Exercises — Errors — absolute, relative, percentage; systematic vs random

WHAT kind: the jaws touching should read $0$ but read $0.03$ cm. Every measurement is pushed up by the same $0.03$ cm — a consistent bias in one direction. That is a systematic (zero) error, not random. WHY we can fix it: a known constant offset can simply be subtracted (see Least Count and Vernier Calipers for how zero errors arise on that instrument). Corrected reading: $4.57-0.03=\boxed{4.54\text{cm}}$.

The readings fall both above and below each other with no fixed direction — a coin-flip nudge. That is random error. Averaging these will shrink it, because the highs and lows partly cancel.

(a) The $\pm$ number is the absolute error: $\Delta L=0.5$ cm (same units as $L$). (b) Relative $=\frac{0.5}{25.0}=0.02$ (dimensionless). (c) Percentage $=0.02\times 100\%=\boxed{2\%}$.

Mean (best estimate, WHY: minimises spread of random scatter): $t_{\text{mean}}=\frac{12.1+12.4+12.0+12.3+12.2}{5}=\frac{61.0}{5}=12.2\text{s}.$ Absolute errors $|t_{\text{mean}}-t_i|$: $0.1,0.2,0.2,0.1,0.0$. Mean absolute error: $\Delta t_{\text{mean}}=\frac{0.1+0.2+0.2+0.1+0.0}{5}=\frac{0.6}{5}=0.12\text{s}.$ Report (error to 1 sig fig, value to same decimal place): $\boxed{t=12.2\pm 0.1\text{s}}$.

Relative $=\frac{\Delta t_{\text{mean}}}{t_{\text{mean}}}=\frac{0.12}{12.2}=0.0098$. Percentage $=0.0098\times 100\%=\boxed{0.98\%\approx 1.0\%}$. WHY relative here: it lets you compare this timing directly against, say, a mass measured to $3\%$ — the percentage strips away the units.

Mean $=\frac{0.42+0.44+0.43+0.43}{4}=\frac{1.72}{4}=0.43$ mm. Absolute errors: $0.01,0.01,0.00,0.00$ → mean $=\frac{0.02}{4}=0.005$ mm. Report: $\boxed{d=0.430\pm 0.005\text{mm}}$. Percentage $=\frac{0.005}{0.43}\times 100\%=\boxed{1.2\%}$.

(a) The constant empty-pan reading $0.15$ g pushes every weighing up by the same amount → systematic zero error. The scatter $\pm 0.02$ g about the run is random. (b) Uncorrected mean $=\frac{8.00+8.02+7.98}{3}=\frac{24.00}{3}=8.00$ g. Subtract the offset: $8.00-0.15=7.85$ g. Absolute errors (unchanged by a constant shift): $|8.00-8.00|=0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.0133\approx 0.01$ g. Report: $\boxed{m=7.85\pm 0.01\text{g}}$. WHY the offset does not change $\Delta m$: subtracting the same number from every reading slides them all together — their spread about the mean is untouched.

A's mean $=\frac{50.31+50.29+50.30}{3}=50.30$ cm; readings tightly clustered → precise, but $+0.30$ cm off truth → inaccurate (a systematic bias). This is the left panel of the target figure. B's mean $=\frac{49.6+50.4+49.9+50.1}{4}=\frac{200.0}{4}=50.00$ cm; readings wider → less precise, but the mean equals the true value → accurate. This is the right panel. Averaging: helps B (random scatter cancels to the truth) but not A (a $+0.30$ bias survives any number of repeats).

Same absolute error does not mean same quality — divide by the size to judge fairly. $P$: $\frac{0.1}{2.0}\times 100\%=5\%$. $Q$: $\frac{0.1}{200.0}\times 100\%=0.05\%$. $Q$ is better; ratio $=\frac{5}{0.05}=\boxed{100\times }$ more precise as a fraction of the thing measured.

