Two students performed the same experiment separately and each one of them recorded two readings of mass which

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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:

(i) 3.01 g,

(ii) 2.99 g
Student B — Readings:

(i) 3.05 g,

(ii) 2.95 g

(i) Results of both the students are neither accurate nor precise.

(ii) Results of student A are both precise and accurate.

(iii) Results of student B are neither precise nor accurate.

(iv) Results of student B are both precise and accurate.

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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.

Note

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 $1 \text{ m} = 100 \text{ cm}$ are exact by definition.

Tip

When in doubt, write the number in scientific notation. $1.20 \times 10^3$ clearly has 3 sig figs, while $1.2 \times 10^3$ 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.

$3.14 \times 2.5 = 7.85$ 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)

Watch out

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 = $152.3 \times 45.0 = 6853.5$ m²

But your width measurement only has 3 sig figs, so you report $6.85 \times 10^3$ 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.

Important

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:

$A = 12.3 \pm 0.05$ (uncertainty in the tenths place)
$B = 4.56 \pm 0.005$ (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:

\Delta(A+B) = \sqrt{(\Delta A)^2 + (\Delta B)^2} \approx \sqrt{0.05^2 + 0.005^2} \approx 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:

$12.3 \times 4.56 = 56.088$ → round to 56.1 (three sig figs, like both inputs)

Why this holds

Let’s use relative uncertainty (percentage error):

$A = 12.3 \pm 0.05$ → relative uncertainty = $\frac{0.05}{12.3} \approx 0.00407$ (0.407%)
$B = 4.56 \pm 0.005$ → relative uncertainty = $\frac{0.005}{4.56} \approx 0.00110$ (0.110%)

For multiplication, relative uncertainties add (approximately):

\frac{\Delta(A \times B)}{A \times B} \approx \sqrt{\left(\frac{\Delta A}{A}\right)^2 + \left(\frac{\Delta B}{B}\right)^2}

Plugging in:

\approx \sqrt{0.00407^2 + 0.00110^2} \approx 0.00422 \text{ (0.422\%)}

Now, the absolute uncertainty in the product:

\Delta(A \times B) \approx 0.00422 \times (12.3 \times 4.56) \approx 0.00422 \times 56.088 \approx 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)
$1000 \times 0.001 = 1$ (sig figs rule: 1000 has 4 sig figs? Actually ambiguous — but if 1000 has 1 sig fig, result is $1 \times 10^0$ )

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.

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 $3.0 \text{ g}$ . This is our true value.
Analyze Student A's Readings

Student A's readings are $3.01 \text{ g}$ and $2.99 \text{ g}$ .
- Assess Accuracy for Student A:
  
  To assess accuracy, we compare the average of the readings to the true value.
  
  Average for Student A $= \frac{3.01 \text{ g} + 2.99 \text{ g}}{2} = \frac{6.00 \text{ g}}{2} = 3.00 \text{ g}$ .
  
  The average reading ( $3.00 \text{ g}$ ) is exactly equal to the true value ( $3.0 \text{ g}$ ). 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 $3.01 \text{ g}$ and $2.99 \text{ g}$ . The difference between them is $3.01 - 2.99 = 0.02 \text{ g}$ . 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 $3.05 \text{ g}$ and $2.95 \text{ g}$ .
- Assess Accuracy for Student B:
  
  Average for Student B $= \frac{3.05 \text{ g} + 2.95 \text{ g}}{2} = \frac{6.00 \text{ g}}{2} = 3.00 \text{ g}$ .
  
  The average reading ( $3.00 \text{ g}$ ) is exactly equal to the true value ( $3.0 \text{ g}$ ). This indicates that Student B's results are accurate.
- Assess Precision for Student B:
  
  The readings are $3.05 \text{ g}$ and $2.95 \text{ g}$ . The difference between them is $3.05 - 2.95 = 0.10 \text{ g}$ . This spread ( $0.10 \text{ g}$ ) is significantly larger than Student A's spread ( $0.02 \text{ g}$ ). 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:
1. Results of both the students are neither accurate nor precise. (Incorrect, Student A is both, Student B is accurate)
2. Results of student A are both precise and accurate. (Correct, matches our conclusion for Student A)
3. Results of student B are neither precise nor accurate. (Incorrect, Student B is accurate)
4. Results of student B are both precise and accurate. (Incorrect, Student B is not precise)

✓Final answer

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) .

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 = $\frac{3.05 + 2.95}{2} = 3.00$ 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).

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 20 on this concept.

CBSE 2026Set ANNUAL1 markMCQ
$Q. How many significant figures does 0.057 have?$
(a) 2
(b) 4
(c) 3
(d) 0
›Reveal solutionSolution
Leading zeros used only to fix the decimal point are never significant. 0.057 has exactly 2 significant figures: 5 and 7.

Rule for counting 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 0.057, the digits are 0, 0, 5, 7. The first two zeros are leading zeros (before the first non-zero digit '5'), so by rule 3 they are not counted. That leaves only 5 and 7 as significant digits.

So 0.057 has 2 significant figures.

