Physics 5054/42 — October/November 2024

Walkthrough

The student is given a full-size diagram of the test-tube (Fig. 1.1) and must measure the height $h$ using a ruler. In a real Paper 4 exam, the candidate places a ruler along the dimension arrow labeled $h$ and reads the value. The mark scheme specifies the value as 15.0 cm. The question asks for the reading to the nearest 0.1 cm, so a trailing zero is required to show the precision.

Key Takeaways

When measuring from a full-size diagram in Paper 4, always use a ruler and read to the precision demanded by the question. A trailing zero is significant and indicates the precision of the measurement.

Common Mistakes

• Forgetting the unit (cm).
• Not writing the trailing zero (e.g. 15 instead of 15.0), which fails the precision requirement.
• Reading the wrong dimension (e.g. reading the diameter instead of the height).

Things to Be Careful About

The question asks for the answer to the nearest 0.1 cm. This means the answer must have one decimal place. 15.0 is correct; 15 is not.

Walkthrough

The student measures the external diameter $d$ of the test-tube from Fig. 1.1. The mark scheme gives the value as 1.5 cm. The question does not specify a decimal precision here, but typically a ruler reading in cm is to 0.1 cm. The value 1.5 cm is consistent with the mark scheme.

Key Takeaways

Always read the dimension asked for. The external diameter is the full width across the outside of the tube.

Common Mistakes

• Reading the internal diameter instead of the external diameter.
• Forgetting the unit.

Things to Be Careful About

Ensure the ruler is aligned with the dimension arrow labeled $d$ at the bottom of the diagram.

Walkthrough

Fig. 1.4 shows a digital balance reading 0.01375 kg. The question asks for the mass to the nearest gram.

1. Convert kg to g: $0.01375 \times 1000 = 13.75$ g.
2. Round to the nearest whole number: 13.75 rounds up to 14 g.

Key Takeaways

Always check the units requested. Digital balances often read in kg; mass in physics problems is often required in g. Rounding rules: 0.5 and above rounds up.

Common Mistakes

• Forgetting to convert kg to g (answering 0.01375).
• Rounding incorrectly (e.g. 13.75 to 13).
• Not following the instruction "to the nearest gram".

Things to Be Careful About

The mark scheme accepts 14. If the candidate uses 13.75 g in the next part, they will get an ECF (error carried forward) mark as long as the calculation is correct.

Walkthrough

Density is mass divided by volume. The question asks for the unit as well.
Using the rounded mass from (g)(i): $m = 14$ g.
Using the volume from (e): $V_G = 5.6625$ cm$^3$ (or 5.66).
$\rho = 14 / 5.6625 = 2.472...$ g/cm$^3$.
The mark scheme accepts any answer between 2.4 and 2.5 inclusive. If a candidate used 13.75 g (the exact balance reading), they would get $13.75 / 5.6625 = 2.428$ g/cm$^3$, which is also in range. The mark scheme allows ECF if the candidate's mass or volume is different, as long as the formula and method are correct.

The unit for density is mass/volume. Since mass is in g and volume in cm$^3$, the unit is g/cm$^3$.

Key Takeaways

Density = mass / volume. Always include the unit. Check if the answer is reasonable (glass density is typically around 2.5 g/cm$^3$).

Common Mistakes

• Using the wrong volume (e.g. $V_E$ or $V_I$ instead of $V_G$).
• Forgetting the unit or writing it incorrectly (e.g. kg/m$^3$ without converting, though g/cm$^3$ is standard here).
• Calculation errors.

Things to Be Careful About

The question explicitly asks for the unit. It must be stated. The mark scheme gives "g/cm$^3$". Using kg/m$^3$ would require converting mass to kg ($0.014$) and volume to m$^3$ ($5.6625 \times 10^{-6}$), giving $2472$ kg/m$^3$. While correct, g/cm$^3$ is simpler and matches the input units. The mark scheme specifically looks for g/cm$^3$.

Walkthrough

The question asks for the symbol for a voltmeter and its correct placement. A voltmeter is always connected in parallel with the component whose potential difference is being measured. Here, we need to measure $V_{XY}$, the p.d. across the $270\ \Omega$ resistor between points X and Y. We draw the standard circuit symbol for a voltmeter (a circle containing the letter V) and connect it with wires to points X and Y, forming a parallel branch across the resistor.

