Alternative to Practical Skills Worked ExamplesIGCSE Physics 0625

(a) Work out the value of one division first. Between the $30$ and $40$ labels there are $10$ spaces, so each small division is $1\,^{\circ}\text{C}$. The thread ends $4$ divisions above $30$.

Record it to the resolution of the scale: $34\,^{\circ}\text{C}$, not $34.0\,^{\circ}\text{C}$ or $34.00\,^{\circ}\text{C}$. The thread ends exactly on a line, so no half-division estimate is needed.

(b) Between the $70$ and $80$ labels there are $5$ spaces, so each division is $2\,\text{cm}^3$. The bottom of the meniscus sits on the second line above $70$.

(c) Mark-scheme style answers (any two): read the scale at the bottom of the meniscus; have your eye level with the meniscus, looking perpendicular to the scale (to avoid parallax); stand the cylinder on a flat, horizontal bench so the surface is level.

Voltmeter. There are $10$ divisions between $1$ and $2$, so one division is $0.1\,\text{V}$. The longer tick half-way is $1.5\,\text{V}$, and the pointer is $2$ divisions beyond it.

Ammeter. There are $10$ divisions between $0.2$ and $0.4$, so one division is $0.02\,\text{A}$. The pointer is $3$ divisions beyond $0.4$.

Record each to the resolution of its scale: $1.7\,\text{V}$ (not $1.70\,\text{V}$) and $0.46\,\text{A}$ (not $0.460\,\text{A}$, and not $4.6\,\text{A}$).

(b) Write the equation, substitute, then round.

The third figure is $9$, so the second figure rounds up: $3.7\,\Omega$. Truncating to $3.6\,\Omega$ loses the mark.

(a) On the enlarged scale each small division is $1\,\text{mm} = 0.1\,\text{cm}$. The longer tick below $26$ is $26.5\,\text{cm}$, and the pointer is $2$ divisions below it.

Quote it to the nearest millimetre: $26.7\,\text{cm}$, not $27\,\text{cm}$.

(b) Extension is the new reading minus the unloaded reading.

(c) Mark-scheme style answers (any two): place the rule close to the pointer; check the rule is vertical (for example with a set square standing on the bench against the rule); view the pointer with the eye level with it, perpendicular to the scale (avoiding parallax); use a set square with one edge against the rule and the other edge touching the pointer to transfer the pointer's position onto the scale.

(a) Each heading needs a quantity (or its symbol) and a unit, written as quantity / unit: $l\,/\,\text{cm}$, $\;V\,/\,\text{V}$, $\;I\,/\,\text{A}$, $\;R\,/\,\Omega$. The units then do not appear in the body of the table.

(b) The rule was read to the nearest mm, so every length is recorded to $0.1\,\text{cm}$. The entry $60$ should be written $60.0$ (cm) to match the others.

(c) Use $R = V/I$ for each row. The first two rows:

The remaining rows:

The currents were recorded to $2$ significant figures, so $2$ s.f. is suitable for $R$ — and it must be the same for every row. Keep trailing zeros: $0.80$ and $4.0$, not $0.8$ and $4$. (Three significant figures in every row, $0.800,\ 1.59,\ 2.41,\ 3.20,\ 3.95$, would also be consistent.)

(a) The ray crosses the scale at $56^{\circ}$ — that is the angle between the ray and the block surface. The angle of incidence is measured from the normal, which is at $90^{\circ}$ on the scale.

(b) The values are not identical, so compare them numerically. The ratio of the smaller to the larger:

At IGCSE, results within about $10\%$ are treated as equal within the limits of experimental accuracy. So: yes, the results support $e = i$, because $32^{\circ}$ and $34^{\circ}$ differ by only about $6\%$, which is within $10\%$.

(c) Mark-scheme style answers (any two): place the pins far apart (at least $5\,\text{cm}$); make the pins vertical / view the bases of the pins when lining them up; use thin pencil lines (sharp pencil); use thin pins.

Read the unloaded meter: each division is $0.1\,\text{N}$ and the pointer is $2$ divisions below zero, so it reads $0.2\,\text{N}$ with no force applied. This is a zero error.

