In this experiment, you will investigate the optical properties of glass jars. You have been provided with two glass jars A and B, each containing water. Each jar has a lid. (a) The diameter of jar A is $D$, as shown in Fig. 2.1. Measure and record $D$. $D$ = \_\_\_\_\_\_ (i) ● Hold the nail next to jar A, as shown in Fig. 2.2. ● Close one eye and look at the nail through the water. The bottom of the nail seen through the water will appear to be wider than the top of the…

(a) D = 8.5 cm · (b)(i) y = 2.6 cm · (b)(ii) 8% · (b)(iii) 6.9 cm · (c) D = 10.2 cm, y = 3.0 cm, (r + y) = 8.1 cm · (d)(i) k = 1.61 and 1.59 · (d)(ii) k quoted to 3 s.f. because D and y were measured to 0.1 cm, limiting r and (r+y) to about 0.1 cm. · (e) Yes; k values agree within 20%. · (f) y = 2.4 cm, r = 4.1 cm · (g)(i) Four uncertainties/limitations listed (see solution). · (g)(ii) Four improvements listed (see solution).

Physics 9702/35 — May/June 2024

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

Q1MediumManipulation, Measurement and ObservationPresentation of Data and ObservationsAnalysis, Conclusions and Evaluation

In this experiment, you will investigate the motion of a loaded metre rule.

(a)

● Set up the apparatus as shown in Fig. 1.1.

Fig. 1.1

● Place the slotted mass in the string loop attached to the springs.
● Pull the slotted mass downwards through a small distance.
● Release the mass. The mass will oscillate.
● Determine the period $T_0$ of the oscillations of the mass.

$T_0$ = ______

● Remove the slotted mass from the string loop attached to the springs.

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Measure time for $n$ oscillations (e.g. $n=10$ ) and calculate

T_0 = \frac{t}{n}

Repeat and take the mean $T_0$ .

Example: for $n=10$ , $t=8.0\ \text{s}$

T_0 = \frac{8.0}{10} = 0.80\ \text{s}

Answer

$T_0 = 0.80\ \text{s}$ (example)

Final answer

T0 = 0.80 s (example)

Detailed explanation

Background Concept

The period $T$ of an oscillation is the time for one complete cycle (e.g. from a maximum displacement back to the same maximum displacement in the same direction).

When using a stopwatch, the reaction time in starting/stopping is a significant fraction of a single period, so a standard method is to time several oscillations and divide by the number:

T = \frac{t}{n}

where $t$ is the total measured time for $n$ complete oscillations.

Understanding the Question

You are asked to determine $T_0$ , the period of oscillations of the slotted mass hanging from the springs (before the metre rule is used). This $T_0$ value will later be used to calculate $(T-T_0)^2$ , so it should be measured carefully and to sensible precision.

Approach

Displace the mass a small distance and release to start oscillations.
Choose a number of oscillations $n$ large enough to reduce reaction-time error (typically $10$ or more).
Measure the time $t$ for $n$ complete oscillations using a fixed reference point.
Calculate $T_0=t/n$ .
Repeat the timing and average $T_0$ .

Step-by-Step Reasoning

Pick a reference point (e.g. the equilibrium position or the lowest point) and count one oscillation when the mass returns to the same point moving in the same direction.
If you time $n=10$ oscillations and obtain $t=8.0\ \text{s}$ , then

T_0 = \frac{8.0\ \text{s}}{10} = 0.80\ \text{s}.

Repeat (e.g. another $t$ value for $10$ oscillations) and take the mean to reduce random scatter.

Key Takeaways

Period is best found by timing many oscillations and dividing by $n$ .
Repeats and averaging improve reliability.

Common Mistakes

Timing just one oscillation (large percentage uncertainty due to reaction time).
Miscounting oscillations (not returning to the same position and direction).
Using a large amplitude so the motion is not close to simple harmonic motion (period may change slightly).

Things to Be Careful About

Use a small displacement so the oscillation period is as constant as possible.
Use consistent start/stop points for every timing.
Quote $T_0$ to a sensible precision (typically $0.01\ \text{s}$ if using a stopwatch and $n\ge 10$ ).

Techniques used

measure the time for multiple oscillations and divide by the number of oscillationsrepeat timings and calculate a mean to reduce random erroruse consistent reference point for counting complete oscillations

(b)

● Set up the apparatus as shown in Fig. 1.2.

Fig. 1.2

● Position the string loop attached to the springs at the 50 cm mark on the rule. This string loop must remain in this position throughout the experiment.
● The distance between the string loop supporting the slotted mass and the end B of the rule is $x$ .

Position the mass so that $x$ is approximately 20 cm.

● Adjust the apparatus so that the rule is parallel to the bench and the springs are vertical.
● Record $x$ .

$x$ = ______

● Pull B downwards through a small distance.
● Release B. The rule will oscillate.
● Determine the period $T$ of the oscillations of the rule.

$T$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Example readings (to suitable precision):

x = 20.0\ \text{cm}

Determine $T$ by timing $n$ oscillations and using $T=t/n$ .

Example:

T = 0.95\ \text{s}

Final answer

x = 20.0 cm, T = 0.95 s (example)

Detailed explanation

Background Concept

In this set-up the metre rule oscillates vertically when end B is displaced and released. You need two measurements:

$x$ : a length read from the rule (distance between the hanging mass position and end B).
$T$ : the oscillation period, found from timing multiple oscillations:

T = \frac{t}{n}.

Understanding the Question

You must:

Set the spring attachment point at the $50\ \text{cm}$ mark and keep it fixed.
Place the slotted mass so that the distance from the mass to end B is approximately $20\ \text{cm}$ .
Adjust the apparatus so the rule is horizontal and the springs are vertical.
Record $x$ and then determine the oscillation period $T$ of the rule.

Approach

Read $x$ directly from the rule scale by identifying the position of the hanging mass relative to end B.
Create small oscillations at B, then time $n$ oscillations and divide by $n$ .

Step-by-Step Reasoning

After positioning the mass near the required location, ensure the rule is parallel to the bench: this reduces unwanted sideways motion and keeps the oscillation close to the intended mode.
Measure $x$ to the nearest millimetre (typically $0.1\ \text{cm}$ ).
Pull end B down slightly and release.
Time, for example, $n=10$ oscillations: if $t=9.5\ \text{s}$ , then

T=\frac{9.5}{10}=0.95\ \text{s}.

