3.1 Measurements and their errors — revision question pack

3 specification points · notes, questions, answers and worked methods

Checked against AQA 7408 section 3.1. Review basis: the qualification registry sourced from the AQA A-level Physics (7408) specification; registry verification recorded 11 July 2026.

How this checking works

Answer all questions in the spaces provided.

3.1.1 · Use of SI units and their prefixes

Explanation

The required base quantities and SI units are mass in kilograms, length in metres, time in seconds, amount of substance in moles, temperature in kelvin and electric current in amperes; candela is excluded.
Derived units combine base units, for example $\text{N}=\text{kg m s}^{-2}$ .
Prefixes required are $\text{T}$ , $\text{G}$ , $\text{M}$ , $\text{k}$ , $\text{c}$ , $\text{m}$ , $\mu$ , $\text{n}$ , $\text{p}$ and $\text{f}$ , used with standard form.
Conversions also include units of the same quantity, such as joules and electronvolts or kilowatt-hours.
Base-quantity definitions and dimensional analysis are not required.

Common SI prefixes step through powers of one thousand around the base unit.

Worked example

Convert an area of $0.36\,\text{mm}^2$ to square metres.

1. $1\,\text{mm}=10^{-3}\,\text{m}$ .
2.Square the whole conversion: $1\,\text{mm}^2=10^{-6}\,\text{m}^2$ .
3. $0.36\times10^{-6}=3.6\times10^{-7}$ .

Answer: The area is 3.6 × 10⁻⁷ m².

Common mistakes

Don't square a measurement and fail to square its prefix conversion.
Don't use grams rather than kilograms as the SI base unit of mass.
Don't write a numerical conversion without retaining the physical unit.

Exam tip

Convert every quantity to a consistent unit system before substitution and show each prefix multiplier.

Tier 1 · Easy

Convert $4.7\,\mu\text{m}$ into metres.
[1 mark]
Total for this question: 1
Convert $3.2\,\text{pF}$ into farads.
[1 mark]
Total for this question: 1

Tier 2 · Standard

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]
Total for this question: 2
A spring exerts a force of $6.4\,\text{mN}$ . Give this force in SI base units.
[2 marks]
Total for this question: 2
The pressure changes by $18\,\text{Pa}$ across a $2.5\,\text{mm}$ filter layer. Calculate the pressure gradient in SI base units.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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]
Total for this question: 4
A metal has density $7.9\,\text{g cm}^{-3}$ . Calculate the mass in kilograms of a cube of this metal with side $2.5\,\text{cm}$ .
[4 marks]
Total for this question: 4
A detector deposits $2.8\,\text{GeV}$ of energy uniformly in a volume of $4.5\times10^{-17}\,\text{m}^3$ . Calculate the energy per unit volume in SI base units. Use $1\,\text{eV}=1.60\times10^{-19}\,\text{J}$ .
[4 marks]
Total for this question: 4
A laser pulse transfers $38\,\text{mJ}$ in $0.75\,\text{ns}$ through a circular spot of diameter $42\,\mu\text{m}$ . Calculate the mean intensity in $\text{W m}^{-2}$ and give its unit in SI base units.
[5 marks]
Total for this question: 5
A sensor crystal contains $4.6\,\text{pC}$ of charge per cubic millimetre. Calculate the charge density in $\text{C m}^{-3}$ and SI base units. Calculate the charge in a crystal volume of $0.38\,\text{cm}^3$ .
[4 marks]
Total for this question: 4

3.1.2 · Limitation of physical measurements

Explanation

Random errors cause scatter and are reduced by repeats and averaging; systematic errors shift results consistently and require correction or removal. Accuracy means closeness to the true value, precision concerns spread, resolution is the smallest detectable change, repeatability keeps method and operator fixed, and reproducibility changes them.
Uncertainty may be absolute, fractional or percentage. For addition and subtraction, combine absolute uncertainties; for multiplication and division, combine percentage uncertainties; for a power $x^n$ , multiply percentage uncertainty by $|n|$ .
Graph points may carry error bars.
Steepest and shallowest acceptable lines give gradient uncertainty, with corresponding intercept limits.
Report value and absolute uncertainty to compatible decimal places and significant figures.

