Physics 9702/32 — May/June 2013

Cambridge AS Level · Advanced Practical Skills · worked solutions for every part, with the mark scheme

questions

marks

120

minutes

Topics Manipulation, Measurement and Observation · Presentation of Data and Observations · Analysis, Conclusions and Evaluation

Q1Manipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and EvaluationFree sample

In this experiment, you will investigate the time for the voltage across a component to decrease after a switch is opened.

You have been provided with a circuit containing a power supply, switch and a component C, as shown in Fig. 1.1.

Throughout the experiment do not disconnect this circuit.

Fig. 1.1

(a)

Assemble the circuit of Fig. 1.2 with the $10.0\ \text{k}\Omega$ resistor clipped into the component holder as resistance $S$ .

Fig. 1.2

DifficultyEasy

Worked solution

Answer

Construct the circuit as in Fig. 1.2, with the $10.0\ \text{k}\Omega$ resistor clipped into the component holder as $S$ , and the voltmeter connected in parallel with correct polarity.

Final answer

Circuit assembled with 10.0 kΩ resistor as S and voltmeter connected in parallel (correct polarity).

Detailed explanation

Background Concept

A circuit diagram shows how components are connected (series or parallel), not their physical layout. A voltmeter measures potential difference across a component, so it must be connected in parallel with that component/branch. A voltmeter also has polarity: the terminal marked $+$ should be at higher potential than the terminal marked $-$ .

Understanding the Question

You are instructed to assemble the provided circuit of Fig. 1.2, specifically placing the $10.0\ \text{k}\Omega$ resistor into the component holder and treating it as resistance $S$ . You must not disconnect the supplied base circuit containing the supply, switch, and component $C$ .

Approach

Identify where the resistor $S$ sits in the circuit (the component holder position).
Place the $10.0\ \text{k}\Omega$ resistor into the holder securely.
Check that the voltmeter is connected across the correct two points (parallel) and that its polarity matches the supply polarity.

Step-by-Step Reasoning

Clip the $10.0\ \text{k}\Omega$ resistor into the component holder labelled $S$ .
Ensure the voltmeter leads go across the same two nodes shown in Fig. 1.2 (i.e. in parallel with the branch containing $C$ ).
Connect the voltmeter $+$ terminal to the side of the circuit nearer the supply positive terminal, and the voltmeter $-$ terminal to the side nearer the supply negative terminal.
Make sure all connections are firm so the reading is stable (loose clips give fluctuating readings).

Key Takeaways

Voltmeters go in parallel; ammeters go in series.
Polarity matters for a d.c. reading (a reversed voltmeter may read negative).

Common Mistakes

Putting the voltmeter in series (gives incorrect/near-zero reading).
Reversing voltmeter polarity and then misreading the sign.
Not fully clipping the resistor into the holder so the resistance is intermittent.

Things to Be Careful About

Do not disconnect the provided circuit (instruction in the stem).
Ensure the component holder contacts the resistor leads properly (metal-to-metal contact).

Techniques used

assemble an electrical circuit following a circuit diagramconnect a voltmeter in parallel with correct polarityensure secure connections using clips/component holders

(b)

(i)

Close the switch and check that the voltmeter reading is between $4\ \text{V}$ and $8\ \text{V}$ .

DifficultyEasy

Worked solution

Answer

Close the switch and confirm that the voltmeter reading lies between $4\ \text{V}$ and $8\ \text{V}$ .

Final answer

Voltmeter reading confirmed between 4 V and 8 V.

Detailed explanation

Background Concept

When a switch is closed in a d.c. circuit, a potential difference from the supply appears across components as determined by the circuit connections. A voltmeter displays this potential difference; readings outside a required range usually mean the supply setting is incorrect, a connection is wrong, or a component is faulty.

Understanding the Question

You are asked to close the switch and simply check that the voltmeter reads between $4\ \text{V}$ and $8\ \text{V}$ before taking timing measurements. This ensures the starting voltage is high enough so that the decay down to $2.0\ \text{V}$ is measurable.

Approach

Close the switch.
Observe the voltmeter reading.
If necessary, adjust only what is permitted by the apparatus (e.g. supply setting) while keeping the circuit connected.

Step-by-Step Reasoning

With the switch closed, the component(s) are connected to the supply so the voltmeter should show a steady value.
Compare that value to the allowed interval $4$ to $8\ \text{V}$ .
If the reading is unstable, check for loose connections or incorrect voltmeter polarity.

Key Takeaways

Practical checks before data collection reduce wasted results.
Stable readings usually require firm electrical connections.

