Edexcel International A Level Physics · Unit 6 · Practical Skills in Physics II
Planning and evaluating experiments (A2)
28 free, high-quality past-paper-style questions on Planning and evaluating experiments (A2) for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.28 new this month · 177 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U6-1-B-01Unit 6 · A2Easy6 marks
A student investigates the deflection of an electron beam passing between two parallel metal plates inside an evacuated glass tube. Electrons leave a heated filament and are accelerated through a potential difference of 3.0 kV. A separate supply of several hundred volts is connected across the plates.
(a)
Identify two hazards in this experiment and describe a precaution for each.
(4)
(b)
Calculate the speed of the electrons as they leave the accelerating region. Use e = 1.60 × 10−19 C and me = 9.11 × 10−31 kg.
(2)
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Answer
Mark
Notes
Part (a)
high voltage / risk of electric shock from the kV supplies
✓ 1
use a current-limited supply; make and change connections only with the supplies switched off; use fully shrouded leads
✓ 1
the evacuated glass tube could implode if knocked
✓ 1
handle the tube gently by its base, clamp it securely in its stand / use a safety screen
✓ 1
also accept as a hazard/precaution pair: the hot filament and tube neck — allow them to cool before touching
—
Part (b)
eV = ½mev2, so v = √(2 × 1.60 × 10−19 × 3.0 × 103 ÷ 9.11 × 10−31)
✓ 1
= 3.2 × 107 m s−1
✓ 1
Exam tip: Pair each hazard with the precaution for that hazard. For high voltages the precautions are about the supply and the connections — current-limited supply, switched off while connecting, shrouded leads — not about goggles.
A 470 µF capacitor, charged to 9.00 V, discharges through a 100 kΩ resistor. A student reads a digital voltmeter (resolution 0.01 V) every 10 s, timing with a stopwatch, and calculates the charge Q = CV. Her table is:
Time
p.d. / V
Q / C
0
9.00
4.23
10
7.3
3.42
20.0
5.88
2.76
30
4.75
2.23
40
3.84
(a)
Identify three faults in the table and state how each should be corrected.
(3)
(b)
Calculate the missing value of charge at t = 40 s, in mC.
(1)
(c)
Explain why a stopwatch is a suitable timer for this experiment.
(1)
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Answer
Mark
Notes
Part (a)
the time column has no unit: heading should be t / s, and 20.0 should be written 20 like the other set times
✓ 1
7.3 is not recorded to the resolution of the voltmeter: it should be 7.28 (two decimal places, like the others)
✓ 1
the charge values are in mC, not C: heading should be Q / mC
✓ 1
Part (b)
470 × 10−6 × 3.84 = 1.80 × 10−3 C = 1.80 mC
✓ 1
Part (c)
the time constant RC = 100 × 103 × 470 × 10−6 = 47 s, so a reaction-time error of about 0.2 s is well under 1% of the time scale of the discharge
✓ 1
Common error: Putting a unit after every number instead of in the heading, or leaving it out altogether. Every heading is quantity / unit, and the unit must match the size of the numbers in the column.
A student uses a Geiger–Müller tube and counter to measure the count rate from a sealed beta source placed 0.150 m from the tube. With the source locked away, the counter records 186 counts in 5.00 minutes. With the source in place, it records 1520 counts in 60.0 s.
(a)
Describe two precautions the student should take when handling the source.
(2)
(b)
Calculate the count rate due to the source alone.
(2)
(c)
Explain why the background count was measured over a much longer time than the count with the source.
(1)
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Answer
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Notes
Part (a)
handle it only with long tongs / hold it at arm's length, pointing away from the body and from other people
✓ 1
keep it out of its lead-lined box for as short a time as possible and return it as soon as the readings are complete
the background count rate is small and decay is random, so a long count is needed to record enough counts for a reliable (precise) average rate
✓ 1
Exam tip: Change minutes to seconds before dividing, and subtract the background rate from the measured rate — never subtract counts recorded over different times.
Two students measure the capacitance of the same capacitor, each by timing its discharge through a resistor whose value is taken from its label. Student A obtains 452 µF, 456 µF, 449 µF and 455 µF. Student B obtains 430 µF, 505 µF, 470 µF and 488 µF. A capacitance meter that has recently been calibrated gives 471 µF.
(a)
Calculate the mean value obtained by each student.
(2)
(b)
State which set of results is more precise. Give a reason.
(1)
(c)
State which mean is more accurate. Give a reason.
(1)
(d)
All of student A's values are low. Suggest a cause.
(1)
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Answer
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Notes
Part (a)
A: 1812 ÷ 4 = 453 µF
✓ 1
B: 1893 ÷ 4 = 473 µF (473.25)
✓ 1
Part (b)
A: its values are spread over only 7 µF (± 3.5 µF), compared with 75 µF (± 37.5 µF) for B
✓ 1
Part (c)
B: its mean (473 µF) is closer to the calibrated value of 471 µF than A's mean (453 µF)
✓ 1
Part (d)
a systematic error: e.g. the actual resistance of her resistor is lower than its labelled value (within its tolerance), so every calculated capacitance is low by the same proportion
✓ 1
Exam tip: Precision is judged from the spread of repeats, accuracy from how close the mean is to the true value. A set can be precise and still inaccurate when a systematic error moves every value the same way.
A student makes a parallel-plate capacitor from two square aluminium plates. She measures the side of each plate with a metre rule (resolution 1 mm) as 0.250 m. The plates are held apart by four identical insulating spacers, and she measures the thickness d of one spacer with a vernier caliper (resolution 0.1 mm). Figure 1 shows an enlarged view of the caliper scales. The capacitance is given by C = ε0A/d, where ε0 = 8.85 × 10−12 F m−1.