Key Takeaways

Identify distances using the correct reference points.
Use multiple oscillations for timing-based measurements.

Common Mistakes

Measuring $x$ from the wrong end of the rule (must be from end B as stated).
Allowing the spring attachment point to shift from the $50\ \text{cm}$ mark.
Large oscillation amplitude causing non-reproducible periods.

Things to Be Careful About

Keep springs vertical to avoid lateral forces and twisting.
Ensure the rule is level before releasing, so the motion is consistent each time.
Count oscillations consistently (same reference point and direction).

Techniques used

measure a length from a specified reference point on a scaleensure the apparatus is level and aligned before taking readingsmeasure the period by timing multiple oscillations

(c)

Change $x$ by moving the mass along the rule. For each value of $x$ , adjust the apparatus so that the rule is parallel to the bench and the springs are vertical, then determine $T$ .

Repeat until you have six sets of values of $x$ and $T$ with $x$ in the range $10\ \text{cm} \leq x \leq 40\ \text{cm}$ .

Record your results in a table. Include values of $(T - T_0)^2$ in your table.

9M Similar

DifficultyMedium

Worked solution

Answer

Record six pairs of $x$ and $T$ with $10\ \text{cm} \le x \le 40\ \text{cm}$ and include a calculated column for $(T-T_0)^2$ .

Example (using $T_0=0.80\ \text{s}$ ):

$x\ /\ \text{cm}$	$T\ /\ \text{s}$	$(T-T_0)^2\ /\ \text{s}^2$
10.0	0.96	0.0256
16.0	0.95	0.0225
22.0	0.93	0.0169
28.0	0.92	0.0144
34.0	0.91	0.0121
40.0	0.90	0.0100

Final answer

Table of x, T and (T−T0)^2 with 6 sets over 10–40 cm (see solution).

Detailed explanation

Background Concept

Good experimental tables allow the reader to see raw measurements and derived quantities clearly and unambiguously.

Here, you measure $x$ and $T$ , and calculate

(T-T_0)^2

Since $T$ and $T_0$ are in seconds, $(T-T_0)^2$ has unit $\text{s}^2$ .

Understanding the Question

You must:

vary $x$ by moving the mass along the rule,
for each $x$ , ensure the rule is horizontal and springs are vertical,
measure the period $T$ of the rule,
repeat until you have six sets of $(x,T)$ with $x$ spanning 10 cm to 40 cm,
present results in a single table, including the derived quantity $(T-T_0)^2$ .

Approach

Choose six $x$ values spread across 10–40 cm (not clustered), e.g. roughly equal steps.
For each $x$ $x$ :
- adjust alignment (rule level, springs vertical),
- time $n$ oscillations and compute $T=t/n$ ,
- optionally repeat timing and average $T$ .
Compute $(T-T_0)^2$ for each row.
Present everything with headings containing quantity and unit.

Step-by-Step Reasoning

Independent variable: $x$ (you choose values across the required range).
Dependent variable: $T$ (measured by oscillation timing).
Derived column: $(T-T_0)^2$ .

For each row, after measuring $T$ and using your previously found $T_0$ :

(T-T_0)^2 = (T-T_0)\times(T-T_0).

Example: if $T=0.93\ \text{s}$ and $T_0=0.80\ \text{s}$ ,

T-T_0 = 0.13\ \text{s},\qquad (T-T_0)^2 = (0.13)^2 = 0.0169\ \text{s}^2.

A correct table typically earns marks for:

one table containing all results,
clear column headings with units,
consistent precision within each column (e.g. $x$ to $0.1\ \text{cm}$ , $T$ to $0.01\ \text{s}$ ),
correctly calculated derived values.

Key Takeaways

Use a wide, sensible range and enough data points (here, six) for a reliable graph.
Derived quantities must be calculated correctly and given appropriate units.

Common Mistakes

Fewer than six sets of readings.
$x$ values not spanning the full 10–40 cm range.
Missing units in table headings.
Inconsistent decimal places within a column (suggests poor measurement/recording quality).
Calculating $(T-T_0)^2$ incorrectly (e.g. squaring $T$ and $T_0$ separately).

Things to Be Careful About

Always keep the spring attachment point fixed at the 50 cm mark as instructed.
Keep oscillation amplitude small and similar for each run.
If you repeat timings, record the mean $T$ (and be consistent about whether the table contains single or mean values).

Techniques used

collect a suitable range of the independent variable with sufficient data pointsrecord measurements with consistent precision and appropriate unitscalculate a derived quantity from measured values and present it in a tablerepeat measurements and use mean values where appropriate

(d)

(i)

Plot a graph of $(T - T_0)^2$ on the $y$ -axis against $x$ on the $x$ -axis.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Plot $(T-T_0)^2$ on the $y$ -axis against $x$ on the $x$ -axis.

Label axes: $x / \text{cm}$ and $(T-T_0)^2 / \text{s}^2$ .
Use a suitable scale (at least half the grid in both directions).
Plot all six points accurately.

Final answer

Graph of (T−T0)^2 (y) against x (x) plotted.

Detailed explanation

Background Concept

Graph plotting is used to reveal and test relationships between variables. A good graph should:

have correctly labelled axes (quantity and unit),
use scales that make good use of the available grid,
show accurately plotted points.

Understanding the Question

You have calculated values of $(T-T_0)^2$ for six different $x$ values. You are asked to plot a graph with:

horizontal axis: $x$ ,
vertical axis: $(T-T_0)^2$ .

This choice of axes is important because it is intended to produce a straight-line graph if the suggested relationship is correct.

Approach

Put the independent variable ( $x$ ) on the $x$ -axis.
Put the dependent/derived variable ( $(T-T_0)^2$ ) on the $y$ -axis.
Choose scales that spread the points out (avoid awkward scales like 3 squares = 0.01 unless necessary).
Plot each point carefully.