Error bars constrain the steepest and shallowest acceptable straight lines.

Worked example

A rectangle measures $(4.20\pm0.05)\,\text{cm}$ by $(2.10\pm0.03)\,\text{cm}$ . Find its area and uncertainty.

1. $A=4.20(2.10)=8.82\,\text{cm}^2$ .
2.Percentage uncertainty is $100(0.05/4.20+0.03/2.10)=2.62\%$ .
3.Absolute uncertainty is $0.0262(8.82)=0.23\,\text{cm}^2$ .

Answer: The area is (8.8 ± 0.2) cm².

Common mistakes

Don't add absolute uncertainties for a multiplication calculation.
Don't call a tightly clustered but offset set of readings accurate.
Don't quote the measured value to more decimal places than its absolute uncertainty.

Exam tip

State whether each operation needs absolute or percentage uncertainty before combining terms.

Tier 1 · Easy

A diameter is measured as $(82.0\pm0.5)\,\text{mm}$ . Calculate its percentage uncertainty.
[2 marks]
Total for this question: 2
Repeated measurements of a time are $14.2$ , $14.6$ , $14.3$ and $14.8\,\text{s}$ . Determine the absolute uncertainty using half the range.
[2 marks]
Total for this question: 2

Tier 2 · Standard

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]
Total for this question: 3
A balance reads $0.15\,\text{g}$ when its pan is empty. Identify this error, state its effect on every reading and describe one correction.
[3 marks]
Total for this question: 3
A calibrated source has a true temperature of $20.0\,^\circ\text{C}$ . Sensor A gives repeated readings $19.8$ , $20.0$ and $20.2\,^\circ\text{C}$ ; sensor B gives $21.0\,^\circ\text{C}$ each time. Determine which sensor is more precise, which is more accurate, and explain why averaging more readings from sensor B will not correct its error.
[3 marks]
Total for this question: 3

Tier 3 · Hard

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]
Total for this question: 5
A graph has best-fit gradient $2.46\,\text{V A}^{-1}$ . The steepest and shallowest acceptable gradients are $2.58\,\text{V A}^{-1}$ and $2.31\,\text{V A}^{-1}$ . Using the larger deviation from the best-fit gradient, determine the gradient with its absolute and percentage uncertainty.
[4 marks]
Total for this question: 4
A straight-line graph is described by $y=mx+c$ , where $m=(4.80\pm0.12)\,\text{V s}^{-1}$ and $c=(1.20\pm0.06)\,\text{V}$ . Determine the $x$ -intercept with its absolute uncertainty, giving the uncertainty to $1$ significant figure.
[4 marks]
Total for this question: 4
An object distance is $(0.420\pm0.003)\,\text{m}$ and the corresponding image distance is $(0.680\pm0.004)\,\text{m}$ . Use $f=\dfrac{uv}{u+v}$ to determine the focal length with its absolute uncertainty.
[5 marks]
Total for this question: 5
A graph has best-fit gradient $1.84$ . The steepest and shallowest acceptable gradients are $1.93$ and $1.70$ . A constant is calculated from $K=4\pi^2/m^2$ , where $m$ is the gradient. Using the larger gradient deviation, determine $K$ with its absolute uncertainty. Deduce whether a reference value of $13.8$ is consistent with the result.
[5 marks]
Total for this question: 5

3.1.3 · Estimation of physical quantities

Explanation

An order of magnitude is the nearest power of ten. A sound estimate chooses plausible approximate inputs, states assumptions such as representative size, density or operating fraction, and uses relevant physics to derive the requested quantity.
Intermediate values usually need only one significant figure.
The final numerical estimate is then compared with neighbouring powers of ten: the boundary between $10^n$ and $10^{n+1}$ is $\sqrt{10}\times10^n\approx3.2\times10^n$ .
Units and dimensions provide an essential reasonableness check.
Estimation is not guessing a remembered number; the awarded reasoning comes from transparent assumptions, a valid physical relationship and a final answer rounded to the nearest order of magnitude.

The nearest-order boundary lies at approximately 3.2 times the lower power of ten.

Worked example

Estimate the order of magnitude of the mass of air in an $8\,\text{m}\times6\,\text{m}\times3\,\text{m}$ room, using air density $1\,\text{kg m}^{-3}$ .