(a)
Use Figure 1 to determine d.
(1)
(b)
Calculate the capacitance.
(1)
(c)
Taking each uncertainty as the resolution of the instrument used, calculate the percentage uncertainties in d, in A and in C. Hence suggest which instrument the student should change, and to what.
(3)
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Answer
Mark
Notes
Part (a)
the vernier zero is between 4 mm and 5 mm on the main scale and vernier line 6 lines up with a main-scale line, so d = 4 + 0.6 = 4.6 mm
the spacer thickness gives the larger uncertainty, so measure it with a micrometer screw gauge (resolution 0.01 mm, about 0.2%)
✓ 1
Exam tip: To read a vernier, take the main-scale reading just before the vernier zero, then find the one vernier line that lines up exactly with a main-scale line: that line number gives the next decimal place.
Figure 1 shows apparatus for investigating the force on a small charged conducting ball hanging between two parallel metal plates. Each plate is 0.5 cm thick. The plates are connected to an EHT (high-voltage) supply, and a centimetre rule lies on the bench directly beneath them.
(a)
State two safety precautions the student should take when using the EHT supply.
(2)
(b)
The p.d. between the plates is 4.0 kV. Use Figure 1 to determine the plate separation, and calculate the electric field strength between the plates.
(2)
(c)
Describe how the student could check that the plates are parallel.
(1)
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Answer
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Notes
Part (a)
use a supply with a current-limiting (high internal) resistance, set to its lowest output before switching on
✓ 1
make or change connections, and adjust the plates or ball, only with the supply switched off; use fully shrouded leads and do not touch the plates while it is on
✓ 1
Part (b)
separation between the inner faces = 8.0 − 3.0 = 5.0 cm
✓ 1
E = V/d = 4.0 × 103 ÷ 0.050 = 8.0 × 104 V m−1
✓ 1
Part (c)
measure the gap between the inner faces near the top and near the bottom (and at each side) with a vernier caliper or with identical spacer blocks, and adjust until all are equal
✓ 1
Common error: Measuring from the outer faces or from the middle of each plate. The field is set up across the gap, so the relevant dimension is the distance between the inner faces.
A small coil is rotated at a steady rate in a uniform magnetic field, and the e.m.f. induced in it is displayed on an oscilloscope. Figure 1 shows the trace and the oscilloscope settings.
(a)
Determine the peak e.m.f.
(1)
(b)
Determine the number of revolutions per second of the coil.
(2)
(c)
The student changes the Y-gain to 0.2 V div−1. Each height on the screen can be read to ± 0.1 div. Calculate the percentage uncertainty in the peak e.m.f. at each setting, and explain why the new setting is better.
(2)
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Answer
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Notes
Part (a)
peak = 1.4 div × 0.5 V div−1 = 0.70 V (or peak-to-peak 2.8 div ÷ 2)
✓ 1
Part (b)
one cycle = 4.0 div × 10 ms div−1 = 40 ms (measure across two whole cycles, 8.0 div, and halve)
✓ 1
f = 1 ÷ 0.040 = 25 s−1 (25 Hz)
✓ 1
Part (c)
at 0.5 V div−1: 0.1 ÷ 1.4 = 7.1%; at 0.2 V div−1 the peak is 0.70 ÷ 0.2 = 3.5 div, so 0.1 ÷ 3.5 = 2.9%
✓ 1
the trace is taller but still fits on the screen (7.0 div peak to peak, screen 8 div), so the same reading uncertainty is a smaller fraction of the height: the reading is more precise
✓ 1
Exam tip: Choose the most sensitive Y-gain that still keeps the whole trace on the screen, and measure a time across several whole cycles before dividing.
A thermistor is calibrated by placing it beside a thermometer in a stirred water bath and recording its resistance at 10 °C intervals. Figure 1 shows the calibration graph. The resistance is measured with an ohmmeter of resolution 0.01 kΩ.
(a)
The thermistor is placed in a liquid and its resistance is 8.0 kΩ. Use Figure 1 to determine the temperature of the liquid.
(1)
(b)
Explain why a calibration graph with many points is needed, rather than a straight line drawn between the readings at 0 °C and 100 °C.
(1)
(c)
The calibration readings are 20.2 kΩ at 10 °C, 12.5 kΩ at 20 °C, 1.27 kΩ at 80 °C and 0.934 kΩ at 90 °C. Calculate the mean sensitivity of the thermistor in each of these intervals, and explain in which interval it can detect the smaller temperature change.
(2)
(d)
State one precaution during the calibration that ensures the thermistor and the thermometer are at the same temperature.
(1)
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Answer
Mark
Notes
Part (a)
30 °C (allow 29–31 °C)
✓ 1
Part (b)
the resistance does not change linearly with temperature (the graph is a curve), so a straight line between two fixed points would give wrong temperatures in between
10–20 °C: the sensitivity is much greater, so the 0.01 kΩ resolution corresponds to only about 0.01 K there, compared with about 0.3 K between 80 °C and 90 °C
✓ 1
Part (d)
stir the water, hold the thermistor right beside the thermometer bulb at the same depth, and wait until both readings are steady before recording
✓ 1
Exam tip: Sensitivity is the change in output per unit change in input — here the gradient of the calibration curve. Where the curve is steep, a small temperature change gives a large, easily read change in resistance.
A student sets up a simple pendulum: a steel sphere on a thread. Using a metre rule, she measures the distance from the point of suspension to the bottom of the sphere as 0.640 m. She measures the diameter of the sphere with a vernier caliper as 31.8 mm. She times 20 oscillations as 31.7 s.