Common Mistakes

Forgetting to close the switch before checking.
Misreading the voltmeter scale or range.
Ignoring an unstable reading that indicates a poor contact.

Things to Be Careful About

Do not disconnect the circuit while troubleshooting (stem instruction).
Ensure the voltmeter is on an appropriate range so the reading is not over-range or low resolution.

Techniques used

operate a switch safely in a d.c. circuitread an analogue/digital voltmeter correctlycheck readings fall within a specified range

(ii)

When the switch is opened the voltmeter reading will gradually decrease.
Take measurements to find the time $t$ for the voltmeter reading to decrease to $2.0\ \text{V}$ after the switch is opened.
Record $t$ .

$t$ = ______

DifficultyMedium-Easy

Worked solution

Working

Open the switch and start timing at the instant the switch is opened.
Stop timing when the voltmeter first reads $2.0\ \text{V}$ .
Repeat and take a mean.

Example (with $S = 10.0\ \text{k}\Omega$ ):

t_1 = 2.9\ \text{s},\quad t_2 = 2.8\ \text{s}

\bar{t} = \frac{t_1+t_2}{2} = \frac{2.9+2.8}{2} = 2.85\ \text{s}\ \approx 2.9\ \text{s}

Answer

$t = 2.9\ \text{s}$ (example; record your measured mean value).

Final answer

t ≈ 2.9 s (example; student-dependent).

Detailed explanation

Background Concept

When a switch is opened in a circuit containing a component that can store energy (often a capacitor), the voltage across it does not drop instantly to zero; instead it decreases over time. The practical skill here is measuring the time interval for the voltage to fall to a specified value.

Understanding the Question

After you close the switch and obtain a starting voltmeter reading between $4$ and $8\ \text{V}$ , you then open the switch. The voltmeter reading gradually decreases. You must measure the time $t$ taken from the moment the switch is opened until the voltmeter reading reaches $2.0\ \text{V}$ .

Approach

Define a clear start event: the instant the switch is opened.
Define a clear stop event: the first time the voltmeter reads $2.0\ \text{V}$ .
Use repeat readings to reduce random reaction-time error.

Step-by-Step Reasoning

diagram-bii-1

With the switch closed, allow the voltmeter reading to stabilise.
Prepare the stopwatch: finger ready on start/stop.
Open the switch and start the stopwatch at the same instant (or as close as possible).
Watch the voltmeter reading decrease.
The moment the display reaches $2.0\ \text{V}$ , stop the stopwatch and record $t$ .
Repeat the timing at least once (better: 3 times) and calculate a mean value to reduce random error.

Key Takeaways

Always define start/stop events precisely.
Repeat timings and average to improve reliability.

Common Mistakes

Starting timing before the switch is fully opened, or stopping late after passing $2.0\ \text{V}$ .
Taking only one reading (poor reliability).
Recording $t$ to an unrealistic precision (e.g. many decimal places) compared with reaction time.

Things to Be Careful About

Use the same criterion each time: stop when the voltmeter first shows $2.0\ \text{V}$ .
If the reading changes quickly near $2.0\ \text{V}$ , repeat more times and use the mean.
Ensure the voltmeter range gives sufficient resolution around $2.0\ \text{V}$ (e.g. not a coarse scale).

Techniques used

measure time intervals using a stopwatch triggered by a circuit eventidentify when a voltmeter reading reaches a stated valuerepeat measurements and calculate a mean time

(c)

Repeat (b) with different resistors in the component holder until you have six sets of values of $S$ and $t$ .
Include values of $\frac{1}{S}$ and $\frac{1}{t}$ in your table.

10M

DifficultyMedium

Worked solution

Answer

Record six sets of $S$ and mean $t$ , then calculate $\frac{1}{S}$ and $\frac{1}{t}$ .

Example table (illustrative):

$S\ /\ \text{k}\Omega$	$t\ /\ \text{s}$	$1/S\ /\ \text{k}\Omega^{-1}$	$1/t\ /\ \text{s}^{-1}$
10.0	2.9	0.100	0.345
15.0	4.0	0.0667	0.250
22.0	5.4	0.0455	0.185
33.0	7.1	0.0303	0.141
47.0	8.8	0.0213	0.114
68.0	10.6	0.0147	0.0943

(Use your measured values of $t$ and your available resistors for $S$ .)

Final answer

Single table with 6 sets of S and t plus calculated 1/S and 1/t (student-dependent).

Detailed explanation

Background Concept

In Paper 3, marks are awarded for quality of data and good presentation as much as for the physics. You must:

take a suitable range of the independent variable (here $S$ ),
obtain repeated readings to improve reliability,
and present all results in one clear table with correct headings and units.