Key Takeaways

• The circuit symbol for a voltmeter is a circle with a 'V' inside.
• A voltmeter must always be connected in parallel with the component across which the potential difference is to be measured.

Common Mistakes

• Drawing an ammeter symbol (circle with an 'A') instead of a voltmeter.
• Connecting the voltmeter in series with the resistor, which would break the circuit or give an incorrect reading due to the voltmeter's very high resistance.

Things to Be Careful About

• Ensure the voltmeter symbol is clearly a circle, not a square (that would be a resistor).
• The connections to X and Y must be drawn clearly to show a parallel connection, not just touching the wire at one point.

Walkthrough

Fig. 2.2 shows an analog voltmeter dial with a range from 0 to 3 V. The scale is divided into major units of 1 V, and each 1 V interval is subdivided into 10 small divisions, meaning each small division represents $0.1\text{ V}$. The pointer is pointing exactly halfway between 2 and 3, at the fifth small division past 2. This gives a reading of $2.5\text{ V}$.

Key Takeaways

• When reading an analog scale, determine the value of each small division first.
• Record the reading to the precision allowed by the scale, including a trailing zero if necessary (e.g., 2.5, not 2).

Common Mistakes

• Reading the scale incorrectly (e.g., reading 2.05 or 25).
• Failing to include the correct unit (V) if the blank requires it, though here the table header already has the unit.

Things to Be Careful About

• Ensure you are reading the correct scale. The dial shows 0 to 3, so the reading is in volts.
• The pointer is exactly on the mark, so 2.5 is the precise reading to one decimal place.

Walkthrough

The table requires the sum of the potential differences across the resistor ($V_{XY}$) and the LED ($V_{YZ}$) for the first row. From part (a)(ii), $V_{XY} = 2.5\text{ V}$. The table gives $V_{YZ} = 2.1\text{ V}$. Adding these together: $2.5 + 2.1 = 4.6\text{ V}$. This sum represents the total potential difference supplied by the power source (e.m.f.) in this series circuit.

Key Takeaways

• In a series circuit, the sum of the potential differences across the components equals the total e.m.f. of the source.
• Data processing often involves simple arithmetic on recorded readings.

Common Mistakes

• Adding the wrong columns or misaligning rows.
• Forgetting that the sum should ideally be constant (equal to the battery e.m.f.) across all rows.

Things to Be Careful About

• Ensure the values are taken from the correct row (the $270\ \Omega$ resistor row).

Walkthrough

The current $I$ flowing through the circuit is the same everywhere in series. We can calculate it using the resistance of the known resistor ($R = 270\ \Omega$) and the potential difference across it ($V_{XY} = 2.5\text{ V}$). Using Ohm's law:

Rounding to two significant figures (consistent with the data), we get $0.0093\text{ A}$. The mark scheme also accepts $0.0092$ if a slightly different reading or rounding was used.

Key Takeaways

• Current in a series circuit is the same through all components.
• Ohm's law $I = V/R$ can be applied to any known resistor to find the circuit current.

Common Mistakes

• Using the wrong voltage (e.g., using $V_{YZ}$ or the total voltage 4.6 V with the 270 Ω resistor).
• Calculation errors or incorrect rounding.

Things to Be Careful About

• The resistance is in ohms ($\Omega$) and voltage in volts (V), so the current is directly in amperes (A). No unit conversion is needed here.

Walkthrough

The resistance of the LED is found using the potential difference across it ($V_{YZ} = 2.1\text{ V}$) and the current flowing through it ($I \approx 0.009259\text{ A}$). Using Ohm's law rearranged for resistance:

Rounding to 3 significant figures gives $227\ \Omega$. The mark scheme accepts a range (226-230) due to rounding differences in the current value used.

Key Takeaways

• Ohm's law $R = V/I$ applies to any component, including non-ohmic devices like LEDs, to find their resistance at a specific operating point.
• LEDs are non-ohmic, meaning their resistance changes with current/voltage.

