Q.Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
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🔒 Start your 14-day free trial to unlock the full solution →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) …
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 …
Concept: Dimensional Analysis
We assume the period T depends on length l, mass m, and gravity g as a power-law relation:
T=klambgc
where k is a dimensionless constant and a,b,c are exponents to be determined.
Writing dimensions: [T]=T, [l]=L, [m]=M, [g]=LT−2.
Substituting into the dimensional equation:
T=La⋅Mb⋅(LT−2)c=MbLa+cT−2c
Equating powers on both sides:
- Mass: b=0
- Length: a+c=0⟹a=−c …
Dimensional analysis reveals that the period of a simple pendulum depends only on its length and gravity, not mass. The result is T=kgl, where k is a dimensionless constant (which turns out to be 2π).
The method of dimensions rests on a powerful principle: any physically meaningful equation must be dimensionally consistent. When we don't know the exact form of a relationship but understand which variables matter, we can deduce the structure of the formula by demanding that both sides have the same dimensions.
For a simple pendulum, we suspect the period T depends on length l, mass m, and acceleration due to gravity g. We assume a power-law relationship:
T=k⋅la⋅mb⋅gc
where k is a dimensionless constant and a, b, c are exponents we need to find.
Step-by-step derivation
1. Write down the dimensions of each quantity
Every physical quantity can be expressed in terms of fundamental dimensions: mass [M], length [L], and time [T].
| Quantity | Symbol | Dimensions |
|---|---|---|
| Period | T | [T] |
| Length | l | [L] |
| Mass | m | [M] |
| Acceleration | g | [LT−2] |
2. Substitute dimensions into the assumed relationship
Replace each variable with its dimensional formula:
[T]=[L]a⋅[M]b⋅[LT−2]c
Simplify the right side:
[T]=[L]a⋅[M]b⋅[L]c⋅[T]−2c
[T]=[M]b⋅[L]a+c⋅[T]−2c
3. Equate the exponents of corresponding dimensions
For dimensional consistency, the exponents of [M], [L], and [T] on both sides must match.
Left side: [M]0⋅[L]0⋅[T]1
Right side: [M]b⋅[L]a+c⋅[T]−2c
This gives us three equations:
- For [M]: b=0
- For [L]: a+c=0
- For [T]: −2c=1
4. Solve the system of equations
From the time equation:
−2c=1⟹c=−21
From the length equation:
a+c=0⟹a=−c=21
The mass exponent is already determined:
b=0 …
Concept: Dimensional Analysis
We assume the time period T depends on l, m, and g as:
T∝lambgc
Step 1: Write dimensions
- [T]=[M0L0T1]
- [l]=[M0L1T0]
- [m]=[M1L0T0]
- [g]=[M0L1T−2]
Step 2: Equate dimensions
[M0L0T1]=[Mb][La+c][T−2c]
Step 3: Solve exponents
- For M: b=0
- For T: 1=−2c⇒c=−21
- For L: 0=a+c⇒a=21 …
Common Mistakes in Deriving the Time Period of a Simple Pendulum by Dimensional Analysis
Mistake 1: Assuming the period must depend on mass, since the bob has mass
What students do wrong:
Because the problem states the period depends on l, m, and g, students expect the mass exponent b to come out non-zero, and get confused (or think they made an arithmetic error) when solving the equations gives b=0.
Correct approach:
Dimensional analysis only tells you what the exponent equations allow -- it doesn't assume the answer. Solving [T]=[L]a[M]b[LT−2]c genuinely forces b=0: the left side, [T], has zero powers of M, so b must be exactly 0. This is a real physical result -- a heavier bob and a lighter bob on identical strings swing with the same period -- not an error.
✓ Rule: Trust the algebra. If the exponent equations give zero for a variable, that variable genuinely doesn't affect the answer.
Mistake 2: Mixing up the exponent equations for L and T
What students do wrong:
After substituting [g]=[LT−2], some students write the length equation as a+c=1 (forgetting the left side, [T], has zero powers of L) or the time equation as −2c=0.
Correct approach:
Write out both sides fully before equating exponents:
[M0L0T1]=[Mb][La+c][T−2c]
Matching M: b=0. Matching L: a+c=0. Matching T: −2c=1.
✓ Tip: Always write the left-hand side's exponents explicitly as M0L0T1 before comparing -- it's easy to silently drop the zero powers.
Mistake 3: Getting the sign of c wrong
What students do wrong:
Solving −2c=1 as c=2 instead of c=−21.
Correct approach:
−2c=1⟹c=−21
Then a=−c=21.
✓ Tip: Double-check by plugging back in: T∝l1/2g−1/2 means T∝l/g, which has the right dimension [T] -- verify this at the end.
Mistake 4: Thinking dimensional analysis gives the constant k
What students do wrong:
After deriving T=kl/g, some students assume k=1 or try to derive k=2π from dimensions alone.
Correct approach:
Dimensional analysis can never determine a dimensionless constant -- it only fixes the exponents of the variables. The value k=2π comes from solving the actual equation of motion (a separate, non-dimensional argument), not from this method.
✓ Rule: Dimensional analysis gives the form of a relationship, never the numerical constant in front of it.
--- …
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. …
- 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. …
- 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. …
- 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. …
- 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. …
- 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. …
- 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. …
- 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. …
- 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). …
- 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. …
- 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). …
- 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. …
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