Step-by-Step Reasoning

Draw axes and choose a scale:
- $x$ should cover at least $10$ to $40\ \text{cm}$ .
- $(T-T_0)^2$ should cover the full range of your calculated values.
Label axes with both symbol and unit:
- $x / \text{cm}$
- $(T-T_0)^2 / \text{s}^2$
Plot each point as a small cross ((\times)) or dot with a circle.
Check that each plotted point matches the table value (common plotting mistakes are swapping axes or mis-reading the scale).

Key Takeaways

Correct axes and sensible scales are essential for marks and for a reliable gradient/intercept later.

Common Mistakes

Missing units on one or both axes.
Using a scale that compresses points into a small region.
Plotting $(T-T_0)$ instead of $(T-T_0)^2$ .

Things to Be Careful About

Do not join dot-to-dot; you will later draw a single best-fit straight line.
Ensure the origin does not need to be included unless it helps the scale; the graph need only cover the data range.

Techniques used

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

(ii)

Draw the straight line of best fit.

1M Similar

(iii)

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

gradient = ______
$y$ -intercept = ______

2M Similar

DifficultyMedium

Worked solution

Working

Using two well-separated points on the best-fit line,

\text{gradient } m = \frac{\Delta (T-T_0)^2}{\Delta x}

Example from line: $m = -5.0 \times 10^{-4}\ \text{s}^2\ \text{cm}^{-1}$ .

At $x=0$ , read $y$ -intercept $c$ .

Example: $c = 3.0 \times 10^{-2}\ \text{s}^2$ .

Answer

gradient $= -5.0 \times 10^{-4}\ \text{s}^2\ \text{cm}^{-1}$ (example)

$y$ -intercept $= 3.0 \times 10^{-2}\ \text{s}^2$ (example)

Final answer

gradient = −5.0×10^−4 s^2 cm^−1, y-intercept = 3.0×10^−2 s^2 (example)

Detailed explanation

Background Concept

For a straight-line graph,

y = mx + c

$m$ is the gradient (slope):

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

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

The units come from the axes: here $y=(T-T_0)^2$ in $\text{s}^2$ and $x$ in $\text{cm}$ , so

[m] = \frac{\text{s}^2}{\text{cm}} = \text{s}^2\ \text{cm}^{-1}.

Understanding the Question

You must use your best-fit line on the graph of $(T-T_0)^2$ (vertical axis) against $x$ (horizontal axis) to find:

the gradient of the line,
the $y$ -intercept.

These will be used in part (e) to find constants $P$ and $Q$ .

Approach

Pick two points on the best-fit line that are far apart (not necessarily actual plotted data points).
Read their coordinates accurately.
Compute $m=\Delta y/\Delta x$ .
Read the intercept where the best-fit line crosses the $y$ -axis (or extrapolate to $x=0$ if needed).

Step-by-Step Reasoning

diagram-1

Choose two points on the line with a large separation in $x$ to reduce percentage reading error.
Suppose two points read from the line are $(x_1, y_1)$ and $(x_2, y_2)$ . Then:

\Delta y = y_2 - y_1,\qquad \Delta x = x_2 - x_1

and

m = \frac{y_2-y_1}{x_2-x_1}.

A negative gradient means $y$ decreases as $x$ increases, so the line slopes downwards left-to-right.
For the $y$ -intercept, look at where the line crosses $x=0$ . If the axis does not include $x=0$ , extend the line back (extrapolation) using the ruler.

Key Takeaways

Gradient must come from the best-fit line, using a large triangle.
Units of gradient and intercept come directly from axis units.

Common Mistakes

Using two neighbouring points (large fractional uncertainty in gradient).
Calculating $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Using plotted points rather than points on the best-fit line (increases effect of scatter).
Forgetting units, especially for the gradient.

Things to Be Careful About

Read coordinates to about half a small square on the graph paper.
Keep consistent units: if $x$ is plotted in cm, the gradient is per cm (do not suddenly quote per metre unless you convert properly).
If extrapolating to find the intercept, extend the line carefully with a ruler; do not guess freehand.

Techniques used

determine the gradient using a large triangle on the best-fit linecalculate gradient as $\Delta y / \Delta x$ with correct unitsread the y-intercept from the best-fit line at $x = 0$

(e)

It is suggested that the quantities $T$ , $T_0$ and $x$ are related by the equation

(T - T_0)^2 = -Px + Q

where $P$ and $Q$ are constants.

Using your answers in (d)(iii), determine the values of $P$ and $Q$ .
Give appropriate units.

$P$ = ______
$Q$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Let $y=(T-T_0)^2$ .

Given:

(T-T_0)^2 = -Px + Q

So comparing with $y=mx+c$ :

m=-P,\qquad c=Q

Hence

P=-m,\qquad Q=c

Using gradient $m=-5.0\times 10^{-4}\ \text{s}^2\ \text{cm}^{-1}$ and intercept $c=3.0\times 10^{-2}\ \text{s}^2$ :

P = 5.0\times 10^{-4}\ \text{s}^2\ \text{cm}^{-1}

Q = 3.0\times 10^{-2}\ \text{s}^2

Answer

$P = 5.0\times 10^{-4}\ \text{s}^2\ \text{cm}^{-1}$ (example)

$Q = 3.0\times 10^{-2}\ \text{s}^2$ (example)

Final answer

P = 5.0×10^−4 s^2 cm^−1, Q = 3.0×10^−2 s^2 (example)

Detailed explanation

Background Concept

If a relationship can be written in the linear form

y = mx + c,

then plotting $y$ against $x$ gives a straight line whose:

gradient is $m$ ,
$y$ -intercept is $c$ .

Comparing your experimental straight-line equation with the theoretical one allows you to identify constants.

Understanding the Question

You are told the suggested relationship is

(T-T_0)^2 = -Px + Q

and you have already found (from the graph in part (d)(iii)):

the gradient of the graph of $(T-T_0)^2$ against $x$ ,
the $y$ -intercept.

You must use these to determine $P$ and $Q$ , including correct units.

Approach

Recognise that your graph is of $y=(T-T_0)^2$ versus $x$ .
Write the equation in the form $y = mx + c$ .
Identify $m$ $m$ and $c$ $c$ by comparison:
- $m=-P$
- $c=Q$
Use the axis units to assign units to $P$ and $Q$ .

