3.1 Measurements and their errors — printable coverage pack

3.1.1 · Use of SI units and their prefixes

The required base quantities are mass, length, time, amount of substance, temperature and electric current, with SI units $\text{kg}$ , $\text{m}$ , $\text{s}$ , $\text{mol}$ , $\text{K}$ and $\text{A}$ respectively.
Derived units are combinations of base units; for example $1\,\text{N}=1\,\text{kg m s}^{-2}$ and $1\,\text{J}=1\,\text{kg m}^2\text{s}^{-2}$ .
Know the multipliers for $\text{T}$ , $\text{G}$ , $\text{M}$ , $\text{k}$ , $\text{c}$ , $\text{m}$ , $\mu$ , $\text{n}$ , $\text{p}$ and $\text{f}$ , and convert the entire measured quantity before substituting.
Write very large or small results in standard form with a unit. A common error is to square a value but not its prefix, such as treating $1\,\text{mm}^2$ as $10^{-3}\,\text{m}^2$ instead of $10^{-6}\,\text{m}^2$ .

1. Convert $4.7\,\mu\text{m}$ into metres.[1 mark]
Answer
- $4.7\times10^{-6}\,\text{m}$
Method: The prefix $\mu$ means $10^{-6}$ , so $4.7\,\mu\text{m}=4.7\times10^{-6}\,\text{m}$ .

1. A heater transfers $2.4\,\text{kW h}$ of energy. Calculate this energy in joules. Use $1\,\text{kW h}=3.60\times10^6\,\text{J}$ .[2 marks]
Answer
- $8.6\times10^6\,\text{J}$
Method: Multiply by the stated conversion factor: $E=2.4\times3.60\times10^6=8.64\times10^6\,\text{J}$ . The input $2.4$ has two significant figures, so $E=8.6\times10^6\,\text{J}$ .

1. A pulse transfers charge $5.4\,\text{nC}$ through a sensor of area $0.36\,\text{mm}^2$ in $12\,\mu\text{s}$ . Calculate the mean current density in $\text{A m}^{-2}$ .[4 marks]
Answer
- $1.3\times10^3\,\text{A m}^{-2}$
Method: Convert every quantity to SI units: $Q=5.4\times10^{-9}\,\text{C}$ , $A=0.36\times10^{-6}=3.6\times10^{-7}\,\text{m}^2$ and $t=12\times10^{-6}\,\text{s}$ . The current is $I=Q/t=(5.4\times10^{-9})/(12\times10^{-6})=4.5\times10^{-4}\,\text{A}$ . Hence the current density is $J=I/A=(4.5\times10^{-4})/(3.6\times10^{-7})=1.25\times10^3\,\text{A m}^{-2}$ , which to two significant figures is $1.3\times10^3\,\text{A m}^{-2}$ .

Random error causes readings to scatter and can be reduced by repeats and averaging; systematic error shifts readings consistently and must be identified and removed or corrected.
Accuracy is closeness to the true value, precision is the spread of readings, resolution is the smallest detectable change, repeatability uses the same method and operator, and reproducibility changes the method or operator.
For sums and differences, add absolute uncertainties. For products and quotients, add fractional or percentage uncertainties; for $y=x^n$ , multiply the percentage uncertainty in $x$ by $|n|$ .
Quote an absolute uncertainty to one significant figure, or sometimes two when its first digit is $1$ , and round the measured value to the same decimal place. Do not imply more precision than the uncertainty supports.
On a graph, use error bars where appropriate and estimate gradient or intercept uncertainty from the steepest and shallowest acceptable lines, not from arbitrary lines through one point.

1. A diameter is measured as $(82.0\pm0.5)\,\text{mm}$ . Calculate its percentage uncertainty.[2 marks]
Answer
- $0.61\%$
Method: Percentage uncertainty is $(0.5/82.0)\times100=0.6098\%$ , which is $0.61\%$ to two significant figures.

