis:
- Student A: Accurate and Precise
- Student B: Accurate but Not Precise
Let's check the given options:
- Results of both the students are neither accurate nor precise. (Incorrect, Student A is both, Student B is accurate)
- Results of student A are both precise and accurate. (Correct, matches our conclusion for Student A)
- Results of student B are neither precise nor accurate. (Incorrect, Student B is accurate)
- Results of student B are both precise and accurate. (Incorrect, Student B is not precise)
Based on the analysis, Student A's results are both precise and accurate, while Student B's results are accurate but not precise. Therefore, the correct option is (ii).
Concept: Accuracy and Precision in Measurements
Accuracy = how close a measurement is to the true value (here, 3.0 g).
Precision = how close repeated measurements are to each other.
Method: The Spread-and-Deviation Check
Steps:
-
Find the average (mean) of each student’s readings.
- Student A: 23.01+2.99=3.00 g
- Student B: 23.05+2.95=3.00 g
-
Check accuracy — compare the mean to the true value (3.0 g).
- Both means are exactly 3.00 g → both are accurate.
-
Check precision — look at the spread between the two readings.
- Student A: readings differ by 3.01−2.99=0.02 g (very close) → precise.
- Student B: readings differ by 3.05−2.95=0.10 g (wider spread) → not precise.
-
Combine the two judgments:
- Student A: accurate (mean = true value) and precise (small spread).
- Student B: accurate (mean = true value) but not precise (large spread).
Final Answer
Correct option: (ii) Results of student A are both precise and accurate.
Key takeaway: Accuracy is about hitting the target (true value), precision is about grouping tightly — you can have one without the other.
Here’s a breakdown of the common mistakes students make on this concept (accuracy vs. precision) and how to avoid each.
Mistake 1: Confusing accuracy with precision
What students do wrong:
They think that if readings are close to each other, they must also be close to the true value. Or they assume that if the average is correct, the readings are automatically precise.
Example of error:
A student sees Student B’s readings (3.05 g and 2.95 g) and thinks “the average is 3.00 g, so it’s accurate” — but they ignore that each individual reading is far from the true value.
How to avoid:
- Accuracy = closeness of a single measurement to the true value.
- Precision = closeness of multiple measurements to each other.
Check each reading against the true value (3.0 g) for accuracy. Check the spread (range) between readings for precision.
- Student A: 3.01 and 2.99 → both near 3.0 (accurate) and very close to each other (precise).
- Student B: 3.05 and 2.95 → each is 0.05 g away from 3.0 (not accurate), but they are 0.10 g apart (still precise — they cluster together).
Key rule:
Precision does not guarantee accuracy, and accuracy does not guarantee precision.
Mistake 2: Using the average to judge accuracy
What students do wrong:
They calculate the mean of the two readings and compare it to the true value. If the mean equals 3.0 g, they call the results “accurate.”
Example of error:
Student B’s mean = 23.05+2.95=3.00 g. A student then marks option (iv) as correct.
How to avoid:
Accuracy is about individual readings, not the average. The true value is 3.0 g.
- Student A’s readings: 3.01 and 2.99 → each is within 0.01 g of 3.0 → accurate.
- Student B’s readings: 3.05 and 2.95 → each is 0.05 g away from 3.0 → not accurate.
The average can be misleading when errors cancel out. Always check each data point.
Mistake 3: Thinking “precise” means “exactly equal”
What students do wrong:
They think precision requires readings to be identical (e.g., both 3.01 g). If they differ even slightly, they call them “not precise.”
Example of error:
A student sees 3.01 and 2.99 and says “they are different, so not precise.”
How to avoid:
Precision refers to low spread — how tightly the readings cluster. A difference of 0.02 g (as in Student A) is very small, so it is precise. Student B’s readings differ by 0.10 g — still a small spread, so they are also precise.
Rule of thumb:
Compare the range to the size of the measurements. If the range is small relative to the values, the data is precise.
Mistake 4: Misreading the options — “neither” vs “both”
What students do wrong:
They pick an option that says “neither accurate nor precise” for Student B, even though Student B’s readings are precise (they cluster together).
How to avoid:
- For Student B:
- Accuracy? No (each reading is 0.05 g off).
- Precision? Yes (readings are close to each other). So “neither” is wrong. The correct description is “precise but not accurate.”
Check the options carefully:
- (i) says both students are neither — false for both.
- (ii) says Student A is both — true.
- (iii) says Student B is neither — false (they are precise).