Step-by-Step Reasoning

Let

y=(T-T_0)^2.

Then the given equation becomes

y = -Px + Q.

Comparing with $y=mx+c$ :

m=-P\Rightarrow P=-m,

c=Q.

Units:

$y$ has unit $\text{s}^2$ .
$x$ has unit $\text{cm}$ (as plotted).
So $P$ must have unit

\frac{\text{s}^2}{\text{cm}} = \text{s}^2\ \text{cm}^{-1}

and $Q$ has unit $\text{s}^2$ .

Key Takeaways

Constants in a linear model come directly from the gradient and intercept.
Units follow from dimensional consistency with the plotted variables.

Common Mistakes

Taking $P$ equal to the gradient instead of the negative of the gradient.
Giving $P$ the wrong units (e.g. forgetting the "per cm" part).
Mixing units: using $x$ in cm for plotting but quoting $P$ in $\text{s}^2\ \text{m}^{-1}$ without converting.

Things to Be Careful About

Sign: because the equation has $-Px$ , if the graph slopes downwards then $P$ should come out positive.
State units explicitly: $P$ in $\text{s}^2\ \text{cm}^{-1}$ and $Q$ in $\text{s}^2$ (if your $x$ axis was in cm).

Techniques used

match a straight-line equation to y = mx + cidentify constants by comparing coefficientsdeduce units of constants from the graph axes

Q2MediumManipulation, Measurement and ObservationAnalysis, Conclusions and EvaluationPresentation of Data and Observations

In this experiment, you will investigate the optical properties of glass jars.

You have been provided with two glass jars A and B, each containing water. Each jar has a lid.

(a)

The diameter of jar A is $D$ , as shown in Fig. 2.1.

Fig. 2.1

Measure and record $D$ .

$D$ = ______

1M Similar

DifficultyEasy

Worked solution

Answer

Measured diameter (example):

$D = 8.5\ \text{cm}$ (recorded to the nearest $0.1\ \text{cm}$ ).

Final answer

D = 8.5 cm

Detailed explanation

Background Concept

A diameter is the straight-line distance across a circle through its centre. When measuring a diameter, the key ideas are:

Avoid parallax: your eye should be directly above the scale marking.
Use appropriate precision: for a standard ruler, you typically record to the nearest $1\ \text{mm} = 0.1\ \text{cm}$ .
Reduce random error: repeat the measurement at different orientations and take a mean.

Understanding the Question

You are given jar A and asked to measure and record its diameter $D$ as shown. There is no calculation required, but the mark depends on a sensible measured value recorded with an appropriate unit and precision.

Approach

Place the jar so the top is accessible.
Use a ruler (or calipers if available) to measure straight across the widest part through the centre.
Read the value at eye level and record it with unit and suitable precision (typically $0.1\ \text{cm}$ ).

Step-by-Step Reasoning

Align the ruler so that it passes through the centre of the circular lid/jar opening.
Ensure the ruler is not tilted (tilt gives an underestimate of the true diameter).
Read the scale at both edges of the jar and take the difference (this can be more reliable than trying to align exactly at zero).
Record, e.g. $D = 8.5\ \text{cm}$ to the nearest $0.1\ \text{cm}$ .

Key Takeaways

Measure across the centre.
Avoid parallax.
Record with unit and consistent precision.

Common Mistakes

Missing unit.
Measuring a chord not through the centre (gives too small $D$ ).
Recording with excessive precision (e.g. $8.532\ \text{cm}$ from a ruler).

Things to Be Careful About

If the jar has a lip or thickness, be consistent about whether you measure outer diameter or inner diameter—use what the diagram implies (usually the full diameter shown).
If using a ruler, reading both edges and subtracting reduces zero-end errors.

Techniques used

measure a length with a ruler or calipers using an appropriate scaletake a diameter reading through the centre to reduce systematic errorrecord a measurement to a justified precision with unit

(b)

(i)

● Hold the nail next to jar A, as shown in Fig. 2.2.

Fig. 2.2

● Close one eye and look at the nail through the water.

The bottom of the nail seen through the water will appear to be wider than the top of the nail, as shown in Fig. 2.3.

Fig. 2.3

● Move the nail away from the jar. The bottom of the nail will appear to become wider until it suddenly disappears. Hold the nail at this point.

● The distance between the nail and jar A is $y$ , as shown in Fig. 2.4.

Fig. 2.4

Measure and record $y$ .

$y$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Measured distance (example):

$y = 2.6\ \text{cm}$ (to the nearest $0.1\ \text{cm}$ ).

Final answer

y = 2.6 cm

Detailed explanation

Background Concept

This experiment relies on refraction through curved glass/water surfaces producing lens-like effects. The instruction to move the nail until the image “suddenly disappears” defines an operational point you can reproduce. Even if you do not fully model the optics, you can still obtain consistent data by:

Keeping the viewing position fixed (one eye closed).
Moving the nail slowly to detect the transition point.
Measuring the distance $y$ consistently from the same reference points.

Understanding the Question

You must:

Hold the nail next to jar A.
Look through the water with one eye.
Move the nail away until the bottom image suddenly disappears.
Measure $y$ , the horizontal distance between the nail and the nearest edge of jar A.

The mark is for a sensible, properly recorded value of $y$ .

Approach

Set up the nail close to the jar and choose a fixed eye position.
Move the nail directly away from the jar (keeping it parallel to the jar side).
Stop at the instant the bottom image disappears.
Measure $y$ with a ruler, ensuring the ruler is horizontal and aligned with the distance shown.

Step-by-Step Reasoning

Closing one eye prevents binocular viewing from changing the perceived disappearance point.
Move the nail slowly so you do not overshoot the disappearance position.
When the bottom disappears, keep the nail still (using a clamp helps, but if not available, hold as steady as possible).
Place a ruler so that its edge touches the jar at the reference edge and read across to the nail position.
Record $y$ to appropriate precision, e.g. $y = 2.6\ \text{cm}$ to the nearest $0.1\ \text{cm}$ .

Key Takeaways

Follow the procedure exactly to define a repeatable measurement condition.
Consistent alignment matters more than “getting a particular value”.