Derived quantities (like $1/S$ and $1/t$ ) should be calculated consistently and recorded to sensible significant figures based on the raw measurements.

Understanding the Question

You must repeat the timing measurement for different resistors placed in the holder so that you obtain six pairs of values $(S, t)$ . You must also include the calculated columns $1/S$ and $1/t$ in your table.

Approach

Choose six different resistor values $S$ spanning as wide a range as available.
For each $S$ , measure $t$ (preferably repeated, then average).
Record $S$ and mean $t$ in a table with units.
Calculate and record $1/S$ and $1/t$ with consistent significant figures.

Step-by-Step Reasoning

Independent variable: $S$ (choose different resistors).
Dependent variable: $t$ (the time to decay to $2.0\ \text{V}$ ).
For each resistor:
- close switch, confirm starting voltage is $4$ – $8\ \text{V}$ ,
- open switch and time to $2.0\ \text{V}$ ,
- repeat to reduce random error, then compute mean $t$ .
Build a single table:
- headings must be in the form “quantity / unit”, e.g. $t / \text{s}$ .
- calculate $1/S$ and $1/t$ with a calculator.
- keep the same number of decimal places (or significant figures) down each calculated column.

Key Takeaways

Good tables: one table, clear headings with units, consistent precision.
Good data: six points, wide range of $S$ , repeats/means for timing.

Common Mistakes

Missing units in headings (e.g. writing just “ $t$ ” instead of “ $t / \text{s}$ ”).
Mixing units (some $S$ in $\Omega$ , some in $\text{k}\Omega$ ) without stating/being consistent.
Incorrect reciprocal calculations (especially forgetting that $1/(10.0\ \text{k}\Omega) = 0.100\ \text{k}\Omega^{-1}$ if using kΩ units).
Inconsistent precision (e.g. $0.1$ , $0.067$ , $0.04545$ in one column).

Things to Be Careful About

Choose a sensible unit for $S$ (often $\text{k}\Omega$ ) and stick to it for all rows and for $1/S$ .
If $t$ is measured to the nearest $0.1\ \text{s}$ , quoting $1/t$ to 4–5 s.f. is usually unjustified.
Ensure you really have six distinct $S$ values and corresponding $t$ values (not repeats of the same resistor).

Techniques used

select a range of resistor values for the independent variablemeasure and record times with repeat readings and a meancalculate reciprocal quantities correctly with consistent significant figuresconstruct a results table with headings and units

(d)

(i)

Plot a graph of $\frac{1}{t}$ on the $y$ -axis against $\frac{1}{S}$ on the $x$ -axis.

DifficultyMedium

Worked solution

Answer

Plot $y = \frac{1}{t}$ (units $\text{s}^{-1}$ ) on the $y$ -axis against $x = \frac{1}{S}$ (units consistent with your table, e.g. $\text{k}\Omega^{-1}$ ) on the $x$ -axis, using a suitable scale and plotting all six points accurately.

Final answer

Graph of 1/t (y-axis) against 1/S (x-axis) plotted with correct labels/units and suitable scales.

Detailed explanation

Background Concept

A graph is used to reveal relationships and allow constants to be determined from gradients and intercepts. For full credit you must:

label axes with both quantity and unit,
choose scales that use at least about half the grid in each direction,
plot points with small, neat crosses/dots accurately.

Understanding the Question

You have calculated $1/S$ and $1/t$ in your results table. You must now plot a graph with $1/t$ on the vertical axis and $1/S$ on the horizontal axis.

Approach

Decide which column is $x$ and which is $y$ .
Choose simple scales (e.g. 1 big square = 0.01) that spread the points.
Plot all six pairs $(1/S, 1/t)$ .

Step-by-Step Reasoning

On the horizontal axis write $1/S$ with units (e.g. $\text{k}\Omega^{-1}$ if you used $S$ in kΩ).
On the vertical axis write $1/t$ with units $\text{s}^{-1}$ .
Mark a scale that covers your full data range.
Plot each point using the table values, checking you have not swapped $x$ and $y$ .

Key Takeaways

Axes must be quantity AND unit.
A good scale makes later gradient calculation more accurate.

Common Mistakes

Plotting $t$ against $S$ instead of $1/t$ against $1/S$ .
Missing units on one or both axes.
Using an awkward scale (e.g. 3 squares = 0.02) that makes reading errors larger.

Things to Be Careful About

Keep units consistent: if $x$ is in $\text{k}\Omega^{-1}$ , your gradient unit will depend on that choice.
Plot all six points; do not omit a point because it seems “off” unless you have a clear experimental reason.