1.Volume is approximately $8(6)(3)=1.4\times10^2\,\text{m}^3$ .
2. $m=\rho V\approx1(1.4\times10^2)=1.4\times10^2\,\text{kg}$ .
3.This is nearer $10^2$ than $10^3$ .

Answer: The mass has order of magnitude $10^2\,\text{kg}$ .

Common mistakes

Don't report a multi-significant-figure value instead of a power-of-ten order of magnitude.
Don't use an assumed value without stating or justifying it.
Don't round every coefficient above one to the next power of ten.

Exam tip

Show the assumption, physics relationship and numerical estimate before stating the nearest power of ten.

Tier 1 · Easy

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]
Total for this question: 1
State the nearest power-of-ten order of magnitude of a time interval of $2.6\times10^{-7}\,\text{s}$ .
[1 mark]
Total for this question: 1

Tier 2 · Standard

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]
Total for this question: 3
A $5\,\text{kg}$ bag contains rice grains of average mass $25\,\text{mg}$ . Estimate how many grains it contains to the nearest order of magnitude.
[3 marks]
Total for this question: 3
A storm drops $12\,\text{mm}$ of rain uniformly over a city of area $25\,\text{km}^2$ . Take the density of water as $1.0\times10^3\,\text{kg m}^{-3}$ . Estimate the mass of rain to the nearest power of ten.
[4 marks]
Total for this question: 4

Tier 3 · Hard

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]
Total for this question: 5
Estimate the order of magnitude of the electrical energy used in one school day by $60$ rooms, each with eight $12\,\text{W}$ lamps operating for $7\,\text{h}$ .
[5 marks]
Total for this question: 5
Estimate the number of molecules in Earth's atmosphere to the nearest power of ten. Model Earth as a sphere of radius $6.4\times10^6\,\text{m}$ with uniform surface pressure $1.0\times10^5\,\text{Pa}$ . Use $g=9.81\,\text{N kg}^{-1}$ , mean atmospheric molar mass $2.9\times10^{-2}\,\text{kg mol}^{-1}$ and $N_A=6.02\times10^{23}\,\text{mol}^{-1}$ .
[6 marks]
Total for this question: 6
Estimate the order of magnitude of the number of electrons passing through a charger during one year. Assume its current is $1.4\,\text{A}$ for $2.0\,\text{h}$ each day and that the charge magnitude of one electron is $1.60\times10^{-19}\,\text{C}$ .
[5 marks]
Total for this question: 5
Estimate the order of magnitude of the force that radiation would exert on a planet if all incident radiation were absorbed. Treat the planet as a disc of radius $5.8\times10^6\,\text{m}$ , take the incident intensity as $7.1\times10^2\,\text{W m}^{-2}$ and use $F=P/c$ with $c=3.00\times10^8\,\text{m s}^{-1}$ .
[5 marks]
Total for this question: 5

Answer key

Answers begin on a new printed page so the question pack can be completed without the solutions alongside it.

3.1.1 · Use of SI units and their prefixes

Tier 1 · Easy

Mark scheme for 3.1.1 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$4.7\times10^{-6}\,\text{m}$	The prefix $\mu$ means $10^{-6}$ , so $4.7\,\mu\text{m}=4.7\times10^{-6}\,\text{m}$ .	1
02.1	$3.2\times10^{-12}\,\text{F}$	The prefix $\text{p}$ represents $10^{-12}$ , so $3.2\,\text{pF}=3.2\times10^{-12}\,\text{F}$ .	1

Tier 2 · Standard

Mark scheme for 3.1.1 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$8.6\times10^6\,\text{J}$	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 $E=8.6\times10^6\,\text{J}$ .	2
02.1	$6.4\times10^{-3}\,\text{kg m s}^{-2}$	The prefix $\text{m}$ means $10^{-3}$ and $1\,\text{N}=1\,\text{kg m s}^{-2}$ . Therefore $6.4\,\text{mN}=6.4\times10^{-3}\,\text{kg m s}^{-2}$ .	2
03.1	$7.2\times10^3\,\text{kg m}^{-2}\text{ s}^{-2}$	Convert the distance: $2.5\,\text{mm}=2.5\times10^{-3}\,\text{m}$ . The gradient is $18/(2.5\times10^{-3})=7.2\times10^3\,\text{Pa m}^{-1}$ . Since $\text{Pa}=\text{kg m}^{-1}\text{ s}^{-2}$ , the base units are $\text{kg m}^{-2}\text{ s}^{-2}$ .	3