Figure: the measured distances (sphere enlarged; not to scale)
(a)
Calculate the length of the pendulum that should be used in T = 2π√(l/g).
(1)
(b)
Describe how she could make the point of suspension well defined and ensure that the metre rule is vertical.
(2)
(c)
Determine a value for g.
(2)
(d)
State where a fiducial mark should be placed, and why.
(1)
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Answer
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Notes
Part (a)
to the centre of the sphere: 0.640 − 0.0159 = 0.624 m (0.6241 m)
✓ 1
Part (b)
clamp the thread between two small flat blocks (or a split cork) held in the clamp, so it pivots at a fixed point
✓ 1
check the rule is vertical with a set square standing on the bench against it (or with a plumb line)
✓ 1
Part (c)
T = 31.7 ÷ 20 = 1.585 s
✓ 1
g = 4π2 × 0.6241 ÷ 1.5852 = 9.81 m s−2
✓ 1
Part (d)
at the centre of the oscillation (the rest position of the sphere), where the sphere moves fastest so the instant it passes is sharply defined
✓ 1
Common error: Using the distance to the bottom (or top) of the sphere as the length. The equation needs the distance to the centre of mass, so add or subtract half the diameter.
A student investigates how the temperature of hot water in an uncovered beaker falls over 20 minutes. She can use either a liquid-in-glass thermometer (resolution 1 °C) with a stopwatch, or a temperature probe (resolution 0.1 °C) connected to a data logger.
(a)
Explain two advantages of using the temperature probe and data logger.
(2)
(b)
Suggest a suitable time interval between readings. Justify your answer.
(1)
(c)
Describe two precautions to reduce the risk of harm from the hot water.
(2)
(d)
The data logger records 78.4 °C at t = 120 s and 71.0 °C at t = 300 s. Calculate the mean rate of cooling over this interval.
(1)
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Answer
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Notes
Part (a)
readings are taken automatically at exactly regular times over the whole 20 minutes, with no reaction-time or reading (parallax) errors
✓ 1
the probe's finer resolution (0.1 °C) means each reading has a smaller uncertainty, and many more readings can be collected
✓ 1
Part (b)
e.g. every 10 s: the temperature changes slowly, and this gives 120 readings over 20 minutes, plenty to show the shape of the cooling curve
✓ 1
Part (c)
stand the beaker on a heat-proof mat, held in a clamp or away from the bench edge so it cannot be knocked over
✓ 1
do not carry the beaker while the water is hot; pour hot water from a kettle carefully and let the water cool before emptying
Exam tip: Justify a data logger by what it does in this experiment — regular readings over a long time, finer resolution, no reaction time — and give a sampling interval that suits how quickly the quantity changes.
A thin glass tube of uniform bore, sealed at the bottom, traps a column of air under a short bead of oil. The tube is clamped upright in a beaker of water with a thermometer, and the water is heated. The length l of the air column is read from a rule fixed beside the tube. Results: θ / °C: 10.0, 20.0, 30.0, 40.0, 50.0, 60.0, 70.0 l / mm: 51.8, 53.7, 55.5, 55.9, 59.2, 61.0, 62.8
(a)
Identify the anomalous reading.
(1)
(b)
Suggest a cause of this reading and how it could be avoided.
(2)
(c)
Explain why the length of the air column can be used in place of its volume.
(1)
(d)
Using the readings at 10.0 °C and 70.0 °C, determine the temperature at which the length of the air column would become zero.
(2)
(e)
State the graph that should be plotted to find this temperature more reliably.
(1)
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Answer
Mark
Notes
Part (a)
55.9 mm at 40.0 °C (the pattern predicts about 57.3 mm)
✓ 1
Part (b)
the reading was taken before the trapped air had reached the temperature of the water, so the air was cooler and shorter than expected
✓ 1
stir the water and wait (e.g. a minute) at each temperature until the length stops changing before reading; then repeat the reading at 40.0 °C
✓ 1
Part (c)
the bore is uniform, so the volume = cross-sectional area × length, and the volume is proportional to the length
l = 0 at 10.0 − 51.8 ÷ 0.1833 = −273 °C (−272.5 °C)
✓ 1
Part (e)
l against θ using all the readings except the anomaly; draw the best-fit line and extend it to l = 0, reading the intercept on the temperature axis
✓ 1
Model sketch
Figure: best-fit line (anomaly left out) extended to l = 0
Common error: Discarding a reading only because it is “wrong” without giving a reason or repeating it. Identify it, suggest a physical cause, repeat if possible, and leave it out of the line.
A student calibrates a Hall probe using the field inside a long solenoid. The solenoid has 600 turns wound evenly over a length of 0.300 m. The flux density at its centre is B = μ0NI/L, where μ0 = 4π × 10−7 H m−1.
(a)
State which length L should be measured, and why.
(1)
(b)
Calculate B at the centre of the solenoid when the current is 2.00 A.
(2)
(c)
Describe how the Hall probe should be zeroed and positioned.
(2)
(d)
At 2.00 A the probe output is 0.126 V. Calculate the sensitivity of the probe.
(1)
(e)
Near a magnet the probe output is 0.088 V. Determine the flux density there.
