Edexcel International A Level Physics · Unit 3 · Practical Skills in Physics I
Planning experiments and measuring instruments
28 free, high-quality past-paper-style questions on Planning experiments and measuring instruments 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 · 167 marks · Updated 26 October 2026 · Created by Evan Kotronis with AI assistance · every answer re-solved before publishing · How questions are checked
1IALP-U3-1-B-01Unit 3 · ASEasy5 marks
A student uses a vernier caliper to measure the diameter of a glass marble. The main scale is marked in millimetres and the vernier scale has 10 divisions. With the jaws closed, the zero of the vernier scale lies slightly to the left of the zero on the main scale, and vernier line 7 lines up with a main-scale line. With the marble between the jaws, the zero of the vernier scale lies between the 15 mm and 16 mm marks, and vernier line 8 lines up with a main-scale line.
Figure: main scale and vernier scale with the marble between the jaws (enlarged)
(a)
State the resolution of the vernier caliper.
(1)
(b)
State the reading with the marble between the jaws.
(1)
(c)
Determine the zero error and hence the diameter of the marble.
(2)
(d)
Calculate the percentage uncertainty in the diameter. Take the uncertainty as half the resolution.
(1)
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Answer
Mark
Notes
Part (a)
0.1 mm
✓ 1
Part (b)
15 + 8 × 0.1 = 15.8 mm
✓ 1
Part (c)
the vernier zero is 0.3 mm to the left of the main zero: zero error = −(10 − 7) × 0.1 = −0.3 mm
✓ 1
diameter = 15.8 − (−0.3) = 16.1 mm
✓ 1
Part (d)
0.05 ÷ 16.1 × 100 = 0.31%
✓ 1
Common error: With a negative zero error every reading is too small, so the correction is added. Always write corrected reading = reading − zero error and keep the sign.
A student wants the thickness of one sheet of card. She stacks 10 identical sheets and closes a micrometer screw gauge on the stack using the ratchet. On the sleeve, the 4 mm mark is visible but the half-millimetre mark after it is not. Thimble division 38 is on the datum line. With the anvil and spindle closed, the zero line of the sleeve is just visible and thimble division 3 is on the datum line (the thimble zero is just below the datum line).
Figure: micrometer scales closed on the stack of 10 sheets (enlarged)
(a)
State the reading on the micrometer for the stack.
(1)
(b)
Determine the thickness of the stack, corrected for the zero error.
(1)
(c)
Calculate the thickness of one sheet.
(1)
(d)
Explain why measuring 10 sheets together gives a better value than measuring one sheet.
(1)
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Answer
Mark
Notes
Part (a)
4 + 38 × 0.01 = 4.38 mm
✓ 1
Part (b)
zero error = +0.03 mm, so 4.38 − 0.03 = 4.35 mm
✓ 1
Part (c)
4.35 ÷ 10 = 0.435 mm
✓ 1
Part (d)
the uncertainty (± 0.005 mm) is the same but the reading is 10 times bigger, so the percentage uncertainty is 10 times smaller (0.11% instead of about 1.1%)
✓ 1
Exam tip: Before adding the thimble reading, check whether a half-millimetre mark is showing below the datum line. Here it is not, so the sleeve reading is 4.00 mm, not 4.50 mm.
A student investigates how much a thin copper wire stretches before it breaks. The wire is clamped at one end of a bench, passes over a pulley at the other end and carries a hanger to which slotted masses are added one at a time.
Figure: arrangement for stretching the wire to breaking point; not to scale
(a)
Identify two hazards in this experiment and describe a precaution for each.
(4)
(b)
Explain why repeat readings of extension for the same load are not appropriate once the wire has started to deform plastically.
(1)
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Answer
Mark
Notes
Part (a)
the wire may snap and whip back
✓ 1
wear safety goggles / stand to the side of the wire, not in line with it
✓ 1
the masses fall when the wire breaks
✓ 1
place a box of sand or foam under the hanger and keep feet clear
✓ 1
Part (b)
plastic deformation is permanent, so the wire has changed and a repeat would not be a measurement on the same wire under the same conditions
✓ 1
Exam tip: Pair each hazard with the precaution that deals with that hazard. A list of general rules — “no running”, “tie hair back” — does not answer the question.
The outside diameter of a length of copper pipe is found with a vernier caliper whose vernier scale has 10 divisions spanning 9 mm. Figure 1 shows the scales (a) when the jaws are closed with nothing between them and (b) when the jaws are closed on the pipe.
(a)
Use Figure 1(a) to determine the zero error of the caliper, including its sign.