Tier 3 · Hard

Mark scheme for 3.1.1 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$1.3\times10^3\,\text{A m}^{-2}$	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 $1.3\times10^3\,\text{A m}^{-2}$ .	4
02.1	$0.12\,\text{kg}$	The density is $7.9\times10^3\,\text{kg m}^{-3}$ and the side is $2.5\times10^{-2}\,\text{m}$ . The volume is $(2.5\times10^{-2})^3=1.5625\times10^{-5}\,\text{m}^3$ . Hence $m=\rho V=(7.9\times10^3)(1.5625\times10^{-5})=0.1234375\,\text{kg}$ , which is $0.12\,\text{kg}$ to two significant figures.	4
03.1	$1.0\times10^7\,\text{kg m}^{-1}\text{ s}^{-2}$	The energy is $2.8\times10^9\times1.60\times10^{-19}=4.48\times10^{-10}\,\text{J}$ . Hence the energy per unit volume is $(4.48\times10^{-10})/(4.5\times10^{-17})=9.96\times10^6\,\text{J m}^{-3}=1.0\times10^7\,\text{J m}^{-3}$ to two significant figures. Since $\text{J}=\text{kg m}^2\text{ s}^{-2}$ , $\text{J m}^{-3}=\text{kg m}^{-1}\text{ s}^{-2}$ .	4
04.1	$3.7\times10^{16}\,\text{W m}^{-2}=3.7\times10^{16}\,\text{kg s}^{-3}$	Convert the data: $E=38\times10^{-3}\,\text{J}$ , $t=0.75\times10^{-9}\,\text{s}$ and $r=21\times10^{-6}\,\text{m}$ . The mean power is $P=E/t=(38\times10^{-3})/(0.75\times10^{-9})=5.0667\times10^7\,\text{W}$ . The spot area is $A=\pi r^2=\pi(21\times10^{-6})^2=1.3854\times10^{-9}\,\text{m}^2$ . Hence $I=P/A=(5.0667\times10^7)/(1.3854\times10^{-9})=3.6571\times10^{16}\,\text{W m}^{-2}$ , or $3.7\times10^{16}\,\text{W m}^{-2}$ to two significant figures. Since $\text{W}=\text{kg m}^2\text{ s}^{-3}$ , $\text{W m}^{-2}=\text{kg s}^{-3}$ .	5
05.1	$4.6\times10^{-3}\,\text{C m}^{-3}=4.6\times10^{-3}\,\text{A s m}^{-3}$ $1.7\times10^{-9}\,\text{C}$	One cubic millimetre is $(10^{-3}\,\text{m})^3=10^{-9}\,\text{m}^3$ , while $4.6\,\text{pC}=4.6\times10^{-12}\,\text{C}$ . The charge density is therefore $(4.6\times10^{-12})/(10^{-9})=4.6\times10^{-3}\,\text{C m}^{-3}$ . Since $\text{C}=\text{A s}$ , its SI base units are $\text{A s m}^{-3}$ . The volume is $0.38\,\text{cm}^3=0.38\times10^{-6}=3.8\times10^{-7}\,\text{m}^3$ , so the charge is $(4.6\times10^{-3})(3.8\times10^{-7})=1.748\times10^{-9}\,\text{C}$ , or $1.7\times10^{-9}\,\text{C}$ to two significant figures.	4

3.1.2 · Limitation of physical measurements

Tier 1 · Easy

Mark scheme for 3.1.2 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$0.61\%$	Percentage uncertainty is $(0.5/82.0)\times100=0.6098\%$ , which is $0.61\%$ to two significant figures.	2
02.1	$\pm0.3\,\text{s}$	The range is $14.8-14.2=0.6\,\text{s}$ . Half the range is $0.6/2=0.3\,\text{s}$ , so the absolute uncertainty is $\pm0.3\,\text{s}$ .	2