Techniques used

choose suitable axis scales that use most of the gridlabel axes with quantity and unitplot points accurately from a results table

(ii)

Draw the straight line of best fit.

DifficultyMedium-Easy

Worked solution

Answer

Draw a single straight line of best fit through the plotted points, with roughly equal scatter of points above and below the line.

Final answer

Straight line of best fit drawn.

Detailed explanation

Background Concept

A best-fit line represents the overall trend of experimental data when random uncertainties cause scatter. For linear relationships, you should draw one straight line that best represents the data rather than connecting points.

Understanding the Question

Having plotted $1/t$ against $1/S$ , you are told to draw the straight line of best fit. This line will be used next to determine gradient and intercept.

Approach

Use a ruler.
Aim for a line that passes through the middle of the cluster of points.
Ensure the line is not forced through every point (it won’t be if there is scatter).

Step-by-Step Reasoning

Place a ruler so that the number (or distribution) of points above and below the line is about balanced.
Draw a thin, continuous straight line across most of the data range (not just between two middle points).

Key Takeaways

Best-fit line is about the trend, not connecting points.
Extending the line over the whole range improves intercept reading.

Common Mistakes

Joining the points dot-to-dot.
Forcing the line through the origin when the data do not support it.
Drawing a short line segment rather than a full best-fit line.

Things to Be Careful About

If one point is clearly an outlier, still draw the best-fit line for the main trend unless instructed otherwise.
Keep the line thin to reduce reading error when finding intercept/gradient.

Techniques used

draw a straight line of best fit by balancing scatteravoid point-to-point joining when a best-fit line is required

(iii)

Determine the gradient and $y$ -intercept of this line.

gradient = ______
$y$ -intercept = ______

DifficultyMedium

Worked solution

Working

Use a large triangle on the best-fit line:

\text{gradient} = \frac{\Delta(1/t)}{\Delta(1/S)}

Read the $y$ -intercept where the line crosses the $y$ -axis.

Example values (from an illustrative straight line):

\text{gradient} = 3.0\ \text{k}\Omega\ \text{s}^{-1}

\text{$y$-intercept} = 0.050\ \text{s}^{-1}

Answer

gradient $= 3.0\ \text{k}\Omega\ \text{s}^{-1}$

$y$ -intercept $= 0.050\ \text{s}^{-1}$

(example; use your graph readings).

Final answer

gradient ≈ 3.0 kΩ s^-1, y-intercept ≈ 0.050 s^-1 (example; student-dependent).

Detailed explanation

Background Concept

For a straight-line graph of the form

y = mx + c

the gradient is

m = \frac{\Delta y}{\Delta x}

and the y-intercept is the value of $y$ when $x = 0$ .

In practical work, the most accurate gradient comes from using a large triangle on the best-fit line (not between two adjacent data points).

Understanding the Question

You must obtain two numerical values from your best-fit line on the graph of $1/t$ (y-axis) against $1/S$ (x-axis):

the gradient,
the y-intercept.

These will be used in part (e) to find constants $a$ and $b$ .

Approach

Choose two well-separated points on the best-fit line (not necessarily plotted points).
Read off their coordinates accurately.
Compute gradient as $\Delta(1/t) / \Delta(1/S)$ .
Read the y-intercept at $1/S = 0$ .

Step-by-Step Reasoning

diagram-diii-1

Pick two points on the best-fit line far apart to make $\Delta x$ large (this reduces percentage reading uncertainty).
Read $x_1 = (1/S)_1$ , $y_1 = (1/t)_1$ and $x_2 = (1/S)_2$ , $y_2 = (1/t)_2$ .
Calculate

\text{gradient} = \frac{y_2 - y_1}{x_2 - x_1}

Determine the y-intercept by extending the best-fit line to the y-axis and reading $y$ at $x=0$ .
Quote units:
- $y$ has units $\text{s}^{-1}$ ,
- $x$ has units set by your table (e.g. $\text{k}\Omega^{-1}$ ),
- so gradient has units $\text{s}^{-1} / \text{k}\Omega^{-1} = \text{k}\Omega\ \text{s}^{-1}$ .

Key Takeaways

Use the best-fit line, not point-to-point.
Gradient is always $\Delta y/\Delta x$ .
Units come from the axes labels.

Common Mistakes

Using two neighbouring points, giving a small triangle and large percentage error.
Swapping $\Delta x$ and $\Delta y$ (inverting the gradient).
Forgetting units, or using inconsistent units (e.g. mixing $\Omega^{-1}$ and $\text{k}\Omega^{-1}$ ).