(1)
(b)
Use Figure 1(b) to determine the reading on the caliper.
(1)
(c)
Determine the outside diameter of the pipe.
(1)
(d)
The corrected diameter comes from two readings, each uncertain by ± 0.05 mm, so its uncertainty is ± 0.10 mm. Calculate the percentage uncertainty in the diameter.
(1)
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Answer
Mark
Notes
Part (a)
vernier zero is to the right of the main zero and line 3 lines up: zero error = +0.3 mm
✓ 1
Part (b)
20 mm + 7 × 0.1 mm = 20.7 mm
✓ 1
Part (c)
20.7 − (+0.3) = 20.4 mm
✓ 1
Part (d)
0.10 ÷ 20.4 × 100 = 0.49%
✓ 1
Exam tip: Decide the sign of a zero error from Figure 1(a) before anything else: vernier zero to the right of the main zero means every reading is too large, so the zero error is positive and is subtracted.
A steel ball bearing has a mass of 2.02 g. Its diameter is measured with a micrometer screw gauge. Figure 1(a) shows the micrometer when the anvil and spindle are closed together: the zero line on the sleeve is just hidden by the thimble. Figure 1(b) shows the micrometer closed on the ball.
(a)
Use Figure 1(b) to determine the reading on the micrometer.
(1)
(b)
Use Figure 1(a) to determine the zero error, including its sign.
(1)
(c)
Determine the diameter of the ball.
(1)
(d)
Calculate the density of the steel.
(2)
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Answer
Mark
Notes
Part (a)
7.5 mm + 37 × 0.01 mm = 7.87 mm
✓ 1
Part (b)
the thimble has not quite reached zero (48 is two divisions short of 50): zero error = −0.02 mm
Common error: Reading a closed micrometer with 48 on the datum line as +0.48 mm. If the sleeve zero is hidden, the thimble is just short of zero, so the zero error is negative: 0.48 − 0.50 = −0.02 mm.
Figure 1 shows how a student has set up an experiment to find how the resistance of a thermistor varies with temperature between room temperature and 90 °C. The water is heated by an electric hot plate.
(a)
Identify two hazards shown in Figure 1 and describe how the arrangement should be changed to deal with each one.
(4)
(b)
Identify a problem with the positions of the thermistor and the thermometer in Figure 1 and explain how to correct it.
(2)
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Answer
Mark
Notes
Part (a)
the beaker is filled almost to the brim, so hot water may spill or boil over and scald
✓ 1
fill the beaker only about two-thirds full, and leave it to cool before moving it / use heat-proof gloves
✓ 1
the mains lead hangs over the edge of the bench, so someone could catch it and pull the hot plate and boiling water onto themselves
✓ 1
route the lead along the back of the bench so nothing hangs over the edge (also accept: the ohmmeter is next to the water, where spills could reach it — move it further away)
✓ 1
Part (b)
the thermistor is on the bottom, next to the hot plate, while the thermometer bulb is near the surface, so they are not at the same temperature
✓ 1
clamp the thermistor and the thermometer bulb side by side in the middle of the water, and stir before each reading
✓ 1
Exam tip: Pair each hazard with the change that removes it, and make both specific to the apparatus shown. “Be careful with hot water” is not a precaution; “fill the beaker only two-thirds full” is.
Figure 1 shows the meters used to measure the current in a resistor and the potential difference across it: an analogue ammeter with a mirror strip behind its needle, and a digital voltmeter.
(a)
State the resolution of each meter.
(2)
(b)
State the reading on the ammeter.
(1)
(c)
Calculate the resistance of the resistor.
(1)
(d)
Explain how the mirror strip helps the student to read the ammeter.
(1)
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Answer
Mark
Notes
Part (a)
ammeter: 0.02 A (1.0 A ÷ 50 divisions)
✓ 1
voltmeter: 0.01 V
✓ 1
Part (b)
0.64 A
✓ 1
Part (c)
R = 3.52 ÷ 0.64 = 5.5 Ω
✓ 1
Part (d)
the eye is moved until the needle hides its own reflection; the line of sight is then perpendicular to the scale, so there is no parallax error
✓ 1
Exam tip: To find the resolution of an analogue scale, divide the gap between two numbered marks by the number of small divisions between them: here 0.2 A ÷ 10 = 0.02 A.
A student is asked to determine the density of the metal in a small solid cylinder. The cylinder is about 38 mm long and about 13 mm in diameter, and its mass is about 13 g. She has a metre rule, a vernier caliper (resolution 0.1 mm), a micrometer screw gauge (resolution 0.01 mm, range 0–25 mm) and a top-pan balance (resolution 0.01 g).