✓Final answer
$(a) 2 -- only the digits 5 and 7 are significant; the leading zeros just place the decimal point.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. How many significant figures are there in 0.0052?$
(a) 4
(b) 1
(c) 3
(d) 2
›Reveal solutionSolution

0.0052 has exactly 2 significant figures (5 and 2); leading zeros are not counted.

Significant figures are the meaningful digits in a measured or calculated quantity - they indicate the precision of a measurement. The rules for counting them: (1) all non-zero digits are significant; (2) zeros between non-zero digits are significant; (3) leading zeros (to the left of the first non-zero digit) are NEVER significant - they only serve to locate the decimal point; (4) trailing zeros after a decimal point ARE significant.

In 0.0052, reading left to right: 0, 0, 0, 5, 2. The three zeros before the '5' are leading zeros, so they are not counted. Only '5' and '2' are significant digits.

✓Final answer
$The correct option is (d) 2 significant figures.$
CBSE 2026Set ANNUAL1 markMCQ
$Q. If π = 3.14, then the value of π^2 is:$
(a) 9.860
(b) 9.9
(c) 9.86
(d) 9.8596
›Reveal solutionSolution

By the rules of significant figures, a result cannot have more significant figures than the least precise value used to calculate it; since pi = 3.14 has 3 significant figures, pi^2 must also be reported to 3 significant figures, i.e., 9.86.

When multiplying or dividing measured quantities, the result should be rounded off to the same number of significant figures as the quantity with the fewest significant figures used in the calculation. This is because a calculation cannot manufacture precision beyond what the original measurement actually had.

Here, pi is given as 3.14, which has 3 significant figures.

pi^2 = 3.14 x 3.14 = 9.8596 (this is the raw arithmetic result, but it has 5 significant figures, more precision than 3.14 actually supports)

Rounding 9.8596 to 3 significant figures:

9.8596 -> look at the 4th significant digit (5, followed by 9) to decide rounding of the 3rd digit -> rounds up to 9.86

So the properly rounded value of pi^2, respecting significant figures, is 9.86.

✓Final answer
$The correct option is (c) 9.86 — since pi = 3.14 has 3 significant figures, pi^2 must be rounded to 3 significant figures too, giving 9.86 (not the raw unrounded 9.8596).$
CBSE 2026Set ANNUAL1 mark
$Q. Fill in the blank: The number of significant figures in 2.005 is ______.$
›Reveal solutionSolution

2.005 has 4 significant figures because zeros sandwiched between non-zero digits always count as significant.

Rules for significant figures: (1) all non-zero digits are significant; (2) zeros between two non-zero digits (captive/sandwiched zeros) are significant; (3) leading zeros are not significant; (4) trailing zeros after a decimal point are significant.

In 2.005: the digit 2 is significant (non-zero), the two 0s lie between the non-zero digits 2 and 5 so they are captive zeros and are significant, and 5 is significant (non-zero). That gives 2, 0, 0, 5 = 4 significant figures.

✓Final answer
$The number of significant figures in 2.005 is 4.$
CBSE 2026Set ANN1 mark
$Q. Find the number of significant figures in the measurement, 0.04597 g.$
›Reveal solutionSolution

Leading zeros are not significant; the digits 4, 5, 9, 7 count, giving 4 significant figures.

Rule: zeros to the left of the first non-zero digit (leading zeros) are NOT significant; they only locate the decimal point. All non-zero digits are significant.

In 0.04597: the two zeros before 4 are leading zeros (not significant). The significant digits are 4, 5, 9, 7.

✓Final answer
$4 significant figures.$
CBSE 2025Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005 are:$
(a) 1
(b) 2
(c) 3
(d) 4
›Reveal solutionSolution
Every digit in 2.005 is significant because zeros sandwiched between non-zero digits always count.

Rules for counting significant figures:

All non-zero digits are significant.

Zeros between two non-zero digits are significant ("captive zeros").

Leading zeros (before the first non-zero digit) are NOT significant.

Trailing zeros after a decimal point ARE significant.

In $2.005$ : the digits are 2, 0, 0, 5.

'2' is a non-zero digit — significant.

The two '0's lie between the non-zero digits 2 and 5 — significant (captive zeros).

'5' is a non-zero digit — significant.

So all 4 digits count.

✓Final answer
$2.005 has 4 significant figures — option (d).$
CBSE 2024Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005 is:$
(a) Two
(b) Three
(c) Four
(d) Infinite
›Reveal solutionSolution
All four digits in 2.005 are significant because the zeros sit between non-zero digits.

Rules for counting significant figures:

All non-zero digits are significant: the digits 2 and 5 count.

Zeros between two non-zero digits (captive zeros) are always significant: both zeros in 2.005 sit between the 2 and the 5, so they count too.

So 2.005 → digits counted: 2, 0, 0, 5 = 4 significant figures.

✓Final answer
$2.005 has four significant figures (option c).$
CBSE 2024Set ANNUAL1 mark
$Q. Number of significant figures in 0.00300 is ____.$
›Reveal solutionSolution

0.00300 has 3 significant figures: the leading zeros used only to fix the decimal point are not counted, but trailing zeros after a decimal point ARE significant.