Common Mistakes

Measuring from the centre of the jar instead of the jar edge (wrong definition of $y$ ).
Changing head/eye position while moving the nail.
Measuring along a slanted line instead of horizontal.

Things to Be Careful About

Keep the nail vertical; tilting can change the apparent disappearance point.
Avoid parallax when reading the ruler.
Ensure the reference point on the jar is consistent (nearest outer surface/edge as implied by the diagram).

Techniques used

follow a described observation procedure to locate a specific conditionmeasure a horizontal distance with a ruler while controlling alignmentrecord a reading with an appropriate precision and unit

(ii)

Estimate the percentage uncertainty in your value of $y$ . Show your working.

percentage uncertainty = ______ %

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Take ruler reading uncertainty as $\pm 0.1\ \text{cm}$ at each end, so

\Delta y = \pm 0.2\ \text{cm}

\%\ \text{uncertainty} = \frac{\Delta y}{y} \times 100 = \frac{0.2}{2.6} \times 100 = 7.7\% \approx 8\%

Answer

$8\%$

Final answer

Detailed explanation

Background Concept

An uncertainty estimate describes the likely range within which the true value lies.

For a ruler with $1\ \text{mm}$ divisions, a single reading is often taken as $\pm 0.5\ \text{mm} = \pm 0.05\ \text{cm}$ .
When measuring a distance between two points (e.g. from jar edge to nail), you effectively make two readings, so a common approach is to add the absolute uncertainties.

Percentage uncertainty is

\%\ \text{uncertainty} = \frac{\text{absolute uncertainty}}{\text{measured value}} \times 100.

Understanding the Question

You have measured $y$ in part (b)(i). Now you must estimate the percentage uncertainty in that $y$ value and show working.

Approach

Decide a reasonable absolute uncertainty in $y$ based on how it was measured (instrument resolution, parallax, judging the nail position).
Divide by your measured $y$ and multiply by $100$ .
Round sensibly (typically 1 s.f. for an uncertainty percentage).

Step-by-Step Reasoning

If you measured $y$ using a ruler by aligning between the jar edge and nail position, there are two endpoints.
A simple conservative estimate is $\pm 0.1\ \text{cm}$ for each endpoint (this also accounts for parallax and difficulty aligning exactly), giving

\Delta y = \pm (0.1 + 0.1)\ \text{cm} = \pm 0.2\ \text{cm}.

Then

\%\ \text{uncertainty} = \frac{0.2}{2.6} \times 100 = 7.7\% \approx 8\%.

(Your numbers will differ if your measured $y$ differs.)

Key Takeaways

Use a justified absolute uncertainty.
Percentage uncertainty compares the size of the uncertainty to the size of the measurement.

Common Mistakes

Using $\Delta y = 0.1\ \text{cm}$ for the whole distance when two endpoints are involved, without justification.
Forgetting to multiply by $100$ .
Giving the uncertainty to too many significant figures (e.g. $7.6923\%$ ).

Things to Be Careful About

If your main uncertainty is not ruler resolution but judging the disappearance point, it may be larger; your estimate should reflect the real limitation.
Use your own measured value of $y$ in the calculation.

Techniques used

estimate an absolute uncertainty from instrument resolution and reading methodcalculate percentage uncertainty using ratio of uncertainty to measured valuequote an uncertainty to a sensible number of significant figures

(iii)

The radius $r$ of jar A is given by

r = \frac{D}{2}

Calculate $(r + y)$ .

$(r + y)$ = ______

1M Similar

DifficultyEasy

Worked solution

Working

r = \frac{D}{2} = \frac{8.5}{2} = 4.25\ \text{cm}

(r+y) = 4.25 + 2.6 = 6.85\ \text{cm} \approx 6.9\ \text{cm}

Answer

$(r+y) = 6.9\ \text{cm}$

Final answer

6.9 cm

Detailed explanation

Background Concept

When you calculate a new quantity from measured values, you should:

Use the given formula correctly.
Keep sufficient figures during intermediate steps.
Round the final value to a precision consistent with the measurements (often to the same decimal place as the least precise measurement when adding).

Here,

r = \frac{D}{2}

and then you add $y$ to get $(r+y)$ .

Understanding the Question

You have measured $D$ and $y$ for jar A. You must calculate:

(r+y)

where $r$ is the radius of jar A.

Approach

Find $r$ from $D/2$ .
Add $y$ .
Quote the result with an appropriate unit and rounding.

Step-by-Step Reasoning

Using example data:

If $D = 8.5\ \text{cm}$ then

r = \frac{8.5}{2} = 4.25\ \text{cm}.

With $y = 2.6\ \text{cm}$ ,

(r+y) = 4.25 + 2.6 = 6.85\ \text{cm}.

Since $y$ and $D$ were recorded to $0.1\ \text{cm}$ , it is reasonable to give $(r+y)$ to $0.1\ \text{cm}$ :

(r+y) \approx 6.9\ \text{cm}.

Key Takeaways

Use $r = D/2$ then add $y$ .
Round sensibly for addition.

Common Mistakes

Forgetting to divide by 2 (using $r=D$ ).
Missing the unit.
Rounding too early (can shift the final digit).

Things to Be Careful About

Keep extra digits in $r$ before adding to $y$ ; round at the end.
Ensure $D$ and $y$ are in the same unit before adding.

Techniques used

compute a derived quantity from measured valuessubstitute into a given formularound the final value to appropriate precision

(c)

Repeat (a), (b)(i) and (b)(iii) using jar B.

$D$ = ______
$y$ = ______
$(r + y)$ = ______

3M Similar

DifficultyMedium-Easy

Worked solution

Working (example)

For jar B:

$D = 10.2\ \text{cm}$ , so

r = \frac{D}{2} = \frac{10.2}{2} = 5.10\ \text{cm}

Measured $y = 3.0\ \text{cm}$ .

(r+y) = 5.10 + 3.0 = 8.10\ \text{cm} \approx 8.1\ \text{cm}

Answer

$D = 10.2\ \text{cm}$

$y = 3.0\ \text{cm}$

$(r+y) = 8.1\ \text{cm}$

Final answer

D = 10.2 cm, y = 3.0 cm, (r + y) = 8.1 cm

Detailed explanation

Background Concept

Good practical work depends on consistency:

Use the same definitions (where $D$ and $y$ are measured from) for both jars.
Record values to consistent precision.
When calculating $(r+y)$ , calculate $r = D/2$ first and then add $y$ .