(a)
Calculate the percentage uncertainty in the length of the cylinder if it is measured (i) with the metre rule and (ii) with the vernier caliper. Take the uncertainty as half the resolution.
(2)
(b)
State the most suitable instrument for the diameter. Give a reason.
(2)
(c)
Her measurements are: length 38.0 mm, diameter 12.60 mm, mass 12.80 g. Calculate the density of the metal.
(2)
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Answer
Mark
Notes
Part (a)
metre rule: 0.5 ÷ 38 × 100 = 1.3%
✓ 1
vernier: 0.05 ÷ 38 × 100 = 0.13%
✓ 1
Part (b)
micrometer screw gauge
✓ 1
the diameter is within its 25 mm range and its smaller resolution gives the smallest percentage uncertainty (about 0.04%)
Common error: Using the diameter in the volume formula instead of the radius, which makes the volume four times too large. Halve the diameter first, and convert mm to m and g to kg before dividing.
A student needs an accurate value for the diameter of a nylon fishing line, which is about 0.30 mm thick.
(a)
Explain, with a calculation, why a micrometer screw gauge is more suitable than a vernier caliper.
(2)
(b)
Describe how she should use the micrometer to obtain a reliable value for the diameter.
(3)
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Answer
Mark
Notes
Part (a)
vernier: 0.05 ÷ 0.30 × 100 = 17%; micrometer: 0.005 ÷ 0.30 × 100 = 1.7% (or 33% and 3.3% using the full resolution)
✓ 1
the micrometer's percentage uncertainty is ten times smaller
✓ 1
Part (b)
check the zero reading with the jaws closed and correct each reading for any zero error
✓ 1
close the jaws using the ratchet, so the soft nylon is not squashed
✓ 1
measure at several points along the line, rotating through 90° at each point, and calculate the mean
✓ 1
Exam tip: “Repeat and average” scores more when it is specific: say where the repeats are taken (along the length, in two perpendicular directions) and why (the line may be non-uniform or not perfectly round).
A student draws a thick line on paper with a soft pencil. Graphite conducts, so the line has a resistance. She plans to investigate how the resistance R of the line depends on its length l, using two crocodile clips as contacts.
(a)
State the independent variable and the dependent variable.
(1)
(b)
State two control variables and how each would be kept constant.
(2)
(c)
Describe how she could measure R for each length, naming the instruments.
(2)
(d)
State the graph she should plot and the result she expects if R is proportional to l.
(1)
(e)
Suggest why repeat readings are useful in this experiment.
(1)
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Answer
Mark
Notes
Part (a)
independent: length l between the clips; dependent: resistance R
✓ 1
Part (b)
width/thickness of the line: draw it once with the same pencil, using a ruler and the same number of strokes along its whole length
✓ 1
temperature: use a small current and switch off between readings
✓ 1
Part (c)
cell in series with an ammeter and the line, with a voltmeter connected across the line between the clips / an ohmmeter (multimeter on a resistance range) connected to the two clips
✓ 1
record V and I and calculate R = V/I (or read R directly); measure l with a ruler (1 mm resolution) between the clip contacts
✓ 1
Part (d)
R against l: a straight line through the origin
✓ 1
Part (e)
the contact between the clips and the paper varies each time they are placed, so readings scatter randomly; repeating and averaging reduces this effect
✓ 1
Model sketch
Figure: one suitable circuit (ammeter in series, voltmeter across the line between the clips); not to scale
Exam tip: For every control variable say how it is kept constant. “Keep the width the same” names the variable; “draw the whole line in one go with a ruler and the same pencil” explains the control.
A student drops a tennis ball from 1.00 m onto a concrete floor and measures the height to which it rebounds, using a vertical metre rule. Five drops give rebound heights of 0.52 m, 0.48 m, 0.55 m, 0.50 m and 0.51 m.
(a)
Explain why the student took several readings.
(1)
(b)
Calculate the mean rebound height and its uncertainty.
(2)
(c)
Calculate the percentage uncertainty in the mean rebound height.
(1)
(d)
Suggest two improvements to the technique that would reduce this uncertainty.