Rules for significant figures: leading zeros (0.00...) before the first non-zero digit are never significant -- they only locate the decimal point. Zeros appearing after the decimal point AND after a non-zero digit ARE significant, since they indicate real measured precision.

In 0.00300: the digits are 3, 0, 0 (after the leading 0.00). All three are significant.

✓Final answer
$The number of significant figures in 0.00300 is 3.$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Number of significant figures in 2.005:$
(a) 2
(b) 3
(c) 4
(d) Infinite
›Reveal solutionSolution
Captive zeros (zeros sandwiched between non-zero digits) always count as significant, so 2.005 has 4 significant figures.

Rules used:

All non-zero digits (2, 5) are significant.

Zeros between two non-zero digits (captive zeros) are significant.

In $2.005$ , reading left to right: $2$ (non-zero, significant), $0$ (between 2 and 0, significant), $0$ (between 0 and 5, significant), $5$ (non-zero, significant).

So all four digits — 2, 0, 0, 5 — are significant.

✓Final answer
$2.005 has 4 significant figures (option c).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. Round off the number 19.95 into three significant figures.$
(a) 20.1
(b) 19.9
(c) 19.5
(d) 20.0
›Reveal solutionSolution

Rounding 19.95 to three significant figures using the 'round half to even' convention gives 20.0.

19.95 has four significant figures (1, 9, 9, 5). To round to three significant figures, we must decide the fate of the third significant digit (the second 9), based on the digit being dropped (a 5, with nothing after it).

Rule for rounding off when the digit to be dropped is exactly 5: if the preceding digit is even, it is left unchanged; if the preceding digit is odd, it is increased by 1 (this is the 'round half to even' convention used for significant figures, and it avoids a systematic upward bias).

Here, the digit before the dropped 5 is 9, which is odd. So it is increased by 1: 9 + 1 = 10. This causes a carry into the digit before it:

19.9|5 becomes 20.0 after the carry propagates

So 19.95 rounded to three significant figures is 20.0.

✓Final answer
$The correct option is (d) 20.0.$
CBSE 2023Set ANNUAL1 mark
$Q. 5.74 g of a substance occupies 1.2 cm^3. Express its density in proper significant figures.$
›Reveal solutionSolution

Density = 5.74/1.2 = 4.7833... g/cm^3, rounded to 2 significant figures = 4.8 g/cm^3.

Rule for significant figures in division/multiplication: the result must be reported with the same number of significant figures as the input quantity that has the FEWEST significant figures. This keeps the calculated result from claiming more precision than the actual measurements support.

Step 1: Count significant figures in each given quantity.

Mass = 5.74 g has 3 significant figures.

Volume = 1.2 cm^3 has 2 significant figures.

Step 2: Compute the raw quotient.

Density = mass/volume = 5.74 / 1.2 = 4.7833... g/cm^3.

Step 3: Round to match the LOWER count of significant figures (2, from the volume).

Density = 4.8 g/cm^3.

Reporting more digits (e.g. 4.78 g/cm^3) would be false precision, since the volume was only known to 2 significant figures.

✓Final answer
$Density = 4.8 g/cm^3 (2 significant figures).$
CBSE 2023Set ANNUAL1 markMCQ
$Q. If radius of circle is 2.12 cm, then express its area must be:$
(a) 14 cm^2
(b) 14.1 cm^2
(c) 14.11 cm^2
(d) 14.1124 cm^2
›Reveal solutionSolution

Rounded to the correct significant figures, the area is 14.1 cm^2.

Area A = πr^2 = 3.14159 × (2.12)^2 = 3.14159 × 4.4944 = 14.1197 cm^2.

Since the given radius 2.12 cm has only 3 significant figures, the answer must be reported to 3 significant figures: 14.1 cm^2. Quoting more digits (14.11 or 14.1124) would falsely imply greater precision than the data allow.

✓Final answer
$(B) 14.1 cm^2.$

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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:

Concept understanding — Significant Figures Calculation

Significant Figures: The Art of Honest Measurement

The Core Idea

The Rules (Memorize These)

1. Non-zero digits are always significant

2. Zeros between non-zero digits are significant

3. Leading zeros are never significant

4. Trailing zeros are significant only if there's a decimal point

5. Exact numbers have infinite sig figs

Why This Matters: Calculations

Multiplication and Division

Addition and Subtraction

A Concrete Example

The Big Picture

Why this formula?

Significant Figures: Why the Rules Work

1. The Fundamental Idea: Uncertainty is the Key

2. Rule for Addition and Subtraction

Why this holds

3. Rule for Multiplication and Division

Why this holds

4. Why These Rules Are Different

5. The Deeper Reason: It’s All About Honest Reporting

Quick Exam Tip

Concept: Accuracy and Precision in Measurements

Method: The Spread-and-Deviation Check

Steps:

Final Answer

Mistake 1: Confusing accuracy with precision

Mistake 2: Using the average to judge accuracy

Mistake 3: Thinking “precise” means “exactly equal”

Mistake 4: Misreading the options — “neither” vs “both”

Quick Summary Table

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:

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