Physics 9702/22 — October/November 2021

Cambridge AS Level · AS Level Structured Questions · worked solutions for every part, with the mark scheme

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Topics Kinematics · Dynamics · Work, Energy and Power · Waves · Physical Quantities and Units · Electric Fields · +4 more

Q1Physical Quantities and UnitsFree sample

Answer all the questions in the spaces provided.

(a)

A unit may be stated with a prefix that represents a power-of-ten multiple or submultiple.

Complete Table 1.1 to show the name and symbol of each prefix and the corresponding power-of-ten multiple or submultiple.

Table 1.1

prefix	power-of-ten multiple or submultiple
kilo (k)	$10^3$
tera (T)
( )	$10^{-12}$

DifficultyEasy

Worked solution

Answer

tera (T):

10^{12}

$10^{-12}$ corresponds to pico (p).

Final answer

tera (T) = 10^12; 10^-12 is pico (p)

Detailed explanation

Background Concept

SI prefixes represent powers of ten and are used to scale units conveniently (e.g. $1\ \text{km} = 10^3\ \text{m}$ ). Each prefix has a name, a symbol, and a specific power-of-ten multiplier.

Understanding the Question

You are given a partial table of prefixes. You must:

fill in the missing power of ten for tera (T)
identify the prefix (name and symbol) that corresponds to $10^{-12}$ .

Approach

Use the standard SI prefix list:

large multiples: kilo $10^3$ , mega $10^6$ , giga $10^9$ , tera $10^{12}$
small submultiples: milli $10^{-3}$ , micro $10^{-6}$ , nano $10^{-9}$ , pico $10^{-12}$ .

Step-by-Step Reasoning

Tera is the prefix used for $10^{12}$ , so its power-of-ten multiple is $10^{12}$ .
The power $10^{-12}$ corresponds to pico, with symbol p.

Key Takeaways

Tera (T) means multiply by $10^{12}$ .
Pico (p) means multiply by $10^{-12}$ .

Common Mistakes

Confusing tera ( $10^{12}$ ) with giga ( $10^9$ ).
Writing the wrong symbol (e.g. P instead of p; case matters).

Things to Be Careful About

Prefix symbols are case-sensitive: T is tera, p is pico.
Ensure the exponent sign is correct for submultiples (negative powers).

Techniques used

recall the SI prefix corresponding to a power of tenmatch a given power of ten to its prefix name and symbol

(b)

In the following list, underline all the units that are SI base units.

ampere coulomb metre newton

DifficultyEasy

Worked solution

Answer

SI base units: ampere, metre.

Final answer

ampere, metre

Detailed explanation

Background Concept

SI base units are the fundamental units defined independently (e.g. metre, kilogram, second, ampere, kelvin, mole, candela). Derived units are combinations of base units (e.g. newton, joule, coulomb).

Understanding the Question

From the list

ampere
coulomb
metre
newton
you must select only those that are SI base units.

Approach

Recall the SI base units list and check each option:

if it is one of the seven base units, select it
otherwise it is derived, so do not select it.

Step-by-Step Reasoning

ampere is an SI base unit (unit of current).
metre is an SI base unit (unit of length).
coulomb is derived since $Q = It$ so $1\ \text{C} = 1\ \text{A s}$ .
newton is derived since $F = ma$ so $1\ \text{N} = 1\ \text{kg m s}^{-2}$ .

Key Takeaways

Base units are fundamental; derived units are combinations of base units.

Common Mistakes

Choosing coulomb because it feels “basic”; it is not a base unit.
Choosing newton as a base unit; it is derived from $\text{kg}$ , $\text{m}$ and $\text{s}$ .

Things to Be Careful About

In MCQ/selection tasks, include only the base units asked for (do not list derived ones even if common).

Techniques used

identify SI base units from a listdistinguish base units from derived units

(c)

The potential difference $V$ between the two ends of a uniform metal wire is given by

V = \frac{4\rho LI}{\pi d^2}

where $d$ is the diameter of the wire,
$I$ is the current in the wire,
$L$ is the length of the wire,
and $\rho$ is the resistivity of the metal.

For a particular wire, the percentage uncertainties in the values of some of the above quantities are listed in Table 1.2.

Table 1.2

quantity	percentage uncertainty
$d$	$\pm 3.0\%$
$I$	$\pm 2.0\%$
$L$	$\pm 2.5\%$
$V$	$\pm 3.5\%$

The quantities listed in Table 1.2 have values that are used to calculate $\rho$ as $4.1 \times 10^{-7}\ \Omega\ \text{m}$ .

For this value of $\rho$ , calculate:

(i)

the percentage uncertainty

percentage uncertainty = ______ $\%$

DifficultyMedium-Easy

Worked solution

Working

From

V = \frac{4\rho LI}{\pi d^2}

\rho = \frac{V\pi d^2}{4LI}

Percentage uncertainty in $\rho$ :

\%\,\Delta\rho = \%\,\Delta V + 2(\%\,\Delta d) + \%\,\Delta L + \%\,\Delta I

= 3.5 + 2(3.0) + 2.5 + 2.0 = 14.0\%

Answer

$14.0\%$

Final answer

14.0 %

Detailed explanation

Background Concept

When a quantity is calculated from measured values, its uncertainty depends on how those measurements combine.