Collecting results from two jars gives you two data points to test whether a constant $k$ is roughly the same.

Understanding the Question

You must repeat the procedure used for jar A on jar B:

Measure the diameter $D$ .
Measure the disappearance distance $y$ .
Calculate $(r+y)$ using $r = D/2$ .

Approach

Perform the same steps as earlier but on jar B.
Keep the same unit (usually cm) and similar precision (usually $0.1\ \text{cm}$ ).

Step-by-Step Reasoning

Measure $D$ of jar B across the centre.
Obtain $y$ using the same “bottom disappears” condition and measure from the jar edge to the nail.
Compute radius:

r = \frac{D}{2}.

Then compute:

(r+y) = r + y.

Using example values:

$D = 10.2\ \text{cm} \Rightarrow r = 5.10\ \text{cm}$
$y = 3.0\ \text{cm}$
$(r+y) = 8.10\ \text{cm} \approx 8.1\ \text{cm}$

Key Takeaways

Repeatability and consistent definitions are essential.
Derived quantities should be calculated and rounded appropriately.

Common Mistakes

Changing the method (e.g. using a different viewing position) for jar B.
Measuring $y$ from a different reference point than for jar A.
Forgetting to halve $D$ when finding $r$ .

Things to Be Careful About

If jar B has a different thickness or shape, you must still measure the diameter as shown (typically the full outer diameter).
Keep intermediate calculations (like $r$ ) to sufficient precision, and round at the end.

Techniques used

repeat the same measurement procedure on a second objectrecord measurements consistently with units and precisioncalculate a derived quantity from measured values

(d)

It is suggested that the relationship between $r$ and $y$ is

\frac{(r + y)}{r} = k

where $k$ is a constant.

(i)

Using your data, calculate two values of $k$ .

first value of $k$ = ______
second value of $k$ = ______

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For jar A:

k = \frac{(r+y)}{r} = \frac{6.85}{4.25} = 1.61

For jar B:

k = \frac{(r+y)}{r} = \frac{8.10}{5.10} = 1.59

Answer

first value of $k = 1.61$

second value of $k = 1.59$

Final answer

k = 1.61 and 1.59

Detailed explanation

Background Concept

If a relationship

\frac{(r+y)}{r} = k

holds with constant $k$ , then different jars should give similar values of $k$ even though $r$ and $y$ change.

Note that

\frac{(r+y)}{r} = \frac{r}{r} + \frac{y}{r} = 1 + \frac{y}{r},

so $k$ is dimensionless (no units).

Understanding the Question

You have two jars (A and B) and you have already found $r$ and $(r+y)$ for each. You must calculate two values of $k$ by substituting each jar’s values into the given expression.

Approach

For each jar:

Use your calculated $(r+y)$ .
Use your radius $r$ .
Compute the ratio $(r+y)/r$ .

Step-by-Step Reasoning

Using example results:

Jar A: $r = 4.25\ \text{cm}$ and $(r+y) = 6.85\ \text{cm}$ .

k_A = \frac{6.85}{4.25} = 1.61.

Jar B: $r = 5.10\ \text{cm}$ and $(r+y) = 8.10\ \text{cm}$ .

k_B = \frac{8.10}{5.10} = 1.59.

These are close, suggesting $k$ might be approximately constant.

Key Takeaways

A constant in an experiment is often tested by calculating it for multiple trials/objects and checking if it stays similar.
Ratios like this often produce a dimensionless constant.

Common Mistakes

Using $D$ instead of $r$ .
Using $(r+y)$ values from the wrong jar.
Adding instead of dividing.

Things to Be Careful About

Use consistent units for $r$ and $(r+y)$ (both cm or both m), though units cancel in the ratio.
Keep enough digits during division and round at the end.

Techniques used

substitute measured values into a given relationshipcalculate a constant from two sets of experimental datacompare calculated constants for consistency

(ii)

Justify the number of significant figures that you have given for your values of $k$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$D$ and $y$ were measured to the nearest $0.1\ \text{cm}$ , so $r$ and $(r+y)$ are only reliable to about $0.1\ \text{cm}$ . Therefore the ratio $k = \dfrac{(r+y)}{r}$ should be quoted to about $2$ – $3$ significant figures (here $3\ \text{s.f.}$ ).

Final answer

k quoted to 3 s.f. because D and y were measured to 0.1 cm, limiting r and (r+y) to about 0.1 cm.

Detailed explanation

Background Concept

Significant figures in a calculated result should reflect the precision of the measured quantities used.

If you measure lengths to the nearest $0.1\ \text{cm}$ , your absolute uncertainty is typically of order $\pm 0.05\ \text{cm}$ to $\pm 0.1\ \text{cm}$ (depending on method).
When values are added/subtracted, the limiting factor is usually the decimal place.
When values are multiplied/divided, the limiting factor is usually the percentage (fractional) uncertainty, which often corresponds to a sensible number of significant figures.

Here $k$ is a ratio, so its precision depends on the relative uncertainties in $r$ and $(r+y)$ .

Understanding the Question

You calculated $k$ twice. Now you must explain why you wrote $k$ to the number of significant figures you chose.

Approach

Identify the precision of the raw measurements ( $D$ and $y$ ).
State how that affects $r$ and $(r+y)$ .
Conclude what is reasonable for $k$ .

Step-by-Step Reasoning

Suppose $D$ and $y$ are recorded to $0.1\ \text{cm}$ . Then $r = D/2$ is also limited to about $0.1\ \text{cm}$ .
$(r+y)$ is a sum, so it is again limited by the least precise decimal place (typically $0.1\ \text{cm}$ ).
For typical values (a few cm), an absolute uncertainty of around $0.1\ \text{cm}$ corresponds to a few percent uncertainty.
A few percent uncertainty usually justifies quoting $k$ to 2 or 3 significant figures (not more).

Therefore quoting, for example, $k = 1.61$ (3 s.f.) is reasonable, whereas $k = 1.61273$ would imply unjustified precision.

Key Takeaways

Don’t overstate precision in derived results.
Justification should refer back to the measurement precision (and/or uncertainty).