Tier 2 · Standard

Mark scheme for 3.1.2 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$(8.8\pm0.2)\,\text{cm}^2$	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 $(8.8\pm0.2)\,\text{cm}^2$ .	3
02.1	A systematic zero error makes every reading $0.15\,\text{g}$ too large; zero the balance or subtract $0.15\,\text{g}$ from each reading.	The offset is repeated with the same sign and size, so it is systematic rather than random. It shifts every measured mass upward by $0.15\,\text{g}$ . The error is removed by zeroing the balance before use or correcting each result by subtracting $0.15\,\text{g}$ .	3
03.1	sensor B is more precise; sensor A is more accurate; B has a systematic error that averaging does not remove	Sensor B is more precise because its readings have no observed spread. Sensor A is more accurate because its readings are centred on the true value of $20.0\,^\circ\text{C}$ , whereas B is consistently $1.0\,^\circ\text{C}$ too high. Sensor B therefore has a systematic offset; repeats and averaging reduce random scatter but do not remove a fixed systematic error.	3

Tier 3 · Hard

Mark scheme for 3.1.2 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$(10.0\pm0.1)\,\text{N kg}^{-1}$	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 $g=(10.0\pm0.1)\,\text{N kg}^{-1}$ .	5
02.1	$(2.46\pm0.15)\,\text{V A}^{-1}$ Percentage uncertainty $=6.1\%$	The deviations from the best gradient are $2.58-2.46=0.12\,\text{V A}^{-1}$ and $2.46-2.31=0.15\,\text{V A}^{-1}$ . Use the larger deviation, so the absolute uncertainty is $0.15\,\text{V A}^{-1}$ . The percentage uncertainty is $(0.15/2.46)\times100=6.0976\%$ , which is $6.1\%$ to two significant figures. Thus the gradient is $(2.46\pm0.15)\,\text{V A}^{-1}$ .	4
03.1	$(-0.25\pm0.02)\,\text{s}$ (accept absolute uncertainty $\pm0.019$ – $0.02\,\text{s}$ )	At the $x$ -intercept, $0=mx+c$ , so $x=-c/m=-1.20/4.80=-0.250\,\text{s}$ . For the division, add fractional uncertainties: $0.06/1.20+0.12/4.80=0.075$ , or $7.5\%$ . The absolute uncertainty is $0.075(0.250)=0.01875\,\text{s}$ , which rounds to $0.02\,\text{s}$ . Quoting the value to the same decimal place gives $x=(-0.25\pm0.02)\,\text{s}$ .	4
04.1	$(0.260\pm0.002)\,\text{m}$ (accept an unrounded absolute uncertainty from $0.0017$ to $0.0020\,\text{m}$ )	Use the independent measured quantities in $1/f=1/u+1/v$ . The focal length is $f=[1/0.420+1/0.680]^{-1}=0.259636\,\text{m}$ . The reciprocal uncertainties are $\Delta(1/u)=\Delta u/u^2=0.003/(0.420)^2=0.0170\,\text{m}^{-1}$ and $\Delta(1/v)=\Delta v/v^2=0.004/(0.680)^2=0.00865\,\text{m}^{-1}$ . Hence $\Delta(1/f)=0.0170+0.00865=0.02566\,\text{m}^{-1}$ , which is $0.666\%$ of $1/f$ . Therefore $\Delta f=f^2\Delta(1/f)=(0.259636)^2(0.02566)=0.00173\,\text{m}$ . Quoting the uncertainty and value compatibly gives $(0.260\pm0.002)\,\text{m}$ .	5
05.1	$K=11.7\pm1.8$ (accept unrounded uncertainty $1.77$ or a consistently rounded equivalent); $13.8$ is not consistent	The gradient deviations are $1.93-1.84=0.09$ and $1.84-1.70=0.14$ , so use $\Delta m=0.14$ . The value is $K=4\pi^2/(1.84)^2=11.6607$ . Because $K\propto m^{-2}$ , its percentage uncertainty is $2(0.14/1.84)\times100=15.217\%$ . Thus $\Delta K=0.15217(11.6607)=1.7745$ , giving $K=11.7\pm1.8$ . The corresponding interval is approximately $9.9$ to $13.4$ , which does not include $13.8$ , so the reference value is not consistent with this uncertainty range.	5

