Physics 1st Paper · Classes 11-12 · Chapter 1 · 1.13

Error Propagation, Significant Figures and Rounding.

To find g with a pendulum you measure two things, the string length l and the period T, and both carry a little error. How much of that error ends up in g? The answer is a surprise: the error in T goes in twice over. This note first shows how errors combine, then how many digits a result may keep, and how to round it.

1. The power is the multiplier

Say a quantity x comes from two measured quantities y and z, as x = yᵐzⁿ. Take logarithms of both sides and differentiate, and you get the relation below. Each measurement's fractional error is multiplied by its power, and the results add. In the worst case every term pulls the same way, and that gives the maximum possible error.

δx/x = m δy/y + n δz/z

The meaning is plain: the higher a quantity's power, the more times its error is multiplied into the result. So the quantity with the largest power is the one to measure with the most care.

Before the formula, pull the pendulum's error budget yourself. Change the errors in l and T and see how much each adds to the total error in g.

In the pendulum formula g = 4π²l/T², l has power 1 and T has power 2 (it sits in the denominator, but for the maximum error the sign is dropped and the term adds). 4π² is a constant with no error. So

δg/g = δl/l + 2δT/T

Even a small error in T arrives in g doubled.

2. The book's examples

Example 1.5(1): l = (0.8 ± 0.01) m and T = (2.2 ± 0.10) s. Then Δg/g = 0.01/0.8 + 2 × 0.10/2.2 = 0.0125 + 0.0909 = 0.1034, which is 10.34%. Notice that l contributes only 1.25%; almost all the rest comes from T.

A higher-skills example runs on the same pattern: length (0.6 ± 0.01) m, period (2.4 ± 0.12) s. Δg/g = 0.01/0.6 + 2 × 0.12/2.4 = 0.0167 + 0.1 = 11.67%.

Example 1.5(2): a capacitor C = (2.5 ± 0.1) μF at V = (25 ± 0.2) V. The charge is Q = CV = 2.5 × 10⁻⁶ × 25 = 6.25 × 10⁻⁵ C. Both powers are 1, so ΔQ/Q = ΔC/C + ΔV/V = 4% + 0.8% = 4.8%. That makes ΔQ = 0.048 × 6.25 × 10⁻⁵ = 0.3 × 10⁻⁵ C, and the answer is Q = (6.25 ± 0.3) × 10⁻⁵ C.

A sphere's volume is V = (4/3)πR³, so R has power 3. In example 1.6(1), R = 5.3 ± 0.1, so ΔV/V = 3ΔR/R = 3 × 0.1/5.3 and the percentage error in the volume is 5.7%. For the same reason, a 2% error in the radius becomes a 3 × 2% = 6% error in the volume.

In a quotient the fractional errors still add; they never cancel. With M = 100 ± 2 kg and V = 100 ± 3 m³, the density is ρ = M/V = 1 kgm⁻³, its percentage error is 2% + 3% = 5%, and so Δρ = 0.05 kgm⁻³.

Example 1.8(2): a sphere's diameter is D = 0.11 cm, measured with an instrument of least count 0.001 cm. Since V ∝ D³, ΔV/V = 3 × 0.001/0.11 = 0.027, a maximum error of 2.7%.

Example 1.8(3): M = 90 ± 0.001 g and V = 9 ± 0.1 cm³, so ρ = 90/9 = 10 gcm⁻³. dρ/ρ = 0.001/90 + 0.1/9 = 0.00001 + 0.01111 = 0.01112, a 1.11% error, and dρ = 0.111 gcm⁻³. The mass hardly contributes here; the volume carries almost all of it.

Now combine them yourself. Set each term's power first, then write the total percentage error.

3. Significant figures

The significant figures of a measurement are the digits you are completely sure of, plus the first uncertain digit after them. A length of 26.5 cm on a metre scale has 2 and 6 for certain and 5 as the last, uncertain digit: three significant figures.

Before the rules, see where the digits come from. Measure one object with three instruments.

Three counting rules come first. Every non-zero digit counts: 1657 has four. A zero between non-zero digits counts: 38075 has five. Zeros to the left of the first non-zero digit do not count; they only hold the decimal point's place: 0.000385 has three, and so does 0.00569.

A fourth rule: zeros on the right after the decimal point count, because someone measured them: 29.570 has five, 2789.00 has six. And moving the decimal point to change the unit leaves the count alone: 408.0 m = 4.080 km, four figures either way.

4. Decimals for adding, figures for multiplying

In addition and subtraction, the result keeps as many decimal places as the number with the fewest. 1.256 + 2.7 = 3.956, but 2.7 has only one decimal place, so the answer is 4.0.

In multiplication and division you count significant figures instead: the answer has as many as the factor with the fewest. 2.312 × 1.5/1.32 = 2.627, but 1.5 has only two figures, so the answer is 2.6.

Example 1.7(1): R = 25.65 Ω and I = 2.38 A, so V = IR = 61.0470 V. 2.38 has three figures, so V = 61.0 V, and the final zero must stay. Example 1.7(2): a floor of 5.22 m × 3.65 m gives 19.053, written 19.1 m²; multiply by the height 3.92 m and the volume 74.68776 is written 74.7 m³.

Six calculations. For each one, choose the rule first, then write the answer.

5. Rounding off

This is the rule for dropping extra digits. Look at the digit after the last one you keep. If it is above 5, add 1 to the digit before: 3.78 ≈ 3.8. If it is below 5, leave the digit before as it is: 2.23 ≈ 2.2. If it is exactly 5, look at the digit before: add 1 if it is odd (2.75 ≈ 2.8), leave it if it is even (3.45 ≈ 3.4).

See the rule on a number line. Drag the dropped digit and watch which end the number goes to. Stop at exactly 5 and try the other number too.

Eight numbers. Round each to the figures asked, and see which case applied.

All the rules together, example 1.7(3): seven readings of a refractive index, 1.54, 1.55, 1.52, 1.51, 1.48, 1.46 and 1.50. The sum is 10.56 and the mean μ = 10.56/7 = 1.51, rounded to three figures. The absolute errors from the mean add to 0.17, so the mean absolute error is 0.17/7 = 0.024. The relative error is 0.024/1.51 = 0.016, that is 1.6%.

6. Check yourself

Three questions in the exam's style. In the first, find the power.

The second is the exactly-5 rule.

The last is the sphere again, with its own numbers.

The whole note on eight cards.

Source: Physics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Physical World and Measurement”, section 1.13.