Common Mistakes

• Using the wrong voltage or current value.
• Failing to round appropriately.

Things to Be Careful About

• Use the unrounded value of $I$ ($0.009259...$) for this calculation to avoid compounding rounding errors. If you use $0.0093$, you get $2.1 / 0.0093 = 225.8 \approx 226\ \Omega$, which is also accepted.

Walkthrough

Looking at the $(V_{XY} + V_{YZ})$ column in Table 2.1, the values are 4.6, 4.6, and (calculated) 4.6. This sum represents the total potential difference supplied by the power source. As the resistance between X and Y changes, the current changes, but the total voltage supplied by the battery remains the same. Therefore, the sum of the potential differences across the resistor and the LED remains constant.

Key Takeaways

• In a simple series circuit with a constant voltage source, the sum of the p.d.s across the components equals the source e.m.f. and is constant.
• Data analysis involves looking for trends and constants in calculated columns.

Common Mistakes

• Saying it 'increases' or 'decreases' without checking the actual data.
• Confusing $(V_{XY} + V_{YZ})$ with $V_{XY}$ alone (which does change).

Things to Be Careful About

• The question asks specifically about $(V_{XY} + V_{YZ})$, not about individual voltages.

Walkthrough

Examine the $I$ and $R_{LED}$ columns in Table 2.1:

• Row 1: $I = 0.0093\text{ A}$, $R_{LED} = 227\ \Omega$
• Row 2: $I = 0.0053\text{ A}$, $R_{LED} = 380\ \Omega$
• Row 3: $I = 0.0046\text{ A}$, $R_{LED} = 430\ \Omega$

As the current $I$ gets smaller (0.0093 → 0.0053 → 0.0046), the resistance $R_{LED}$ gets larger (227 → 380 → 430). This shows that the LED is a non-ohmic component; its resistance is not constant but depends on the current flowing through it. Specifically, at lower currents, its resistance is higher.

Key Takeaways

• LEDs are non-ohmic devices; their resistance changes with the operating point (current/voltage).
• When analyzing data, look at how one variable changes in relation to another.

Common Mistakes

• Saying 'resistance is constant' (that would be for an ohmic resistor at constant temperature).
• Getting the relationship backwards (saying resistance decreases as current decreases).

Things to Be Careful About

• The question asks to 'describe how the change in current affects $R_{LED}$'. You must state the direction of the change for both (e.g., 'as current decreases, resistance increases').

Walkthrough

To measure the distance $x$ from the centre line of the lens to the screen on Fig. 3.1:

1. Align a metric ruler with the line representing $x$.
2. Read off the distance between the dashed vertical line of the lens and the vertical line of the screen.
3. The measured distance is $5.1\text{ cm}$.

Key Takeaways

• Always measure printed lengths carefully with a transparent or rigid ruler.
• State measurements to the precision requested (nearest $0.1\text{ cm}$).

Walkthrough

The diagram is drawn to a scale of one-third full size ($1:3$). Therefore, to find the actual focal length $f$:

Substitute $x = 5.1\text{ cm}$:

Key Takeaways

• When an image or diagram is scaled to $\frac{1}{n}$ of full size, the actual dimension is found by multiplying the measured length by $n$.

Common Mistakes

• Dividing by 3 instead of multiplying by 3.

Walkthrough

Parallax error occurs when an observer views a scale from an angle, causing the pointer or mark to appear displaced relative to the scale markings.
To avoid parallax error when reading a metre rule:

• Ensure the line of sight is perpendicular (at right angles / $90^\circ$) to the scale at the point of measurement.
• View the scale at eye level or directly from above.

Key Takeaways

• Avoiding parallax requires perpendicular viewing to the scale.

Walkthrough

To find the exact position of best focus for a real image formed on a screen:

• The candidate moves the screen slowly back and forth across the approximate image location.
• By observing when the edges of the image become clear and sharp before beginning to blur again in either direction, the exact position of maximum sharpness can be determined.

Key Takeaways

• Fine adjustments by moving backwards and forwards allow precise identification of the point of sharp focus.

Common Mistakes

• Giving vague answers such as "look closely" or "use a better lens".