(2)
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Answer
Mark
Notes
Part (a)
the top of the rebound is reached only for an instant and is hard to judge, so readings vary randomly; a mean is more reliable
✓ 1
Part (b)
mean = 2.56 ÷ 5 = 0.512 m
✓ 1
uncertainty = half the range = (0.55 − 0.48) ÷ 2 = 0.035 m
✓ 1
Part (c)
0.035 ÷ 0.512 × 100 = 6.8%
✓ 1
Part (d)
film the bounce with a video camera in front of the rule and find the highest point frame by frame
✓ 1
place the camera (or eye) level with the top of the bounce to avoid parallax, and read the bottom of the ball each time; check the rule is vertical with a set square
✓ 1
Common error: Suggesting “use a more precise ruler”. The resolution (1 mm) is not the problem — the scatter is far larger. Improve the technique that causes the scatter.
A small steel ball rolls back and forth in a curved dish. A student uses a hand-held stopwatch to time 10 complete oscillations, three times. Her results are 13.6 s, 13.9 s and 13.4 s.
(a)
Explain why she times 10 oscillations rather than one.
(1)
(b)
Describe how a fiducial mark should be used, and state where it should be placed.
(2)
(c)
Calculate the period of oscillation and its percentage uncertainty.
(3)
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Answer
Mark
Notes
Part (a)
the reaction-time error at start and stop is the same, but it is a much smaller fraction of a longer time, so the percentage uncertainty is smaller
✓ 1
Part (b)
a fixed reference line (or pin) beside the dish, against which each pass of the ball is timed / counted
✓ 1
at the lowest point / centre of the oscillation, where the ball moves fastest, so the instant it passes is sharply defined
✓ 1
Part (c)
mean time for 10 oscillations = 40.9 ÷ 3 = 13.6 s (13.63 s)
Exam tip: The percentage uncertainty in the period is the same as in the time for 10 oscillations: dividing by an exact number (10) does not change it.
A student checks a spring balance (newton meter) by hanging known masses from it. With nothing attached it reads 0.12 N. With masses of 0.100 kg, 0.200 kg and 0.300 kg it reads 1.10 N, 2.08 N and 3.06 N. g = 9.81 N kg−1.
(a)
Calculate the true weight of the 0.200 kg mass.
(1)
(b)
State the type of error shown by the balance and give its size. Justify your answer using the readings.
(2)
(c)
The balance reads 4.35 N when an object hangs from it. Determine the mass of the object.
(2)
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Answer
Mark
Notes
Part (a)
0.200 × 9.81 = 1.96 N
✓ 1
Part (b)
a systematic (zero) error of +0.12 N
✓ 1
every reading is 0.12 N above the true weight: 1.10 − 0.98, 2.08 − 1.96, 3.06 − 2.94 all equal 0.12 N
✓ 1
Part (c)
corrected weight = 4.35 − 0.12 = 4.23 N
✓ 1
mass = 4.23 ÷ 9.81 = 0.431 kg
✓ 1
Model sketch
Figure: readings lie on a line parallel to the ideal line, 0.12 N above it (zero error)
Exam tip: A constant difference between the reading and the true value at every point is the signature of a zero error. If the difference grew with the load, the scale itself would be wrong.
A student determines the refractive index of a transparent liquid in a thin-walled rectangular tank. She directs a narrow ray of light from a ray box onto the side of the tank at angle of incidence i, traces the ray in the liquid, and measures the angle of refraction r with a protractor (resolution 1°).
Figure: the ray entering the liquid through the thin tank wall (seen from above)
(a)
Explain why she should use large angles of incidence rather than small ones.
(2)
(b)
Describe how she should use a graph of her results to find the refractive index.
(2)
(c)
For one ray, i = 50° and r = 35°. Calculate the refractive index of the liquid.
(2)
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Answer
Mark
Notes
Part (a)
the uncertainty in each angle is the same (± 0.5°)
✓ 1
so it is a smaller percentage of a large angle: e.g. 1.0% at 50° but 5.0% at 10°
✓ 1
Part (b)
plot sin i against sin r for at least five angles
✓ 1
the line should pass through the origin; the gradient is the refractive index n
✓ 1
Part (c)
n = sin 50° ÷ sin 35° = 0.766 ÷ 0.574
✓ 1
= 1.34
✓ 1
Exam tip: A gradient from several readings is better than one pair of angles: it averages out random errors in tracing and reading the rays.
Figure 1 shows how a student finds the diameter of a wooden rod with a 30 cm ruler. The rod lies across the ruler and a set square is pushed against each side of the rod, with its vertical edge standing on the ruler.
(a)
Use Figure 1 to determine the diameter of the rod.
(1)
(b)
Explain why the set squares are used.
(1)
(c)
Each position is read to ± 0.5 mm. Calculate the percentage uncertainty in the diameter.