3.1.3 · Estimation of physical quantities

Tier 1 · Easy

Mark scheme for 3.1.3 Tier 1 · Easy
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10^2\,\text{kg}$	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
02.1	$10^{-7}\,\text{s}$	$2.6\times10^{-7}\,\text{s}$ lies below the geometric boundary $\sqrt{10}\times10^{-7}=3.162\times10^{-7}\,\text{s}$ , so it is nearer $10^{-7}\,\text{s}$ than $10^{-6}\,\text{s}$ .	1

Tier 2 · Standard

Mark scheme for 3.1.3 Tier 2 · Standard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10^{25}$ molecules	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.	3
02.1	$10^5$ grains	One grain has mass $25\,\text{mg}=2.5\times10^{-5}\,\text{kg}$ . The estimated number is $5/(2.5\times10^{-5})=2\times10^5$ . This is nearer $10^5$ than $10^6$ , so the order of magnitude is $10^5$ grains.	3
03.1	$10^8\,\text{kg}$	The area is $25(10^3\,\text{m})^2=2.5\times10^7\,\text{m}^2$ and the depth is $1.2\times10^{-2}\,\text{m}$ . The water volume is $(2.5\times10^7)(1.2\times10^{-2})=3.0\times10^5\,\text{m}^3$ . Its mass is $(1.0\times10^3)(3.0\times10^5)=3.0\times10^8\,\text{kg}$ . The boundary is about $3.2\times10^8\,\text{kg}$ ( $\sqrt{10}\times10^8\,\text{kg}$ ), so the nearest order of magnitude is $10^8\,\text{kg}$ .	4

Tier 3 · Hard

Mark scheme for 3.1.3 Tier 3 · Hard
Question	Answers	Additional comments/Guidelines	Mark
01.1	$10^{10}\,\text{W}$	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 $10^{10}\,\text{W}$ .	5
02.1	$10^8\,\text{J}$	There are $60\times8=480$ lamps, so their total power is $480\times12=5760\,\text{W}$ . The operating time is $7\times3600=25200\,\text{s}$ . The energy is approximately $5760\times25200=1.45152\times10^8\,\text{J}$ , whose nearest order of magnitude is $10^8\,\text{J}$ .	5
03.1	$10^{44}$ molecules	Earth's surface area is $4\pi R^2=4\pi(6.4\times10^6)^2=5.15\times10^{14}\,\text{m}^2$ . Pressure times area gives the atmosphere's weight: $W=pA=(1.0\times10^5)(5.15\times10^{14})=5.15\times10^{19}\,\text{N}$ . Its mass is $m=W/g=5.25\times10^{18}\,\text{kg}$ , giving $n=m/M=(5.25\times10^{18})/(2.9\times10^{-2})=1.81\times10^{20}\,\text{mol}$ . The number of molecules is $nN_A=(1.81\times10^{20})(6.02\times10^{23})=1.09\times10^{44}$ , whose nearest order of magnitude is $10^{44}$ molecules.	6
04.1	$10^{25}$ electrons	The operating time is $2.0\times3600\times365=2.628\times10^6\,\text{s}$ . The charge transferred is $Q=It=1.4(2.628\times10^6)=3.6792\times10^6\,\text{C}$ . The number of electrons is $N=Q/e=(3.6792\times10^6)/(1.60\times10^{-19})=2.2995\times10^{25}$ . This is below the boundary $\sqrt{10}\times10^{25}=3.162\times10^{25}$ , so its nearest order of magnitude is $10^{25}$ electrons.	5
05.1	$10^8\,\text{N}$	The planet intercepts radiation over the disc area $A=\pi R^2=\pi(5.8\times10^6)^2=1.057\times10^{14}\,\text{m}^2$ , not over its whole spherical surface. The intercepted power is $P=IA=(7.1\times10^2)(1.057\times10^{14})=7.50\times10^{16}\,\text{W}$ . For complete absorption, $F=P/c=(7.50\times10^{16})/(3.00\times10^8)=2.50\times10^8\,\text{N}$ . This lies below $\sqrt{10}\times10^8\,\text{N}$ and is therefore nearer $10^8\,\text{N}$ than $10^9\,\text{N}$ .	5