For products and quotients,

if $Q = \frac{ab}{c}$ then the fractional (or percentage) uncertainties add:

\frac{\Delta Q}{Q} \approx \frac{\Delta a}{a} + \frac{\Delta b}{b} + \frac{\Delta c}{c}

For powers,

if $Q = a^n$ then the percentage uncertainty is multiplied by $n$ :

\%\,\Delta Q = |n|\,(\%\,\Delta a)

These rules are what Cambridge typically expects at AS for “uncertainty in a derived quantity”.

Understanding the Question

You are given the relationship:

V = \frac{4\rho LI}{\pi d^2}

and the percentage uncertainties in $d, I, L,$ and $V$ . You calculate $\rho$ using these measured quantities, and you must find the percentage uncertainty in the calculated resistivity $\rho$ .

Approach

Rearrange the given equation to make $\rho$ the subject.
Identify how each measured quantity appears in $\rho$ (e.g. $d^2$ means the uncertainty in $d$ is doubled).
Add the relevant percentage uncertainties.

Step-by-Step Reasoning

Rearrange for resistivity:

\rho = \frac{V\pi d^2}{4LI}

Now apply the percentage-uncertainty rules:

$V$ is multiplied: add $3.5\%$ .
$d^2$ is multiplied: add $2 \times 3.0\% = 6.0\%$ .
$L$ is in the denominator (division): still add its percentage uncertainty, $2.5\%$ .
$I$ is in the denominator: add $2.0\%$ .

\%\,\Delta\rho = 3.5 + 6.0 + 2.5 + 2.0 = 14.0\%

Key Takeaways

For multiplication/division, add percentage uncertainties.
For powers, multiply the percentage uncertainty by the power.
Denominator quantities still contribute positively to the total percentage uncertainty.

Common Mistakes

Subtracting the percentage uncertainties for quantities in the denominator (they should be added).
Forgetting to double the uncertainty contribution from $d^2$ .
Using absolute uncertainties when the question clearly provides percentage uncertainties.

Things to Be Careful About

The constants $4$ and $\pi$ are exact here, so they contribute no uncertainty.
Use consistent rounding: the total here is naturally quoted as $14.0\%$ (or $14\%$ depending on marking tolerance), but keep a sensible number of significant figures.

Techniques used

rearrange an equation to express the required quantityadd percentage uncertainties for products and quotientsapply power rules for uncertainties in squared quantities

(ii)

the absolute uncertainty.

absolute uncertainty = ______ $\Omega\ \text{m}$

DifficultyMedium-Easy

Worked solution

Working

Percentage uncertainty in $\rho = 14.0\%$ , so fractional uncertainty $= 0.140$ .

\Delta \rho = 0.140 \times 4.1 \times 10^{-7} = 5.74 \times 10^{-8}\ \Omega\ \text{m}

Answer

$5.7 \times 10^{-8}\ \Omega\ \text{m}$

Final answer

5.7 × 10^-8 Ω m

Detailed explanation

Background Concept

An absolute uncertainty $\Delta x$ tells you the size of the uncertainty in the same units as the quantity. If you know the percentage uncertainty, then

\%\,\Delta x = \frac{\Delta x}{x} \times 100

\Delta x = \left(\frac{\%\,\Delta x}{100}\right)x

Understanding the Question

You have calculated $\rho = 4.1 \times 10^{-7}\ \Omega\ \text{m}$ using measured values. Using the percentage uncertainty from part (i), you must calculate the absolute uncertainty in $\rho$ .

Approach

Convert the percentage uncertainty to a decimal (fractional uncertainty).
Multiply this by the value of $\rho$ .
Quote the absolute uncertainty with appropriate significant figures and correct units.

Step-by-Step Reasoning

From (i),

\%\,\Delta\rho = 14.0\%

Convert to fractional uncertainty:

\frac{\Delta\rho}{\rho} = \frac{14.0}{100} = 0.140

Then

\Delta\rho = 0.140 \times 4.1 \times 10^{-7} = 5.74 \times 10^{-8}\ \Omega\ \text{m}

Rounded suitably (typically to 2 s.f.):

\Delta\rho \approx 5.7 \times 10^{-8}\ \Omega\ \text{m}

Key Takeaways

Absolute uncertainty $=$ (percentage uncertainty as a fraction) $\times$ value.
Always include the unit for absolute uncertainty.

Common Mistakes

Forgetting to divide the percentage by $100$ .
Giving the absolute uncertainty in percent instead of $\Omega\ \text{m}$ .
Rounding too aggressively (e.g. to 1 s.f. without reason).

Things to Be Careful About

Keep the power of ten consistent: $0.140 \times 4.1 \times 10^{-7}$ should give an answer of order $10^{-8}$ .
The absolute uncertainty is usually quoted to 1 or 2 significant figures; match that with the value of $\rho$ if you were asked to state it with uncertainty.

Techniques used

convert a percentage uncertainty to a fractional uncertaintycalculate absolute uncertainty from fractional uncertainty

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