- (iv) says Student B is both — false (they are not accurate).
Correct answer: (ii)
Quick Summary Table
| Student | Accurate? | Precise? | Common Mistake |
|---|---|---|---|
| A (3.01, 2.99) | Yes | Yes | Calling them “not precise” because readings differ |
| B (3.05, 2.95) | No | Yes | Calling them “accurate” because average is 3.00 |
Final tip:
Always define accuracy and precision before looking at the numbers. Write them down if needed:
- Accuracy = closeness to true value (check each reading).
- Precision = closeness of readings to each other (check the spread).
Showing the 12 most recent of 14 on this concept.
- AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQQ.The number of significant figures present in 0.010200×103 is (A) 7 (B) 5 (C) 6 (D) 10
›Reveal solutionSolution
Significant-figure rules: leading zeros never count, but trailing zeros after a decimal point always count. 0.010200 has 5 significant figures, and multiplying by 103 doesn't change that count.
Concept and Intuition
Significant figures reflect the precision of a measurement. Zeros purely used to fix the decimal point's position (leading zeros in a number <1) carry no information about precision and are excluded. But once you're past the first nonzero digit, EVERY digit — including trailing zeros written after a decimal point — is considered a deliberately recorded (significant) digit.
Step-by-Step Solution
- Write out 0.010200: digits are 0,0,1,0,2,0,0 after the decimal point.
- The two leading zeros (before the '1') are NOT significant — they only locate the decimal point.
- Starting from the first nonzero digit '1' through to the end: 1,0,2,0,0 — these ARE all significant (trailing zeros after a decimal point count) — that's 5 digits.
- Multiplying by 103 just rescales the number (equivalent to moving the decimal point); it introduces no new measured digits and removes none, so the significant-figure count stays 5.
- Answer: 5 significant figures.
Common Mistakes
- Miscounting the leading zeros as significant.
- Thinking multiplying by a power of 10 changes the number of significant figures.
✓Final answerThe correct option is (B) — 5.
ANSWER: B
- AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQQ.From the point of view of significant figures, which of the following is/are correct?(i) 11.3 cm+4 cm=15.3 cm(ii) 4.53 m−1.2 m=3.3 m(iii) 5.45 kg−3.2 kg=2.25 kg(iv) 84.8 cm+48.6 cm=133 cm (A)(ii) only (B)(iv) only (C)(i) &(iii) only (D)(ii) &(iv) only
›Reveal solutionSolution
The addition/subtraction rule keeps only as many decimal places as the least
precise term; checking all four, only statement (ii) obeys it. Answer: (A).
Concept and Intuition
When adding or subtracting measured quantities, the result cannot be more
precise than the least precise input. Concretely: round the final sum/ difference to the same number of decimal places as the term with the fewest decimal places (not the fewest significant figures — decimal places are what
matter here).
Step-by-Step Solution
- (i) 11.3 cm+4 cm: raw sum =15.3. "4 cm" has 0 decimal places, so the answer must be rounded to 0 decimals: 15 cm. The given value 15.3 keeps a spurious decimal — statement (i) is incorrect.
- (ii) 4.53−1.2: raw difference =3.33. "1.2" has 1 decimal place (fewer than 4.53's 2), so round to 1 decimal: 3.3. This matches the given value exactly — statement (ii) is correct.
- (iii) 5.45−3.2: raw difference =2.25. "3.2" has 1 decimal place, so round to 1 decimal: 2.3. The given value 2.25 keeps an extra decimal — statement (iii) is incorrect.
- (iv) 84.8+48.6: raw sum =133.4. Both terms have 1 decimal place, so the sum should also keep 1 decimal: 133.4. The given value 133 drops the decimal — statement (iv) is incorrect.
- Only (ii) obeys the significant-figures addition rule → option (A).
Common Mistakes
- Applying the multiplication rule (fewest significant figures) instead of the correct addition/subtraction rule (fewest decimal places).
- Not noticing that "4 cm" is effectively a 0-decimal-place measurement.
✓Final answerThe correct option is (A) — (ii) only.
ANSWER: A
- AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQQ.Which of the following have same number of significant figures? (A) 0.0025 (B) 0.0430 (C) 5005 (D) 500.0 (E) 2.003. The correct answer is (A) A & B only (B) B, C & D only (C) C, D & E only (D) A, C & D only
›Reveal solutionSolution
Counting significant figures carefully (leading zeros never count; trailing zeros count only with a decimal point or between nonzero digits) shows C, D, E all have 4 significant figures.