(1)
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Answer
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Notes
Part (a)
the length of the winding itself (from the first turn to the last), not the length of the tube, because the field depends on the number of turns per metre of winding
✓ 1
Part (b)
B = 4π × 10−7 × 600 × 2.00 ÷ 0.300
✓ 1
= 5.03 × 10−3 T = 5.03 mT
✓ 1
Part (c)
zero the meter with no current in the solenoid (and away from magnets and other current-carrying wires)
✓ 1
place the probe at the centre of the solenoid, with its flat face perpendicular to the axis; rotate it slightly to obtain the maximum reading
✓ 1
Part (d)
0.126 ÷ 5.03 × 10−3 = 25.1 V T−1 (25 mV per mT)
✓ 1
Part (e)
0.088 ÷ 25.1 = 3.5 × 10−3 T (3.5 mT)
✓ 1
Exam tip: Calibration means using a known value to find how an instrument responds. Here the solenoid's calculated field fixes the probe's sensitivity, which then turns any later output into a flux density.
A 1000 µF capacitor is charged to 12.0 V and then discharged through a 47 kΩ resistor. The p.d. across the capacitor is read on a digital voltmeter of resolution 0.01 V and the time is measured with a stopwatch.
(a)
Calculate the time constant of the circuit.
(1)
(b)
Suggest how often readings should be taken and for how long. Justify your answer.
(2)
(c)
Calculate the p.d. after five time constants, and the percentage uncertainty in reading it. Take the uncertainty as half the resolution.
(2)
(d)
Explain why the voltmeter must have a very high resistance.
(1)
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Answer
Mark
Notes
Part (a)
RC = 47 × 103 × 1000 × 10−6 = 47 s
✓ 1
Part (b)
every 10 s (or every 5 s at the start), giving at least 10–15 readings
✓ 1
for about 3 time constants (about 150 s): over this range the p.d. falls to about 5% of its starting value, so the shape of the decay is clear
✓ 1
Part (c)
V = 12.0 e−5 = 0.081 V
✓ 1
0.005 ÷ 0.081 × 100 = 6.2% — so readings this late in the discharge add little useful information
✓ 1
Part (d)
it is connected in parallel with the resistor, so a low-resistance voltmeter would provide a second discharge path, reducing the effective resistance and the time constant (a systematic error)
✓ 1
Exam tip: Spread readings over the part of the change that carries information. For a decay, 2–3 time constants with closer readings early on is usually best: late readings are tiny and have large percentage uncertainties.
A rubber bung of mass 0.025 kg is whirled in a horizontal circle on a string that passes through a smooth vertical tube held in the hand. A 0.100 kg mass hangs from the lower end of the string, supplying the tension. The radius, measured from the top of the tube to the centre of the bung, is 0.500 m. The student times 20 revolutions as 14.6 s.
Figure: side view at the instant the bung is level with the top of the tube; not to scale
(a)
Calculate the centripetal force on the bung, and compare it with the weight of the hanging mass.
(3)
(b)
Suggest one reason for the difference.
(1)
(c)
Suggest an improvement using additional apparatus, and explain how it helps.
(1)
(d)
State one safety precaution.
(1)
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Answer
Mark
Notes
Part (a)
T = 14.6 ÷ 20 = 0.730 s
✓ 1
F = m4π2r/T2 = 0.025 × 4π2 × 0.500 ÷ 0.7302
✓ 1
= 0.93 N, a little less than the weight 0.100 × 9.81 = 0.98 N
✓ 1
Part (b)
friction between the string and the tube / the radius changing during the timing because the bung's speed varies
✓ 1
Part (c)
attach a small crocodile clip (or paper clip) to the string just below the tube and keep it just below the tube throughout, so the radius stays constant / film the motion from above against a scale to find the period frame by frame
✓ 1
Part (d)
wear safety goggles and keep other people well clear of the circle, in case the bung comes off the string
✓ 1
Exam tip: Measure the radius to the centre of the bung — the relevant dimension — and time many revolutions, as with any repeating motion.
Trolley A, of mass 0.500 kg, moves along a horizontal runway towards trolley B, also of mass 0.500 kg, which is at rest. They collide and stick together. A motion sensor connected to a data logger records the displacement of trolley A every 0.05 s. Figure 1 shows the results.
(a)
Use Figure 1 to determine the velocity of trolley A before the collision and the velocity of the pair after it.
(2)
(b)
Calculate the total momentum before and after the collision, and the percentage of the momentum that is lost.
(2)
(c)
Suggest the main cause of this loss, and describe how the runway could be adjusted, using the motion sensor, to reduce it.
(1)
(d)
Give one advantage of using the motion sensor and data logger rather than a stopwatch and marks on the runway.
(1)
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Answer
Mark
Notes
Part (a)
before: (0.80 − 0.20) ÷ 1.00 = 0.60 m s−1
✓ 1
after: (1.08 − 0.80) ÷ 1.00 = 0.28 m s−1
✓ 1
Part (b)
before: 0.500 × 0.60 = 0.30 kg m s−1; after: 1.000 × 0.28 = 0.28 kg m s−1
✓ 1
loss = 0.020 ÷ 0.300 × 100 = 6.7%
✓ 1
Part (c)
friction on the trolleys, which acts throughout the 1 s sections used for the velocities: raise one end of the runway slightly and adjust until a trolley given a push moves at constant velocity, shown by equal displacements in equal 0.05 s intervals on the data logger
✓ 1
Part (d)
many readings at exactly regular 0.05 s intervals with no reaction time, so the velocity just before and just after the collision can both be found from the gradients
✓ 1
Exam tip: Find a velocity from the gradient of a straight section of a displacement–time graph, using two points far apart on that section, never a single reading of s ÷ t from the origin.
An uncharged capacitor is charged from a 6.0 V supply through a resistor. A current sensor connected to a data logger records the charging current every 0.2 s. Figure 1 shows the results.
(a)
Use Figure 1 to determine the resistance of the resistor.