Things to Be Careful About

Read coordinates to about half a small square if possible.
Extend the line cleanly to the y-axis to read the intercept (do not guess without extension).
Keep significant figures sensible (often 2–3 s.f. for gradients from hand graphs).

Techniques used

determine gradient using a large triangle on the best-fit linecalculate gradient as \Delta y / \Delta x with correct unitsread or calculate the y-intercept from the best-fit line

(e)

The quantities $t$ and $S$ are related by the equation

\frac{1}{t} = \frac{a}{S} + ab

where $a$ and $b$ are constants.

Using your answers from (d)(iii), determine the values of $a$ and $b$ .
Give appropriate units.

$a$ = ______
$b$ = ______

DifficultyMedium-Easy

Worked solution

Working

Given

\frac{1}{t} = \frac{a}{S} + ab

For a graph of $y=\frac{1}{t}$ against $x=\frac{1}{S}$ :

y = ax + ab

So gradient $= a$ and y-intercept $= ab$ .

Using example values from (d)(iii):

a = 3.0\ \text{k}\Omega\ \text{s}^{-1}

b = \frac{\text{intercept}}{a} = \frac{0.050}{3.0} = 1.7 \times 10^{-2}\ \text{k}\Omega^{-1}

Answer

$a = 3.0\ \text{k}\Omega\ \text{s}^{-1}$

$b = 1.7 \times 10^{-2}\ \text{k}\Omega^{-1}$

(example; use your gradient and intercept).

Final answer

a = gradient; b = (y-intercept)/(gradient) with corresponding units (student-dependent).

Detailed explanation

Background Concept

If experimental data produce a straight line when plotting $y$ against $x$ , you can compare the equation to

y = mx + c

and identify

$m$ (gradient) as the coefficient of $x$ ,
$c$ (y-intercept) as the constant term.

Units follow from the graph axes: if $y$ is in $\text{s}^{-1}$ and $x$ is in $\text{k}\Omega^{-1}$ , then $m$ has units $\text{s}^{-1}/\text{k}\Omega^{-1} = \text{k}\Omega\ \text{s}^{-1}$ .

Understanding the Question

You are given the relationship

\frac{1}{t} = \frac{a}{S} + ab

and you have already found the gradient and y-intercept from the graph of $1/t$ against $1/S$ . You must use those values to determine the constants $a$ and $b$ , including appropriate units.

Approach

Rewrite the given equation in the same structure as $y = mx + c$ by identifying $y$ and $x$ .
Equate the gradient to $a$ .
Equate the intercept to $ab$ .
Rearrange $b = (ab)/a$ .
Assign units from the axis units.

Step-by-Step Reasoning

Let
- $y = 1/t$ ,
- $x = 1/S$ .
Then

\frac{1}{t} = a\left(\frac{1}{S}\right) + ab

which matches $y = mx + c$ with:

gradient $m = a$ ,
intercept $c = ab$ .

So:

$a$ is simply the gradient you measured.
$b$ is found from

b = \frac{ab}{a} = \frac{\text{intercept}}{\text{gradient}}

Units:

intercept has units of $y$ , i.e. $\text{s}^{-1}$ ,
gradient has units $\text{k}\Omega\ \text{s}^{-1}$ (if $x$ is $\text{k}\Omega^{-1}$ ),
hence

[b] = \frac{\text{s}^{-1}}{\text{k}\Omega\ \text{s}^{-1}} = \text{k}\Omega^{-1}

Key Takeaways

Identify $y$ and $x$ from the plotted graph, then match to $y = mx + c$ .
$a$ comes from the gradient; $b$ comes from intercept divided by gradient.
Always state units and ensure they are consistent with your axis choices.

Common Mistakes

Swapping $a$ and $ab$ (thinking intercept is $b$ directly).
Using $b = a/(ab)$ instead of $b = (ab)/a$ .
Giving units for $a$ and $b$ that do not match the axes used (especially if using $\Omega^{-1}$ vs $\text{k}\Omega^{-1}$ ).

Things to Be Careful About

If you plotted $1/S$ in $\Omega^{-1}$ instead of $\text{k}\Omega^{-1}$ , your numerical value of $a$ changes by a factor of $10^3$ and so do the units.
Quote $a$ and $b$ to a sensible number of significant figures consistent with your gradient/intercept readings from the graph.

Techniques used

match a straight-line graph to y = mx + cequate gradient and intercept to constants in a given equationdetermine units of constants from graph axis units

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Q2Manipulation, Measurement and Observation · Analysis, Conclusions and Evaluation · Presentation of Data and Observations20M

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