Concept and Intuition
The rules for significant figures: (i) all non-zero digits are significant;
(ii) zeros between non-zero digits are significant;
(iii) leading zeros (before the first non-zero digit) are never significant;
(iv) trailing zeros are significant only if the number has a decimal point.
Step-by-Step Solution
- (A) 0.0025: leading zeros (0.00) don't count; significant digits are "2","5" → 2 s.f.
- (B) 0.0430: leading zeros don't count; "4","3","0" — the trailing zero counts because there's a decimal point → 3 s.f.
- (C) 5005: all digits, including the internal zeros, are significant → 4 s.f.
- (D) 500.0: has a decimal point, so all four digits (5,0,0,0) are significant → 4 s.f.
- (E) 2.003: all four digits significant → 4 s.f.
- Matching count of 4: C, D, E.
Common Mistakes
- Treating trailing zeros in 500.0 as ambiguous the way they would be in a bare integer "5000" (without a decimal point) — the explicit decimal point here removes the ambiguity and makes them significant.
- Undercounting (B) by ignoring that its trailing zero (after the decimal point) is significant.
✓Final answerThe correct option is (C) — C, D & E only.
ANSWER: C
- AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQQ.Observe the following I) 0.0063 II) 132.00 III) 1004 The number of significant figures in I, II and III is respectively (A) 4, 3, 5 (B) 4, 5, 4 (C) 4, 3, 4 (D) 2, 5, 4
›Reveal solutionSolution
Applying the standard significant-figure rules to 0.0063, 132.00, and 1004 gives 2, 5, and 4 significant figures respectively.
Concept and Intuition
Significant figures rules: (1) all non-zero digits are significant; (2) zeros between two non-zero digits ("captive" zeros) are significant; (3) leading zeros (before the first non-zero digit) are never significant — they only locate the decimal point; (4) trailing zeros are significant only if the number contains a decimal point.
Step-by-Step Solution
- 0.0063: the leading zeros (0.00) merely position the decimal point and are not significant; only 6 and 3 count → 2 significant figures.
- 132.00: this number has a decimal point, so the trailing zeros after "132" are significant; digits are 1,3,2,0,0 → 5 significant figures.
- 1004: the zeros are sandwiched between the non-zero digits 1 and 4 (captive zeros), so they are significant; digits are 1,0,0,4 → 4 significant figures.
- So the counts, in order I, II, III, are 2, 5, 4.
Common Mistakes
- Counting leading zeros in 0.0063 as significant (giving a wrong count of 4 or 5).
- Forgetting that trailing zeros after a decimal point (as in 132.00) are always significant.
✓Final answerThe correct option is (D) — 2, 5, 4.
ANSWER: D
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.The number of significant figures in the simplification of 0.050.501(0.312−0.03) is (A) 1 (B) 3 (C) 2 (D) 5
›Reveal solutionSolution
The least-precise data (0.05, 0.03) carry one significant figure, so the answer has 1 significant figure.
Concept and Intuition
For × and ÷, the final result keeps the same number of significant figures as the quantity with the fewest. Leading zeros only fix the decimal point and are never significant, so 0.05 and 0.03 each have exactly one significant figure.
Step-by-Step Solution
- Evaluate the value: 0.312−0.03=0.282 and 0.050.501=10.02, giving 10.02×0.282≈2.83.
- Count significant figures in the data: 0.501 has 3, 0.05 has 1, 0.312 has 3, 0.03 has 1.
- The operation is dominated by division and multiplication, so the result is limited by the smallest count, which is 1 (from 0.05 and 0.03).
- Hence the simplification is reported to 1 significant figure (≈3).
Common Mistakes
- Counting the leading zeros in 0.05 or 0.03 as significant.
- Reporting the raw calculator value (2.83, i.e. 3 figures) without applying the least-significant-figure rule.
✓Final answerThe correct option is (A) — 1 significant figure.
ANSWER: A
NoteThis solution was worked out by our team and cross-checked by a second independent solve. The official answer key for this question could not be confirmed, so please cross-verify with the official paper where possible.
- AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQQ.Consider the following A) 0.0025 B) 500.0 C) 2.0034 Number of significant figures in A, B and C respectively, are (A) 5, 4, 4 (B) 2, 4, 2 (C) 4, 3, 2 (D) 2, 4, 5
›Reveal solutionSolution
This tests the significant-figure rules for leading zeros, trailing zeros after a decimal point, and captive zeros; the counts are 2, 4, 5.