(1)
(b)
Use the area under the graph to estimate the charge stored on the capacitor. Hence calculate the capacitance.
(2)
(c)
Determine the time constant from Figure 1 and use it to check your value of the capacitance.
(2)
(d)
The data logger could sample at 1, 5 or 50 readings per second. Explain why 5 per second was a sensible choice.
(1)
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Answer
Mark
Notes
Part (a)
at t = 0 the capacitor is uncharged, so all 6.0 V is across the resistor: R = 6.0 ÷ 4.0 × 10−3 = 1.5 kΩ
✓ 1
Part (b)
area ≈ 12 large squares, each 1 s × 0.5 mA = 0.5 mC, so Q ≈ 6.0 mC (allow 5.6–6.4 mC)
✓ 1
C = Q/V = 6.0 × 10−3 ÷ 6.0 = 1.0 × 10−3 F (allow 930–1070 µF)
✓ 1
Part (c)
the current falls to 0.37 × 4.0 = 1.5 mA at t ≈ 1.5 s, so RC = 1.5 s (allow 1.4–1.6 s)
✓ 1
C = 1.5 ÷ 1500 = 1.0 × 10−3 F, which agrees with the value from the area
✓ 1
Part (d)
the charging is practically complete after about 5RC ≈ 7.5 s, so 5 per second gives about 40 readings with about 7 in the first time constant, enough to show the curve; 1 per second gives too few early readings, and 50 per second gives no extra useful information
✓ 1
Exam tip: When counting squares, decide first what one large square is worth in the units of the axes (here 1 s × 0.5 mA = 0.5 mC), then count whole squares and combine the part squares.
A student investigates how the period T of a mass–spring oscillator depends on the mass m on the spring. For each mass she times what she counts as 20 oscillations, once. Theory gives T2 = 4π2(m + ms)/k, where k is the spring constant and ms is a constant mass that allows for the moving spring. Figure 1 shows her graph of T2 against m and the best-fit line she drew.
(a)
Identify the anomalous point. Show that it is consistent with her having counted only 19 oscillations while recording the time as if for 20.
(2)
(b)
Use the gradient of the line to determine k.
(2)
(c)
Use Figure 1 to determine ms.
(1)
(d)
Suggest two improvements to the procedure that would prevent such a counting error, or show it up at once.
(2)
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Answer
Mark
Notes
Part (a)
the point at m = 0.400 kg (T2 = 0.60 s2)
✓ 1
the line gives T2 ≈ 0.66 s2 at 0.400 kg; with 19 oscillations the calculated T is 19/20 of the true value, so T2 = 0.66 × (19/20)2 = 0.60 s2, as plotted
the line meets the m axis at m = −ms, so ms = 0.020 kg (allow 0.015–0.025 kg; or ms = T2-intercept ÷ gradient ≈ 0.03 ÷ 1.58 = 0.019 kg)
✓ 1
Part (d)
place a fiducial mark (a pin in a clamp) at the equilibrium position and count “zero” as the mass first passes it in a chosen direction, counting each later pass in that direction
✓ 1
time 20 oscillations three times for each mass and average: a miscount gives one time that disagrees with the others and can be repeated
✓ 1
Common error: Starting the stopwatch and counting “one” at the same instant, which records one oscillation too few. Count “zero” as you start the stopwatch.
A student investigates how the pressure p of a fixed mass of air in a sealed syringe depends on its volume V. A pressure gauge is connected to the syringe, and the air is at room temperature, 293 K. Theory predicts that p is proportional to 1/V. Figure 1 shows the graph she drew of p against 1/V, with V in cm3.
(a)
Identify three faults in the graph and state how each should be corrected.
(3)
(b)
Explain why the plunger should be moved slowly and the reading taken after a short wait.
(1)
(c)
One of her readings was V = 24.0 cm3 and p = 100 kPa. Calculate the number of molecules of air in the syringe.
(2)
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Answer
Mark
Notes
Part (a)
the 1/V axis has no unit: label it 1/V / cm−3
✓ 1
the 1/V scale is not linear (equal spaces stand for steps of 0.005 and 0.010): use a uniform scale
✓ 1
the points are joined dot to dot: draw a single best-fit straight line
✓ 1
also accept: the points use only the bottom third of the grid — use a p scale of 0 to 140 kPa; or the 1/V axis does not start at 0, so proportionality (a line through the origin) cannot be tested
—
Part (b)
compressing the air quickly does work on it and raises its temperature; waiting lets it return to room temperature, so the temperature is kept constant as the theory requires
Exam tip: Before plotting, check each axis: quantity / unit as the label, a uniform scale in easy steps (1, 2 or 5), and points spread over at least half the grid in each direction.
A mass hangs from a spring attached to a vibration generator, which is driven by a signal generator. A student varies the driving frequency f and measures the amplitude A of the oscillations of the mass. Figure 1 shows her readings.
(a)
Explain why her readings do not allow the resonant frequency to be found precisely.
(1)
(b)
Describe how she should choose further readings.
(2)
(c)
Describe how the amplitude could be measured accurately.
(1)
(d)
State one variable that must be kept constant, and how this is done.