Area $A=L\times W=50.0{\text{cm}}^2$ (a product → relative errors add). $\frac{\Delta A}{A}=\frac{\Delta L}{L}+\frac{\Delta W}{W}=\frac{0.1}{10.0}+\frac{0.1}{5.0}=0.01+0.02=0.03$. Percentage $=\boxed{3\%}$, and $\Delta A=0.03\times 50.0=1.5{\text{cm}}^2$, so $A=50.0\pm 1.5{\text{cm}}^2$. WHY add relatives, not absolutes: for a product each factor stretches the result proportionally, so it is the fractional wobble of each that feeds through.

$g$ has $L$ to power $1$ and $T$ to power $2$ (in the denominator). $\pi$ is exact, contributing nothing. $\frac{\Delta g}{g}=\frac{\Delta L}{L}+2\frac{\Delta T}{T}=\frac{0.01}{1.00}+2\times \frac{0.02}{2.00}=0.01+0.02=\mathrm{0.03.}$ Percentage in $g=\boxed{3\%}$. WHY the factor $2$ on $T$: $T$ appears squared, so a $1\%$ wobble in $T$ becomes a $2\%$ wobble in $g$. Squaring doubles the sensitivity.

Value $=12.5-12.0=0.5$ cm. For a difference the absolute errors add: $\Delta =0.1+0.1=0.2$ cm. Percentage $=\frac{0.2}{0.5}\times 100\%=\boxed{40\%}$. Comment: subtracting two nearly-equal numbers is dangerous — the result is tiny but the errors pile up, exploding the percentage error. Avoid "difference of large near-equal quantities" when you can.

Diameter: mean $=\frac{2.00+2.02+1.98}{3}=2.00$ cm; abs errors $0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.01\overline{3}$ cm, which we quote to $1$ sig fig as $\Delta d=0.01$ cm. $\frac{\Delta d}{d}=\frac{0.01}{2.00}=0.0050$ (and $\Delta r/r=\Delta d/d$ since $r=d/2$ scales top and bottom the same). Length: mean $=5.00$ cm; abs errors $0.00,0.02,0.02$ → mean $=\frac{0.04}{3}=0.01\overline{3}\to \Delta \ell =0.01$ cm; $\frac{\Delta \ell }{\ell }=\frac{0.01}{5.00}=0.0020$. Mass: $\frac{\Delta m}{m}=\frac{0.5}{50.0}=0.010$. Propagate $\rho =\frac{m}{\pi r^2\ell }$ → $r$ has power $2$, $\ell$ power $1$, $m$ power $1$: $\frac{\Delta \rho }{\rho }=\frac{\Delta m}{m}+2\frac{\Delta d}{d}+\frac{\Delta \ell }{\ell }=0.010+2(0.0050)+0.0020=\mathrm{0.0220.}$ Percentage $\approx \boxed{2.2\%}$. Central value: $r=1.00$ cm, $V=\pi (1.00{)}^2(5.00)=15.708{\text{cm}}^3$, $\rho =\frac{50.0}{15.708}=3.183{\text{g/cm}}^3$. $\Delta \rho =0.022\times 3.183=0.070$, so $\rho =3.18\pm 0.07{\text{g/cm}}^3$. Rounding note: every mean absolute error was rounded to $1$ sig fig ($0.01$ cm) before being fed into the propagation, matching the L2 rule — consistent all the way through.

Look at each term's contribution to $\frac{\Delta \rho }{\rho }=0.0220$: mass $0.010$, diameter $2(0.0050)=0.0100$, length $0.0020$. Mass and diameter tie at $0.0100$ each — but the diameter's contribution is inflated by its power $2$, so halving $\Delta d$ removes twice as much per unit of raw fractional error. Halving $\Delta d$ drops its contribution from $0.0100$ to $0.0050$, giving $\frac{\Delta \rho }{\rho }=0.010+0.0050+0.0020=0.0170\approx 1.7\%.$ Halving $\Delta m$ instead would drop mass's $0.010$ to $0.005$, giving the same $0.0170$ here — but the diameter is the one whose leverage grows with its power, so in any cylinder where $r$'s fractional error is not tiny, diameter is the term to watch. See the contribution bars below.