A zero error shifts every reading by the same amount in the same direction, so it is systematic. Repeating and averaging cannot remove it, so A is wrong. Taking care to avoid parallax cannot remove it either, so D is wrong.

Each reading is $0.2\,\text{N}$ too high, so subtract $0.2\,\text{N}$ (C has the correction the wrong way round). The corrected forces are:

Better still, the meter should be zeroed (adjusted to read $0$ with no load) before any readings are taken.

(a) The four graph marks: axes labelled with quantity and unit ($F\,/\,\text{N}$ up, $d\,/\,\text{cm}$ across — the quantity named first goes on the $y$-axis); sensible scales (steps of $1$, $2$ or $5$ units, points filling at least half the grid in each direction); accurate plots as small neat crosses; a thin, straight best-fit line with the points evenly balanced on either side.

The best-fit line is drawn through the trend of all the points. It is not forced through the origin and does not simply join the first and last points.

(b) Extend the line to $d = 0$ and read where it meets the $F$-axis: it crosses about two-thirds of a small square ($0.1\,\text{N}$ each) above $0.6\,\text{N}$.

(c) Rearrange the moment equation for $W$ and substitute.

The intercept was read to $2$ significant figures, so give $W$ to $2$ significant figures.

Physics check: even with the load at the pivot, the force meter still supports part of the rule's own weight, so the line should not pass through the origin.

(a) Choose two points on the line (not data points), far apart, so the triangle covers more than half the line. Here: $(0.10\,\text{m},\ 0.6\,\Omega)$ and $(0.90\,\text{m},\ 4.9\,\Omega)$. Draw the triangle on the graph and label the two readings.

Gradient = change in $y$ ÷ change in $x$.

The unit is the $y$-unit divided by the $x$-unit, $\Omega/\text{m}$: this wire has about $5.4\,\Omega$ of resistance for every metre.

(b) Resistance = gradient × length.

(c) Rearrange for length.

(d) Yes. The line is straight and passes through the origin (it misses by less than one small square, $0.2\,\Omega$), so $R$ is directly proportional to $l$. Both conditions must be stated.

(a) Look at the drop in each $50\,\text{s}$ interval: $6.5$, $4.5$, $1.5$, $7.0$, $3.5$, $3.5\,^{\circ}\text{C}$. The drops should get steadily smaller; the pair $1.5$ then $7.0$ is out of pattern. The anomaly is $70.5\,^{\circ}\text{C}$ at $t = 150\,\text{s}$ — its cross lies well above the smooth curve through the other points.

From the trend it should be about half-way between $72.0$ and $63.5$, roughly $67.5$–$68\,^{\circ}\text{C}$.

(b) Ignore it when drawing the best-fit curve (and in calculations); if possible, repeat the experiment to check that reading.

(c) Average rate = temperature drop ÷ time taken.

(d) The rate of cooling decreases as the water cools — from $0.11\,^{\circ}\text{C/s}$ in the first $100\,\text{s}$ to $0.070\,^{\circ}\text{C/s}$ in the last $100\,\text{s}$. The hotter the water (compared with the room), the faster it loses thermal energy.

(a) Independent variable (the one the student changes): object distance $u$. Dependent variable (the one measured): image distance $v$.

(b) Control variable: use the same lens (same focal length) throughout. Also acceptable: keep the object, lens and screen centres at the same height on the same axis. Apparatus names on their own ("the rule") are not variables.

(c) A real image forms on the screen only when $u$ is greater than the focal length, so start above $15\,\text{cm}$ — for example $u = 20, 25, 30, 35, 40, 45\,\text{cm}$. That gives at least five values over a wide range. Values just above $15\,\text{cm}$ would need the screen very far away.

(d) For each $u$, move the screen slowly backwards and forwards until the image is as sharp as possible, then measure $v$ from the centre of the lens to the screen, with the eye perpendicular to the rule.

(e) In a darkened room the image on the screen has better contrast (brighter against a dark background), so it is easier to judge when it is sharpest.

(a) The beakers started at different temperatures, so comparing final temperatures alone is misleading. Compare the change in temperature over the same time.