(1)
(e)
State one safety precaution.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
the readings are 2 Hz apart, and the largest amplitudes (11.7 mm at 6 Hz, 9.9 mm at 8 Hz) show the peak lies between them, where there are no readings, so neither the maximum amplitude nor its frequency can be located
✓ 1
award for any answer linking the wide 2 Hz spacing to the missing readings near the peak; accept “between 4 Hz and 8 Hz”; do not award “too few readings” alone
—
Part (b)
first sweep the frequency slowly to find roughly where the amplitude is greatest (between 6 Hz and 8 Hz)
✓ 1
then take readings at small intervals across that region, e.g. every 0.1–0.2 Hz (accept any interval of 0.5 Hz or less), with several readings on each side of the peak to define its shape
✓ 1
Part (c)
any one of: fix a vertical metre rule close behind the mass, with a pointer on the mass, view at eye level and measure the total height of the motion, then halve it; film the motion against the rule and step through frame by frame; use a motion sensor below the mass with a data logger
✓ 1
a rule alone, without a way of avoiding parallax or of reading the moving mass, does not score
—
Part (d)
the amplitude of the driving vibration: keep the signal generator output setting the same, checking the p.d. across the vibration generator with an oscilloscope or a.c. voltmeter at each frequency
✓ 1
also accept: the same mass and the same spring throughout, not changed or re-hung between readings; the variable and the method are both needed for the mark
—
Part (e)
any one of: clamp the stand firmly to the bench (or use a counterweight) so the vibrating system cannot topple; keep feet clear of the hanging mass, or place a box of padding beneath it, in case it falls; wear eye protection in case the stretched spring breaks or flies off near resonance
✓ 1
Exam tip: Readings should be closest together where the quantity changes most quickly: around a peak, a turning point or the start of a decay. Evenly spaced readings can step right over the feature you need.
A student measures the spectrum of the radiation from a lamp filament using a spectrometer fitted with an infrared-sensitive light sensor. Figure 1 shows the relative intensity she recorded at each wavelength. Wien's displacement law is λmaxT = 2.898 × 10−3 m K.
(a)
Use Figure 1 to estimate the wavelength at which the intensity is greatest.
(1)
(b)
Calculate the temperature of the filament.
(1)
(c)
The student estimates that λmax can be read to ± 50 nm. Calculate the uncertainty in the temperature.
(2)
(d)
Suggest how the readings could be changed to reduce this uncertainty.
T ∝ 1/λmax, so T also has a 5% uncertainty: ± 0.05 × 2898 = ± 1.4 × 102 K (allow ± 145 K or ± 150 K)
✓ 1
Part (d)
take extra readings at much smaller intervals (e.g. every 20 nm) between about 800 nm and 1200 nm, near the peak, so its position can be located more precisely
✓ 1
Part (e)
do not touch the lamp while it is on or just after — the glass is hot; let it cool before moving it (or do not stare at the bright filament)
✓ 1
Common error: Forgetting to convert nanometres to metres before using Wien's law: 1000 nm = 1.000 × 10−6 m.
Liquid in a U-shaped glass tube, open at both ends, oscillates when one side is pushed down and released. Theory predicts that the period is T = 2π√(L/2g), where L is the total length of the liquid column measured along the tube.
(a)
Plan an experiment to test this relationship and determine a value for g. Your plan should include the measurements, the instruments, the techniques, the control of variables, a safety point and how the results will be analysed.
(6)
(b)
For L = 0.480 m, the time for 20 oscillations is 19.7 s. Calculate a value for g.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
Indicative content, one mark each:
—
vary L by adding measured amounts of liquid, for at least six values; measure L along the centre line of the tube with a length of thread (or flexible tape), then a metre rule (1 mm)
✓ 1
start the oscillation the same way each time, by blowing gently into one side (or pressing a bulb) and releasing, with small amplitude
✓ 1
use a fiducial mark at the equilibrium level of the liquid on one side and time 20 oscillations with a stopwatch, or record with a video camera / data logger with a pressure sensor
✓ 1
repeat each timing and calculate the mean; keep the same liquid, the same tube and the same temperature
✓ 1
plot T2 against L: a straight line through the origin confirms the relationship; gradient = 2π2/g, so g = 2π2/gradient
✓ 1
safety: clamp the glass tube securely so it cannot fall and break; mop up spills
✓ 1
Part (b)
T = 19.7 ÷ 20 = 0.985 s
✓ 1
g = 2π2L/T2 = 2π2 × 0.480 ÷ 0.9852
✓ 1
= 9.77 m s−2
✓ 1
Exam tip: Rearrange the equation before you plan the graph: squaring gives T2 = (2π2/g)L, so T2 against L is a straight line through the origin and g comes from the gradient.
A student determines the specific heat capacity c of a cooking oil. An immersion heater connected to a 12.0 V supply carries a current of 3.50 A and is placed in 0.250 kg of oil in a beaker. A temperature probe connected to a data logger records the temperature. The graph of temperature against time is a straight line over the first 5 minutes, with gradient 0.0960 K s−1.
(a)
Calculate the specific heat capacity of the oil.
(2)
(b)
Some energy is lost to the surroundings. State and explain the effect on the value of c.
(2)
(c)
The temperature probe has a zero error of +1.0 °C. Explain why this does not affect the value of c.
(1)
(d)
Suggest one improvement, using additional apparatus, that would make the value of c more accurate. Explain how it helps.
(1)
(e)
Describe two safety precautions.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
P = 12.0 × 3.50 = 42.0 W, and P = mc × (Δθ/Δt)
✓ 1
c = 42.0 ÷ (0.250 × 0.0960) = 1750 J kg−1 K−1
✓ 1
Part (b)
the temperature rises more slowly than it would with no losses, so the gradient is smaller
✓ 1
the calculated value of c is too large
✓ 1
Part (c)
every temperature is shifted by the same 1.0 °C, so the line moves up but its gradient (rate of temperature rise) is unchanged
✓ 1
Part (d)
e.g. insulate the beaker with lagging and a lid, reducing energy losses / stir the oil continuously so the probe reads the mean temperature / use a joulemeter to measure the energy supplied directly
✓ 1
Part (e)
make sure the heater is fully immersed in the oil before it is switched on, and switch it off before removing it
✓ 1
hot oil: stand the beaker on a heat-proof mat in a stable position, do not heat the oil above about 60 °C, let it cool before handling
✓ 1
Common error: Saying heat losses make c “inaccurate” without the direction. Work it through: losses mean a smaller temperature rise for the same energy, so the calculated c is too large.