1. The sides of a rectangular card are $(4.20\pm0.05)\,\text{cm}$ and $(2.10\pm0.03)\,\text{cm}$ . Determine its area and absolute uncertainty.[3 marks]
Answer
- $(8.8\pm0.2)\,\text{cm}^2$
Method: The area is $A=4.20\times2.10=8.82\,\text{cm}^2$ . For a product, add percentage uncertainties: $(0.05/4.20+0.03/2.10)\times100=2.62\%$ . The absolute uncertainty is $0.0262\times8.82=0.231\,\text{cm}^2$ . Quoting the uncertainty to one significant figure and the area to the same decimal place gives $(8.8\pm0.2)\,\text{cm}^2$ .

1. A pendulum has length $(0.842\pm0.002)\,\text{m}$ . The time for $20$ oscillations is $(36.4\pm0.2)\,\text{s}$ . Use $g=4\pi^2L/T^2$ to calculate $g$ with its absolute uncertainty.[5 marks]
Answer
- $(10.0\pm0.1)\,\text{N kg}^{-1}$
Method: One period is $T=36.4/20=1.82\,\text{s}$ ; dividing both the value and absolute uncertainty by $20$ leaves its fractional uncertainty equal to $0.2/36.4$ . The value is $g=4\pi^2(0.842)/(1.82)^2=10.03\,\text{N kg}^{-1}$ . Because $g\propto LT^{-2}$ , its percentage uncertainty is $(0.002/0.842+2\times0.2/36.4)\times100=1.34\%$ . The absolute uncertainty is $0.0134\times10.03=0.134\,\text{N kg}^{-1}$ , rounded to $0.1\,\text{N kg}^{-1}$ , so $g=(10.0\pm0.1)\,\text{N kg}^{-1}$ .

An order of magnitude is the nearest power of ten; first estimate each input to roughly one significant figure, then state the final result as $10^n$ with an appropriate unit.
A defensible estimate states its assumptions explicitly, such as a representative dimension, density, occupancy or operating fraction.
Use known physics to derive further estimates, checking dimensions and whether the result is physically reasonable before choosing the nearest order of magnitude.
Do not confuse an order-of-magnitude answer with ordinary rounding: values below about $3.2\times10^n$ are nearer $10^n$ , while larger values are nearer $10^{n+1}$ .

1. Estimate the order of magnitude of the mass of air in a room measuring $8\,\text{m}\times6\,\text{m}\times3\,\text{m}$ . Take the density of air as $1\,\text{kg m}^{-3}$ .[1 mark]
Answer
- $10^2\,\text{kg}$
Method: The room volume is about $8\times6\times3=144\,\text{m}^3$ . Its air mass is therefore about $1\times144=1.44\times10^2\,\text{kg}$ , whose nearest order of magnitude is $10^2\,\text{kg}$ .

1. Estimate the order of magnitude of the number of water molecules in $250\,\text{cm}^3$ of water. Use density $1.0\,\text{g cm}^{-3}$ , molar mass $18\,\text{g mol}^{-1}$ and $N_A=6.0\times10^{23}\,\text{mol}^{-1}$ .[3 marks]
Answer
- $10^{25}$ molecules
Method: The water mass is about $250\times1.0=250\,\text{g}$ . The amount is $250/18\approx14\,\text{mol}$ , so the number of molecules is about $14\times6.0\times10^{23}=8.4\times10^{24}$ . This is nearer $10^{25}$ than $10^{24}$ , so the order of magnitude is $10^{25}$ molecules.

1. Estimate the order of magnitude of the total electrical power drawn by domestic kettles in a country of population $6.8\times10^7$ . Assume $2.5$ people per household, a $3\,\text{kW}$ kettle in each household, and that $4\%$ of kettles are operating at one time.[5 marks]
Answer
- $10^{10}\,\text{W}$
Method: The number of households is approximately $(6.8\times10^7)/2.5=2.72\times10^7$ . The number of kettles operating is $0.04\times2.72\times10^7=1.09\times10^6$ . Their total power is $(1.09\times10^6)(3\times10^3)=3.26\times10^9\,\text{W}$ . Since this is just above $3.2\times10^9\,\text{W}$ , the nearest order of magnitude is $10^{10}\,\text{W}$ .