Common Mistakes

Saying “because the calculator gives it” (not a physics justification).
Quoting $k$ to the same decimal places as a length (significant figures matter more for ratios).

Things to Be Careful About

If your measured $y$ is small, the percentage uncertainty can be large, so fewer significant figures may be justified.
If you repeated readings and averaged, you may justify slightly better precision (but still limited by instrument resolution and the subjective disappearance point).

Techniques used

identify the limiting precision from raw measured quantitieschoose an appropriate number of significant figures for a derived valuelink rounding to measurement uncertainty

(e)

It is suggested that the percentage uncertainty in the values of $k$ is 20%.

Using this uncertainty, explain whether your results support the relationship in (d).

1M Similar

DifficultyMedium

Worked solution

Working

Using the two values $k_1 = 1.61$ and $k_2 = 1.59$ :

\%\ \text{difference} = \frac{|k_1-k_2|}{\frac{(k_1+k_2)}{2}}\times 100 = \frac{0.02}{1.60}\times 100 \approx 1.3\%

Since $1.3\% < 20\%$ , the values agree within the stated uncertainty.

Answer

Yes, the results support the relationship because the two values of $k$ are consistent within $\pm 20\%$ .

Final answer

Yes; k values agree within 20%.

Detailed explanation

Background Concept

To decide if results support a relationship with an expected constant, you compare values allowing for uncertainty.

If the percentage uncertainty in each value of $k$ is about $20\%$ , then two values that differ by much less than $20\%$ are considered consistent (agreement within experimental uncertainty).

A simple way is to compute the percentage difference between the two measured values and check if it is smaller than the uncertainty.

Understanding the Question

You have two experimentally determined values of $k$ (from the two jars). You are told to assume the percentage uncertainty in $k$ is $20\%$ . You must use this to decide whether your measured $k$ values are close enough to treat $k$ as constant.

Approach

Find how different the two $k$ values are (absolute difference).
Convert that into a percentage difference (commonly relative to the mean value).
Compare with $20\%$ and state a conclusion.

Step-by-Step Reasoning

Using example values:

$k_1 = 1.61$ , $k_2 = 1.59$ .
Absolute difference:

|k_1-k_2| = 0.02.

Mean value:

\bar{k} = \frac{1.61+1.59}{2} = 1.60.

Percentage difference:

\frac{0.02}{1.60}\times 100 = 1.25\% \approx 1.3\%.

Since $1.3\%$ is much smaller than $20\%$ , the two results are consistent with a constant $k$ .

Key Takeaways

A relationship is “supported” if the results agree within the stated uncertainty.
Always state a clear comparison and conclusion.

Common Mistakes

Saying “they are close” without quantifying or linking to $20\%$ .
Comparing to $20\%$ incorrectly (e.g. subtracting 20 from the values).

Things to Be Careful About

Use your own two values of $k$ .
If your two values differ by more than about $20\%$ , you should conclude they do not support the relationship (or that the data are inconclusive due to large uncertainty).

Techniques used

compare two experimental values using a stated percentage uncertaintycalculate a percentage differencejudge consistency within uncertainty bounds

(f)

● View the nail through the lens as shown in Fig. 2.5.

Fig. 2.5

● Increase the distance between the nail and the lens until the bottom of the nail seen through the lens disappears.
● Measure and record the distance $y$ between the nail and the surface of the lens.

$y$ = ______

● Use your second value of $k$ to determine a value of $r$ for the lens.
Give an appropriate unit.

$r$ = ______

1M Similar

DifficultyMedium

Worked solution

Working (example)

Measured for the lens: $y = 2.4\ \text{cm}$ .

Using second value of $k$ (e.g. $k = 1.59$ ):

\frac{(r+y)}{r} = k \Rightarrow 1 + \frac{y}{r} = k \Rightarrow r = \frac{y}{k-1}

r = \frac{2.4}{1.59-1} = \frac{2.4}{0.59} = 4.07\ \text{cm} \approx 4.1\ \text{cm}

Answer

$y = 2.4\ \text{cm}$

$r = 4.1\ \text{cm}$

Final answer

y = 2.4 cm, r = 4.1 cm

Detailed explanation

Background Concept

If the same optical effect is assumed to follow

\frac{(r+y)}{r} = k,

then once you know $k$ (from jar B, the “second value”) you can determine an unknown radius $r$ from a new measurement of $y$ .

Rearranging:

\frac{r+y}{r} = 1 + \frac{y}{r} = k \Rightarrow \frac{y}{r} = k-1 \Rightarrow r = \frac{y}{k-1}.

Understanding the Question

You must:

View the nail through the lens and increase the distance until the bottom disappears.
Measure and record the distance $y$ between the nail and the surface of the lens.
Use your second value of $k$ to calculate $r$ for the lens and give an appropriate unit.

Approach

Measure $y$ at the disappearance condition, just as before.
Use the equation for $k$ and rearrange it to solve for $r$ .
Substitute your measured $y$ and your calculated $k$ .

Step-by-Step Reasoning

Measure $y$ carefully from the lens surface to the nail position (top-down arrangement).
With known $k$ :

\frac{r+y}{r} = k.

Rearrange:

r+y = kr \Rightarrow y = (k-1)r \Rightarrow r = \frac{y}{k-1}.

Example substitution:

If $y = 2.4\ \text{cm}$ and $k = 1.59$ ,

r = \frac{2.4}{0.59} = 4.07\ \text{cm} \approx 4.1\ \text{cm}.

Unit: since $y$ was measured in cm, $r$ is also in cm.

Key Takeaways

Rearrangement skill: isolate the required variable before substituting.
Keep units consistent; here $k$ is dimensionless.

Common Mistakes

Using $r = \dfrac{y}{1-k}$ (sign error) giving a negative value.
Using the first $k$ instead of the second value as instructed.
Measuring $y$ from the wrong reference point (not from the lens surface).

Things to Be Careful About

If $k$ is close to 1, then $(k-1)$ is small and $r$ becomes very sensitive to measurement errors in $k$ and $y$ .
Use your own measured $y$ and your own second $k$ value in the calculation.