Concept and Intuition
Significant figures reflect the precision of a measurement: leading zeros (before the first non-zero digit) are never significant since they only locate the decimal point; zeros between non-zero digits (captive zeros) are always significant; trailing zeros are significant only if the number has an explicit decimal point.
Step-by-Step Solution
- A=0.0025: the leading zeros (0.00) merely fix the decimal position and are not significant; only "25" counts → 2 significant figures.
- B=500.0: the number is written with an explicit decimal point, so ALL digits including the trailing zeros are significant → "5","0","0","0" → 4 significant figures.
- C=2.0034: every digit from the first non-zero digit onward is significant, including the captive zeros between 2 and 34 → "2","0","0","3","4" → 5 significant figures.
- So (A, B, C) = (2, 4, 5), matching option (D).
Common Mistakes
- Treating 500.0 as having only 1 or 3 significant figures by ignoring the decimal point rule.
- Miscounting captive zeros in 2.0034 as insignificant.
✓Final answerThe correct option is (D) — 2, 4, 5.
ANSWER: D
- AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQQ.The number of significant figures in 0.03240 is (A) 5 (B) 4 (C) 6 (D) 3
›Reveal solutionSolution
This tests the significant-figures rules for leading vs. trailing zeros: leading zeros are placeholders (never significant), while a zero written after the decimal point past the first nonzero digit is a measured, significant digit.
Concept and Intuition
Significant figures represent the precision of a measurement. Zeros purely used to fix the decimal point's position (leading zeros, e.g. in 0.032) carry no information about precision and are excluded. But once you've reached the first nonzero digit, every digit after it — including trailing zeros written explicitly after a decimal point — is understood to have been measured and counted.
Step-by-Step Solution
- Write out 0.03240 digit by digit: 0,0,3,2,4,0.
- The first two zeros (before the 3) are leading zeros — not significant, since they only locate the decimal point.
- Starting from the first nonzero digit 3, all subsequent digits count: 3,2,4,0.
- The final 0 is written explicitly after the decimal expansion, so per convention it IS significant (it signals precision to that digit).
- Total significant figures =4.
Common Mistakes
- Miscounting the leading zeros as significant, which would wrongly give 6.
- Dropping the trailing zero as "just a zero," which would wrongly give 3.
✓Final answerThe correct option is (B) — 4.
ANSWER: B
- AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQQ.The length of the side of a cube is 1.2×10−2 m. Its volume up to correct significant figures is (A) 1.732×10−6 m3 (B) 1.73×10−6 m3 (C) 1.70×10−6 m3 (D) 1.7×10−6 m3
›Reveal solutionSolution
(1.2×10−2)3=1.728×10−6, rounded to the 2 significant figures of the given data ⇒1.7×10−6 m3.
Concept and Intuition
A calculated quantity can be no more precise than the least precise measurement used to compute it. Since the side length 1.2×10−2 m has only 2 significant figures, the volume must also be reported to 2 significant figures, regardless of how many digits the raw calculation produces.
Step-by-Step Solution
- V=a3=(1.2×10−2)3=1.728×10−6 m3 (raw calculation).
- The given side length has 2 significant figures (1.2).
- Rounding 1.728 to 2 significant figures gives 1.7.
- So V=1.7×10−6 m3.
Common Mistakes
- Reporting all the calculator digits (1.732 or 1.73) without rounding to match the input precision.
- Rounding to 3 significant figures instead of 2.
✓Final answerThe correct option is (D) — 1.7×10−6 m3.
ANSWER: D
- AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQQ.The most accurate measurement among the following is (A) 20×10−3 m (B) 200×10−4 m (C) 2×10−2 m (D) 0.02 m
›Reveal solutionSolution
The number of significant figures determines measurement precision/accuracy; 200×10−4 m carries 3 significant figures, the most among the four choices, making it the most accurate.
Concept and Intuition
When a quantity is expressed in standard scientific notation a×10n, every digit in the coefficient a is treated as significant (this is precisely why scientific notation removes the ambiguity of trailing zeros). More significant figures in a measurement imply a smaller least count / finer precision of the measuring instrument, and hence greater accuracy.
Step-by-Step Solution
- 20×10−3 m — coefficient "20" has 2 significant figures.
- 200×10−4 m — coefficient "200" has 3 significant figures (since all digits in a scientific-notation mantissa are significant).
- 2×10−2 m — coefficient "2" has 1 significant figure.
