Q.Each side of a cube is measured to be $7.203\ \text{m}$. What are the total surface area and the volume of the cube to appropriate significant figures?

Showing the 12 most recent of 14 on this concept.

1. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.The number of significant figures present in $0.010200 \times 10^3$ 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 $10^3$ 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

   1. Write out $0.010200$: digits are $0,0,1,0,2,0,0$ after the decimal point.
   2. The two leading zeros (before the '1') are NOT significant — they only locate the decimal point.
   3. 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. …

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2. AP EAPCET 2026Set ap-2026-05-19-FN1 markMCQ
   Q.From the point of view of significant figures, which of the following is/are correct?

   (i) $11.3\ \text{cm} + 4\ \text{cm} = 15.3\ \text{cm}$

   (ii) $4.53\ \text{m} - 1.2\ \text{m} = 3.3\ \text{m}$

   (iii) $5.45\ \text{kg} - 3.2\ \text{kg} = 2.25\ \text{kg}$

   (iv) $84.8\ \text{cm} + 48.6\ \text{cm} = 133\ \text{cm}$ (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

   1. (i) $11.3\ \text{cm} + 4\ \text{cm}$: 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.
   2. (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.
   3. (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. …

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3. AP EAPCET 2026Set ap-2026-05-20-AN1 markMCQ
   Q.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

   1. (A) 0.0025: leading zeros (0.00) don't count; significant digits are "2","5" $\to$ 2 s.f.
   2. (B) 0.0430: leading zeros don't count; "4","3","0" — the trailing zero counts because there's a decimal point $\to$ 3 s.f.
   3. (C) 5005: all digits, including the internal zeros, are significant $\to$ 4 s.f.
   4. (D) 500.0: has a decimal point, so all four digits (5,0,0,0) are significant $\to$ 4 s.f. …

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4. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.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

   1. 0.0063: the leading zeros (0.00) merely position the decimal point and are not significant; only 6 and 3 count → 2 significant figures.
   2. 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. …

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5. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.The number of significant figures in the simplification of $\dfrac{0.501}{0.05}(0.312 - 0.03)$ 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 $\times$ and $\div$, 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

   1. Evaluate the value: $0.312-0.03 = 0.282$ and $\dfrac{0.501}{0.05} = 10.02$, giving $10.02 \times 0.282 \approx 2.83$.
   2. Count significant figures in the data: $0.501$ has 3, $0.05$ has 1, $0.312$ has 3, $0.03$ has 1.
   3. The operation is dominated by division and multiplication, so the result is limited by the smallest count, which is $1$ (from $0.05$ and $0.03$).
   4. Hence the simplification is reported to $1$ significant figure ($\approx 3$).

   Common Mistakes

   • Counting the leading zeros in $0.05$ or $0.03$ as significant. …

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6. AP EAPCET 2025Set eng-2025-05-22-AN1 markMCQ
   Q.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

   1. $A = 0.0025$: the leading zeros (0.00) merely fix the decimal position and are not significant; only "25" counts → 2 significant figures.
   2. $B = 500.0$: 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. …

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7. AP EAPCET 2025Set eng-2025-05-26-FN1 markMCQ
   Q.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

   1. Write out $0.03240$ digit by digit: $0,0,3,2,4,0$.
   2. The first two zeros (before the $3$) are leading zeros — not significant, since they only locate the decimal point.
   3. Starting from the first nonzero digit $3$, all subsequent digits count: $3,2,4,0$. …

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8. AP EAPCET 2024Set eng-2024-05-21-AN1 markMCQ
   Q.The length of the side of a cube is $1.2 \times 10^{-2}$ m. Its volume up to correct significant figures is (A) $1.732 \times 10^{-6}\ \mathrm{m^3}$ (B) $1.73 \times 10^{-6}\ \mathrm{m^3}$ (C) $1.70 \times 10^{-6}\ \mathrm{m^3}$ (D) $1.7 \times 10^{-6}\ \mathrm{m^3}$

   ›Reveal solutionSolution

   $(1.2\times10^{-2})^3 = 1.728\times10^{-6}$, rounded to the 2 significant figures of the given data $\Rightarrow 1.7\times10^{-6}\ \mathrm{m^3}$.

   Concept and Intuition

   A calculated quantity can be no more precise than the least precise measurement used to compute it. Since the side length $1.2\times10^{-2}$ 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

   1. $V=a^3=(1.2\times10^{-2})^3 = 1.728\times10^{-6}\ \mathrm{m^3}$ (raw calculation).
   2. The given side length has 2 significant figures ($1.2$).
   3. Rounding $1.728$ to 2 significant figures gives $1.7$. …

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9. AP EAPCET 2023Set ap-2023-05-22-AN1 markMCQ
   Q.The most accurate measurement among the following is (A) $20 \times 10^{-3}$ m (B) $200 \times 10^{-4}$ m (C) $2 \times 10^{-2}$ m (D) $0.02$ m

   ›Reveal solutionSolution

   The number of significant figures determines measurement precision/accuracy; $200\times10^{-4}$ 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 $a \times 10^n$, 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

   1. $20\times10^{-3}$ m — coefficient "20" has 2 significant figures.
   2. $200\times10^{-4}$ m — coefficient "200" has 3 significant figures (since all digits in a scientific-notation mantissa are significant).
   3. $2\times10^{-2}$ m — coefficient "2" has 1 significant figure.
   4. $0.02$ m — has 1 significant figure (the leading zeros are not significant). …

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10. AP EAPCET 2023Set eng-2023-05-17-FN1 markMCQ
    Q.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

    1. Write out the digits of $4.870$: $4$, $8$, $7$, $0$.
    2. $4$ is non-zero — significant.
    3. $8$ is non-zero — significant.
    4. $7$ is non-zero — significant. …

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11. AP EAPCET 2022Set ap-2022-07-11-AN1 markMCQ
    Q.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

    1. Write out the digits of $5.6200$: $5, 6, 2, 0, 0$.
    2. $5$ and $6$ are non-zero digits — significant.
    3. $2$ is a non-zero digit between other digits — significant.
    4. 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). …

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12. AP EAPCET 2022Set eng-2022-07-04-FN1 markMCQ
    Q.If $N_A$, $N_B$ and $N_C$ are the number of significant figures in $A = 0.001204$ m, $B = 43120000$ m and $c = 1.200$ m respectively then (A) $N_A = N_B = N_C$ (B) $N_A > N_B > N_C$ (C) $N_A < N_B < N_C$ (D) $N_A > N_B < N_C$

    ›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 $N_A=N_B=N_C=4$.

    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

    1. $A=0.001204$ 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. $N_A=4$.
    2. $B=43120000$ m: No decimal point is shown, so trailing zeros are ambiguous and conventionally not counted. Significant digits are $4,3,1,2$. $N_B=4$. …

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