Techniques used

measure a distance to a defined optical conditionrearrange an equation to make the required variable the subjectsubstitute values and calculate with correct unit

(g)

(i)

Describe four sources of uncertainty or limitations of the procedure for this experiment.

For any uncertainties in measurement that you describe, you should state the quantity being measured and a reason for the uncertainty.

4M Similar

DifficultyMedium

Worked solution

Answer

$y$ measurement (ruler alignment/parallax): difficult to align ruler exactly with the jar edge and nail; eye not directly above scale gives parallax.
Judging disappearance point (value of $y$ ): the point where the bottom of the nail “suddenly disappears” is subjective and may be overshot when moving the nail.
Eye position not fixed: small changes in viewing angle/position change the refraction path, altering the disappearance point and hence $y$ .
Nail position/orientation: nail may not be vertical or may move while measuring, changing the effective distance and the observed disappearance condition.

Final answer

Four uncertainties/limitations listed (see solution).

Detailed explanation

Background Concept

In practical experiments, uncertainty comes from:

Instrument limits (resolution, parallax).
Human judgement (deciding when an event occurs).
Uncontrolled variables (geometry not fixed, changing viewing angle).
Systematic effects (apparatus not matching ideal assumptions).

Good answers name the quantity affected and give a physical reason.

Understanding the Question

You must describe four uncertainties/limitations in this procedure. For measurement uncertainties, you must:

state what quantity has uncertainty (e.g. $y$ or $D$ ), and
state why (e.g. parallax, subjective judgement, misalignment).

Approach

Pick four distinct issues covering:

Measuring distances ( $D$ , $y$ ) with a ruler.
The subjective optical criterion (disappearance).
Geometry/position control (eye position, nail position).
Apparatus limitations (jar shape/thickness, water level consistency).

Step-by-Step Reasoning

Examples of creditworthy limitations:

Parallax / alignment when measuring $y$
- Quantity: $y$ .
- Reason: ruler may not be exactly horizontal between jar edge and nail; eye not directly above scale causes parallax.
Subjective judgement of disappearance
- Quantity: $y$ .
- Reason: “suddenly disappears” is judged by the observer; moving the nail can overshoot the exact point; different observers may choose different points.
Eye position changes
- Quantity: $y$ (and hence $k$ ).
- Reason: refraction depends on the ray path; changing viewing angle changes whether the bottom is visible.
Nail not held fixed/vertical
- Quantity: $y$ .
- Reason: if the nail tilts or moves, the distance to the jar edge changes and the viewed image changes.

Other valid limitations (any four total are needed):

Jar not perfectly cylindrical / varying glass thickness: affects refraction so the idealised relationship may not hold exactly.
Water level/meniscus changes: if the water level is not constant, the optical path changes.
Only one reading of $y$ for each jar: insufficient repeats to assess random scatter.

Key Takeaways

Uncertainty statements must be specific: quantity + reason.
Aim for distinct limitations, not repeats of the same idea.

Common Mistakes

Vague statements like “human error” with no quantity or cause.
Repeating the same point four times (e.g. parallax in different words).
Listing improvements instead of limitations (that belongs in (g)(ii)).

Things to Be Careful About

Make sure each of the four points is clearly different.
Tie the limitation directly to how it affects $y$ , $D$ , or the constancy of $k$ .

Techniques used

identify random and systematic sources of uncertaintylink each uncertainty to a measured quantity and a physical causestate limitations in terms of control of variables and repeatability

(ii)

Describe four improvements that could be made to this experiment. You may suggest the use of other apparatus or different procedures.

4M Similar

DifficultyMedium

Worked solution

Answer

Fix the nail in a clamp stand (and/or use a slider) so the nail stays vertical and the distance can be increased smoothly without wobble.
Fix the viewing position (e.g. use a viewing tube or mark a fixed eye position) to keep the line of sight constant for each reading.
Improve measurement of $y$ by placing a ruler fixed to the bench and using a set square to ensure the measured distance is perpendicular to the jar/lens surface and reduces parallax.
Repeat and average: take several readings of $y$ for each jar/lens (approaching the disappearance point from both directions) and use the mean; discard clear anomalies.

Final answer

Four improvements listed (see solution).

Detailed explanation

Background Concept

Improvements are actions that reduce uncertainty or increase validity. Typical improvement types:

Control geometry (fix positions, ensure perpendicular measurements).
Reduce human judgement effects (standardise method, approach from both directions).
Use better instruments (calipers, mounted scale).
Repeat measurements to reduce random error and identify anomalies.

Understanding the Question

You must suggest four improvements to the experiment. These can involve different apparatus or procedures. The best improvements directly address the limitations from (g)(i).

Approach

Choose four distinct improvements that target:

nail stability,
eye position stability,
distance measurement accuracy,
reliability through repeats.

Step-by-Step Reasoning

Examples of strong improvements:

Clamp the nail vertically
- Use a clamp stand to hold the nail. This stops wobble and ensures the same orientation each time, making the disappearance point more reproducible.
Fix the observer position
- Use a viewing tube or place the eye at a marked position relative to the jar, so the viewing angle is constant.
Reduce parallax and misalignment in $y$
- Fix a ruler to the bench (tape it down) and use a set square against the jar/lens surface to measure the perpendicular distance to the nail.
Repeat and average
- Take multiple readings of $y$ for each jar and average. Approach the disappearance point from both nearer and farther positions to reduce bias from overshooting.

Other acceptable improvements (any four total):

Use vernier calipers to measure $D$ more precisely.
Use a camera/phone at a fixed position to decide the disappearance point more objectively.
Keep water level constant (top up to a marked level) and wait for the water to be still.

Key Takeaways

Improvements should be specific and linked to a particular limitation.
Repeats and better control of positions usually give the biggest gain in reliability.

Common Mistakes

Suggesting an improvement that does not change the main uncertainty (e.g. “use a larger ruler” without addressing judgement/alignment).
Giving four variations of “take more readings” without other practical changes.

Things to Be Careful About

Improvements must be realistic with school-lab equipment.
State clearly what each improvement changes and which uncertainty it reduces.

Techniques used

propose procedural changes that reduce specific uncertaintiessuggest additional apparatus to control geometry and alignmentincrease reliability by repeat readings and averaging