- 0.02 m — has 1 significant figure (the leading zeros are not significant).
- Comparing sig figs: 2, 3, 1, 1 respectively — option (B) has the most (3), making it the most precisely/accurately expressed measurement.
Common Mistakes
- Assuming all four options represent numerically equal values and are therefore "equally accurate" — accuracy here is about precision of expression (sig figs), not numeric equivalence.
- Miscounting trailing zeros in a scientific-notation mantissa as insignificant.
✓Final answerThe correct option is (B) — 200×10−4 m.
ANSWER: B
- AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQQ.The number of significant figures in 4.870 m is (A) 3 (B) 4 (C) 2 (D) 1
›Reveal solutionSolution
A trailing zero after a decimal point is always significant, so 4.870 has 4 significant figures.
Concept and Intuition
Significant-figure rules: all non-zero digits are significant; zeros between non-zero digits are significant; trailing zeros are significant only if there is a decimal point present, because such a zero represents a deliberately measured/reported digit of precision (not just a placeholder).
Step-by-Step Solution
- Write out the digits of 4.870: 4, 8, 7, 0.
- 4 is non-zero — significant.
- 8 is non-zero — significant.
- 7 is non-zero — significant.
- The trailing 0 comes after a decimal point, so it is a measured digit of precision, not a placeholder — significant.
- Total significant figures =4.
Common Mistakes
- Discarding the trailing zero as "just a zero" — this rule only applies to trailing zeros in a number without a decimal point (e.g., 4870 m is ambiguous/could be 3 or 4 sig figs), not to 4.870.
✓Final answerThe correct option is (B) — 4.
ANSWER: B
- AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQQ.The number of significant figures in the quantity 5.6200 J is (A) 3 (B) 5 (C) 2 (D) 4
›Reveal solutionSolution
Tests the rules of significant figures; trailing zeros after a decimal point count as significant, so 5.6200 has 5 significant figures.
Concept and Intuition
Significant figures represent the precision of a measured quantity. The rules state: (i) all non-zero digits are significant;
(ii) zeros between non-zero digits are significant;
(iii) trailing zeros in a number with a decimal point are significant (they were deliberately recorded to show precision);
(iv) leading zeros are never significant.
Step-by-Step Solution
- Write out the digits of 5.6200: 5,6,2,0,0.
- 5 and 6 are non-zero digits — significant.
- 2 is a non-zero digit between other digits — significant.
- The two trailing zeros come after the decimal point, so by rule they are also significant (they indicate the measurement was precise to that last digit).
- Total significant digits = 5.
Common Mistakes
- Wrongly dropping trailing zeros after a decimal point, thinking they are "just placeholders" — they are not; they carry precision information.
- Confusing this with trailing zeros in a number without a decimal point (e.g., 5600), which would be ambiguous/not necessarily significant.
✓Final answerThe correct option is (B) — 5.
ANSWER: B
- AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQQ.If NA, NB and NC are the number of significant figures in A=0.001204 m, B=43120000 m and c=1.200 m respectively then (A) NA=NB=NC (B) NA>NB>NC (C) NA<NB<NC (D) NA>NB<NC
›Reveal solutionSolution
Applying the significant-figure rules (leading zeros never count, zeros between non-zero digits always count, trailing zeros count only with a decimal point) gives NA=NB=NC=4.
Concept and Intuition
Significant figures reflect measurement precision. The rules: (i) all non-zero digits are significant;
(ii) zeros sandwiched between non-zero digits are significant;
(iii) leading zeros (before the first non-zero digit) are never significant;
(iv) trailing zeros are significant only if the number has an explicit decimal point.
Step-by-Step Solution
- A=0.001204 m: The zeros before "1" are leading zeros — not significant. The digits 1,2,0,4 remain; the "0" between "2" and "4" is between non-zero digits, so it counts. NA=4.
- B=43120000 m: No decimal point is shown, so trailing zeros are ambiguous and conventionally not counted. Significant digits are 4,3,1,2. NB=4.
- C=1.200 m: A decimal point is present, so the trailing zeros ARE significant. Digits 1,2,0,0 all count. NC=4.
- Comparing: NA=NB=NC=4.
Common Mistakes
- Assuming trailing zeros in B (no decimal point) are automatically significant — they are not, by the standard convention.
- Missing that the internal zero in A (between 2 and 4) is significant, undercounting NA.
✓Final answerThe correct option is (A) — NA=NB=NC.
ANSWER: A
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