Two temperature-measuring systems are available. System X is a thermocouple with a sensitivity of 41 µV K−1, read with a millivoltmeter of resolution 0.1 mV. System Y is an electronic sensor with a sensitivity of 10.0 mV K−1, read with a voltmeter of resolution 0.01 V. A student needs to detect temperature changes of 0.5 K.
(a)
Define the sensitivity of a measuring instrument.
(1)
(b)
Calculate the smallest temperature change that each system can detect.
(2)
(c)
Suggest how system Y could be modified to detect changes of 0.5 K.
(1)
(d)
With the modified system from (c), system Y gives 0.012 V in melting ice, although its output should be zero at 0 °C. Explain what type of error this is and how the student should deal with it.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the change in the instrument's output (reading) per unit change in the quantity being measured
✓ 1
Part (b)
X: 0.1 × 10−3 ÷ 41 × 10−6 = 2.439 K, so 2.44 K (rounded up: a 2.4 K change gives less than 0.1 mV)
✓ 1
Y: 0.01 ÷ 0.0100 = 1.0 K
✓ 1
Part (c)
read its output with a voltmeter of resolution 0.001 V (or use a more sensitive range), so it can detect 0.1 K changes
✓ 1
Part (d)
a systematic (zero) error, equivalent to +1.2 K, which affects every reading by the same amount and is not reduced by repeating
✓ 1
subtract 0.012 V from every reading before converting (or calibrate the sensor at fixed points and use the calibration line)
✓ 1
Exam tip: Smallest detectable change = resolution of the meter ÷ sensitivity of the sensor. Make the units match first: convert µV and mV to V.
A student plans to investigate how the peak e.m.f. induced in a coil depends on the speed of a bar magnet falling through it. The magnet, 30 mm long, is released from rest at a height h above the coil and falls through a vertical plastic tube that passes through the coil. The coil is connected to a data logger.
Figure: the magnet is released from rest a height h above the coil; not to scale
(a)
Calculate the speed of the magnet as it reaches the coil when h = 0.200 m, and the time it takes to move its own length at this speed.
(2)
(b)
Explain why a data logger is needed and suggest a suitable sampling rate.
(2)
(c)
State two control variables and how each is kept constant.
(2)
(d)
State the graph that should be plotted to test whether the peak e.m.f. is proportional to the speed of the magnet.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
v = √(2gh) = √(2 × 9.81 × 0.200) = 1.98 m s−1
✓ 1
time = 0.030 ÷ 1.98 = 0.015 s (15 ms)
✓ 1
Part (b)
the e.m.f. pulse lasts only about 15 ms, far too short to read on a meter by eye
✓ 1
sample at 1000 per second or more, so there are at least 15 readings in each pulse and the peak is not missed
✓ 1
Part (c)
the same magnet, released with the same pole downwards each time
✓ 1
the same coil (same number of turns), clamped in the same position around the tube
✓ 1
Part (d)
peak e.m.f. against √h (because v = √(2gh)): a straight line through the origin if peak e.m.f. ∝ v
✓ 1
Exam tip: Before choosing a sampling rate, estimate how long the event lasts. A rate that gives only one or two samples in the pulse will miss the true peak.
In a fine-beam tube, electrons are accelerated from rest through a p.d. of 250 V and then move in a circle in the uniform magnetic field of a pair of Helmholtz coils. The flux density is 1.07 mT. A low-pressure gas in the tube makes the path visible. A scale is fixed in front of the tube, as shown in Figure 1. The specific charge of the electron is given by e/m = 2V/(B2r2), where r is the radius of the path.
(a)
State one hazard of this experiment and a precaution to reduce the risk.
(1)
(b)
Use Figure 1 to determine the radius of the circular path.
(1)
(c)
The scale is in front of the beam. Explain how the student should read the positions of the edges of the beam to avoid parallax error.
(1)
(d)
Calculate the specific charge of the electron.
(1)
(e)
Each scale reading has an uncertainty of ± 0.1 cm, and B has an uncertainty of ± 1%. Ignoring the uncertainty in V, calculate the percentage uncertainty in e/m.
(2)
(f)
Suggest two changes to the procedure that would reduce the percentage uncertainty in e/m.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
the 250 V accelerating supply can give an electric shock: make all connections with the supplies off and use shrouded leads (or: the evacuated glass tube could implode — handle it only by its base and do not knock it)
✓ 1
Part (b)
diameter = 12.0 − 2.0 = 10.0 cm, so r = 5.00 cm
✓ 1
Part (c)
for each edge, look at the scale perpendicularly, with the eye directly in front of that edge (line of sight parallel to the axis of the coils), e.g. move a sliding pointer along the scale until it lines up with the beam when viewed straight on
reduce the coil current (smaller B) or raise V so the circle is as large as the tube allows: the same ± 0.2 cm is then a smaller fraction of the diameter
✓ 1
any one of: read each edge several times (e.g. with sliding pointers, re-set each time) and average; measure the diameter at several values of V and use a graph; measure B more precisely, e.g. with a calibrated Hall probe or a coil-current meter of finer resolution
✓ 1
Exam tip: When a quantity is squared in the equation, double its percentage uncertainty. Here both r and B are squared, so they dominate the uncertainty in e/m.
