Nuclear radius R has a dependence on the mass number (A) as $R = 1.3 \times 10^{-15} A^{1/3}$ m. For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre.

Nuclear radius R has a dependence on the mass number (A) as R = 1.3 × 10⁻¹⁵ A1/3 m. For a nucleus of mass numb

Numerical · Q16

R = 1.3 \times 10^{-15} A^{1/3}

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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. …

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Step 1. $A^{1/3} = 125^{1/3} = 5$ (since $5^3 = 125$ ).

Step 2. $R = 1.3\times10^{-15} \times 5 = 6.5\times10^{-15}\ \text{m}$ .

Step 3. Applying the chapter's own order-of-magnitude convention (section 1.9: $A\times10^n$ with $0.5\le A<5$ ), $6.5\times10^{-15}\ \text{m}$ must be renormalized since $6.5$ is outside $[0.5,5)$ : $6.5\times10^{-15} = 0.65\times10^{-14}\ \text{m}$ , giving order of magnitude $-14$ . …

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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 …
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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 — …

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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: …
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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. …
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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).

…

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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. …
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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.

…

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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.

…

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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. …

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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.

…

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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.

…

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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. …
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Q.Nuclear radius R has a dependence on the mass number (A) as R=1.3×10−15A1/3R = 1.3 \times 10^{-15} A^{1/3}R=1.3×10−15A1/3 m. For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre.

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Q.Nuclear radius R has a dependence on the mass number (A) as $R = 1.3 \times 10^{-15} A^{1/3}$ m. For a nucleus of mass number A=125, obtain the order of magnitude of R expressed in metre.

Unlock everything free for 14 days