Conclusion: over $600\,\text{s}$ the insulated beaker B cooled by only $13.0\,^{\circ}\text{C}$ compared with $24.0\,^{\circ}\text{C}$ for A, so the insulation reduces the rate of cooling. (Average rates: $0.040\,^{\circ}\text{C/s}$ for A and $0.022\,^{\circ}\text{C/s}$ for B.)

(b) Beaker A itself cooled faster when it was hotter: it dropped $9.5\,^{\circ}\text{C}$ in the first $200\,\text{s}$ but only $6.5\,^{\circ}\text{C}$ in the last $200\,\text{s}$. So the rate of cooling depends on temperature.

A started $6.0\,^{\circ}\text{C}$ hotter, so part of its larger drop is due to its higher temperature rather than the lack of insulation. For a fair test of the insulation, both beakers must start at the same temperature.

(c) Put a lid on both beakers (reducing heat loss by evaporation and convection from the surface). The rates of cooling of both beakers would decrease.

(a) Draw a clear horizontal construction line from $L = 29.3\,\text{cm}$ across to the best-fit line, then a vertical line down to the $m$-axis. Read the mass where it meets the axis.

(b) Student Y. The lengths were measured to the nearest mm ($3$ significant figures), and a value read from a graph cannot be more precise than the data or the graph scale allow. $216.31\,\text{g}$ claims a precision of $0.01\,\text{g}$ that the method cannot give. More significant figures do not make an answer more accurate.

(c) Not correct. The line is straight, but direct proportion also needs the line to pass through the origin. This line meets the $L$-axis at $12.0\,\text{cm}$ — the unstretched length of the spring.

What is proportional to $m$ is the extension, $e = L - 12.0\,\text{cm}$.

(d) Use a large triangle on the line: at $m = 0$, $L = 12.0\,\text{cm}$; at $m = 300\,\text{g}$, $L = 36.0\,\text{cm}$.

For each $100\,\text{g}$ added, the extension increases by:

(a) Take the difference of each pair of readings first, then share it among the six balls.

Same idea for the volume.

(b) Density = mass ÷ volume. Using the totals avoids carrying a rounded value forward.

Two significant figures are suitable: the volume reading is only known to about $\pm 1\,\text{cm}^3$, so $7.56$ claims too much precision.

(c) Compare with the data-book value.

The values differ by about $3\%$, well within $10\%$, so yes, they agree within experimental accuracy.

(d) Compare the uncertainties. The volume change is $5\,\text{cm}^3$ read on a scale with $1\,\text{cm}^3$ divisions — an uncertainty of about $1$ in $5$, or $20\%$. The mass change of $37.8\,\text{g}$ read to $0.1\,\text{g}$ is under $1\%$. So the volume measurement dominates.

Improvements (any two): use many more ball bearings (e.g. $30$ would displace about $25\,\text{cm}^3$, reducing the uncertainty to about $4\%$); use a narrower measuring cylinder with smaller divisions; measure the diameter of the balls with a micrometer or calipers and calculate the volume from a formula; repeat the whole measurement and calculate the mean density.

Circuit (MP1). Draw a voltmeter (circle with V) connected in parallel across P and Q — one lead to each terminal. The ammeter stays in series.

Method (MP2). Connect one resistor between P and Q. Close the switch, read the voltmeter $V$ and the ammeter $I$, then open the switch. Calculate $R = V/I$.

Change the independent variable (MP3). Repeat with $n = 2, 3, 4, 5$ and $6$ resistors in parallel — six values of the independent variable.

Control variables (MP4). Use resistors of the same value; keep the power-supply setting the same; keep the resistors at the same temperature by opening the switch between readings so the resistors (and wires) do not heat up.

Table (MP5). Columns with quantities and units that match the method: $n$, $\;V\,/\,\text{V}$, $\;I\,/\,\text{A}$, $\;R\,/\,\Omega$.

Conclusion (MP6). Plot a graph of $R$ against $1/n$. If it is a straight line through the origin, $R$ is inversely proportional to $n$. (Alternatively, check whether $R \times n$ is constant, within about $10\%$.) This is the pattern expected, since $n$ identical resistors $R_0$ in parallel give $R_0/n$.

Extra point (MP7). Any one: repeat each set of readings and take the mean $R$; check the meters read zero before starting; choose a supply setting that keeps the current below the ammeter's full-scale reading.