To measure the specific latent heat of vaporisation L of water, a student stands a beaker of water containing an electric heater on a top-pan balance. When the water is boiling steadily she starts a stopwatch and records the balance reading every minute. She then repeats the experiment at a higher heater power. Figure 1 shows her results for powers of 36.0 W and 56.0 W.
(a)
Use Figure 1 to determine the rate of loss of mass, in kg s−1, at each power.
(2)
(b)
The rate of energy transfer to the surroundings, h, is the same at both powers, so P = L(Δm/Δt) + h. Determine L.
(2)
(c)
Calculate h.
(1)
(d)
Explain why the readings are started only when the water is boiling steadily.
(1)
(e)
Describe two safety precautions.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
36.0 W: (250.0 − 242.0) g in 10.0 min: 8.0 × 10−3 ÷ 600 = 1.33 × 10−5 kg s−1
✓ 1
56.0 W: (250.0 − 236.8) g in 10.0 min: 13.2 × 10−3 ÷ 600 = 2.20 × 10−5 kg s−1
✓ 1
Part (b)
subtracting: 56.0 − 36.0 = L × (2.20 − 1.33) × 10−5, so 20.0 = L × 8.67 × 10−6
✓ 1
L = 2.31 × 106 J kg−1
✓ 1
Part (c)
h = 36.0 − 2.31 × 106 × (8.0 × 10−3 ÷ 600) = 36.0 − 30.8 = 5.2 W
✓ 1
Part (d)
before then, some of the energy supplied is still raising the temperature of the water, so the rate of loss of mass is not yet steady and is not due to vaporisation alone
✓ 1
Part (e)
make sure the heater is fully immersed before it is switched on, and switch it off and let it cool before removing it
✓ 1
stand the beaker on a heat-proof mat in a stable position away from the bench edge, and keep hands and face out of the steam; let the water cool before moving the beaker
✓ 1
Exam tip: Using two powers and subtracting removes the unknown energy loss: it is the same in both runs, so it cancels. Look for this whenever losses cannot be measured directly.
A student plans to investigate how the peak e.m.f. ε0 induced in a small search coil depends on the frequency f of the alternating current in a long solenoid. Theory predicts that ε0 is proportional to f if the amplitude of the current stays the same. Figure 1 shows the apparatus and two oscilloscope traces from a trial run.
(a)
Plan the investigation. Your plan should include the measurements, how they are made, the control of variables, the range and number of readings, a safety point and how the results will be used.
(6)
(b)
Use the traces in Figure 1 to determine ε0 at each frequency, and show that the results are consistent with ε0 ∝ f. Also use the 400 Hz trace to check that frequency.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
Indicative content, one mark each:
—
vary f with the signal generator over a wide range, e.g. 50 Hz to 500 Hz, for at least six values; find each frequency from the oscilloscope time-base (time for several cycles) rather than from the dial
✓ 1
measure the peak-to-peak height of the trace and halve it, multiplying by the Y-gain; choose the Y-gain so the trace fills most of the screen
✓ 1
keep the current amplitude constant: adjust the signal generator output so the ammeter reading is the same at every frequency (the solenoid's opposition to the current changes with frequency)
✓ 1
keep the same search coil fixed at the centre of the solenoid, coaxial with it (its plane perpendicular to the axis), so the flux linkage per unit current is unchanged
✓ 1
safety: the solenoid can get warm — switch off between readings and check it is not hot before touching
✓ 1
plot ε0 against f: a straight line through the origin confirms proportionality
✓ 1
also accept: repeat each reading and average; twist the search-coil leads together to reduce stray pick-up
120 ÷ 15 = 8.0 and 400 ÷ 50 = 8.0, so ε0/f is the same (0.30 mV Hz−1)
✓ 1
400 Hz trace: one cycle = 2.5 div × 1.0 ms div−1 = 2.5 ms, so f = 1 ÷ 2.5 × 10−3 = 400 Hz
✓ 1
Exam tip: In a plan, say how each control variable is kept constant, not only what it is. Here the current must be checked and reset at every frequency, because the signal generator's output current changes as f changes.
Three groups of students each make five determinations of absolute zero, using a constant-volume gas thermometer and extrapolating a graph of pressure against temperature. Figure 1 shows their values.
(a)
For group Q, determine the mean value and its uncertainty, taking the uncertainty as half the range.
(2)
(b)
State which group's results are the most precise, and which group's mean is the most accurate. Give a reason for each.
(2)
(c)
Group P's mean is −257.6 °C. Explain whether the difference from −273 °C can be accounted for by random error, and suggest a cause.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
mean = (−281 − 266 − 275 − 270 − 273) ÷ 5 = −273.0 °C
✓ 1
half range = (−266 − (−281)) ÷ 2 = 7.5 °C, so −273 ± 8 °C (allow ± 7.5 °C)
✓ 1
Part (b)
P is the most precise: its values are spread over only 3 °C
✓ 1
Q is the most accurate: its mean is equal to the accepted value of −273 °C (R's mean is −285.4 °C, P's −257.6 °C)
✓ 1
Part (c)
P's uncertainty is only ± 1.5 °C but the difference is 15.4 °C, so random error cannot explain it: there is a systematic error that shifts every value the same way
✓ 1
e.g. part of the trapped air (in the tube to the pressure gauge) is outside the water bath, or the thermometer or gauge has a zero error
✓ 1
Common error: Calling a set “accurate” because its values are close together. Closeness of repeats is precision; accuracy is how close the mean is to the accepted value.
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