(2)
(d)
Name a more suitable instrument and calculate the percentage uncertainty it would give. Take its uncertainty as half its resolution.
(1)
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Answer
Mark
Notes
Part (a)
50 mm − 27 mm = 23 mm (allow 22–24 mm)
✓ 1
Part (b)
the curved edge of the rod is not on the scale; the vertical edges of the set squares touch the widest points and carry their positions straight down to the scale, so they can be read without parallax
Exam tip: A length found from two scale readings has two reading uncertainties. With a ruler, each end is uncertain by half a millimetre, so the length is uncertain by 1 mm.
A student finds the density of a small stone by displacement. A measuring cylinder holds 50.0 cm3 of water. When the stone (mass 37.8 g) is lowered in, the water rises. Figure 1 shows a close-up side view, in cross-section, of the new water surface and three possible positions A, B and C of the student's eye.
(a)
State which eye position, A, B or C, should be used. Give a reason.
(1)
(b)
State the volume reading shown in Figure 1.
(1)
(c)
Explain whether a reading taken from position A would be too large or too small.
(1)
(d)
Calculate the density of the stone.
(2)
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Answer
Mark
Notes
Part (a)
B: the eye is level with the bottom of the meniscus, so the line of sight is perpendicular to the scale and there is no parallax
✓ 1
Part (b)
64.0 cm3
✓ 1
Part (c)
too large: from A the line of sight slopes down to the meniscus and crosses the scale on the front wall above the true level, where the numbers are larger
✓ 1
Part (d)
V = 64.0 − 50.0 = 14.0 cm3
✓ 1
ρ = 37.8 ÷ 14.0 = 2.70 g cm−3 (2700 kg m−3)
✓ 1
Exam tip: For water, read the bottom of the curved surface with the eye level with it. The scale is on the front wall but the lowest point of the meniscus is at the centre, so any tilt of the line of sight shifts the reading.
A student plans to investigate how the speed of a trolley at the bottom of a ramp depends on the height h from which it is released. Figure 1 shows the arrangement. A card on the trolley passes through a light gate on the bench, and the data logger shows the time for which the card blocks the beam.
(a)
State the independent and dependent variables.
(1)
(b)
State two variables that should be controlled and how each would be kept constant.
(2)
(c)
Use the data in Figure 1 to calculate the speed of the trolley at the light gate.
(1)
(d)
Explain why a light gate is better than a hand-held stopwatch for this measurement.
(1)
(e)
If no energy is dissipated, v2 = 2gh. State the graph the student should plot and the result that would confirm this relationship.
(1)
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Answer
Mark
Notes
Part (a)
independent: release height h; dependent: speed of the trolley at the light gate
✓ 1
Part (b)
the trolley's mass: use the same trolley and card with no added load
✓ 1
the ramp surface and its position / the distance of the light gate from the foot of the ramp: do not move the ramp or the gate between runs; release from rest each time without pushing
✓ 1
Part (c)
v = 0.0500 ÷ 0.0625 = 0.800 m s−1
✓ 1
Part (d)
the card blocks the beam for only about 0.06 s, far shorter than human reaction time (about 0.2 s), so a stopwatch could not time it; the light gate has no reaction time
✓ 1
Part (e)
plot v2 against h: a straight line through the origin with gradient 2g (about 19.6 m s−2)
✓ 1
Exam tip: Convert the card length to metres before dividing: 50.0 mm = 0.0500 m. A speed of 800 m s−1 for a laboratory trolley should tell you a unit has been missed.
A student checks a digital thermometer at two fixed points: a stirred mixture of ice and water (0.0 °C) and pure water boiling at standard atmospheric pressure (100.0 °C). The graph shows the readings at these two points, joined by a straight line, together with the line on which the reading would equal the true temperature.
(a)
State the zero error of the thermometer.
(1)
(b)
Explain why subtracting 1.5 °C from every reading would not correct the thermometer.
(1)
(c)
The thermometer reads 50.5 °C. Determine the true temperature.
(2)
(d)
The student uses the thermometer to measure the temperature rise of a block. The readings go from 20.7 °C to 44.7 °C. Determine the true temperature rise.
(1)
(e)
Explain why the student's correction should not be used for readings above 100 °C.
(1)
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Answer
Mark
Notes
Part (a)
+1.5 °C
✓ 1
Part (b)
at 100 °C the reading is 2.5 °C too low, not 1.5 °C too high: the error changes with temperature, so there is a scale (gradient) error as well as a zero error
✓ 1
Part (c)
gradient = (97.5 − 1.5) ÷ 100 = 0.96
✓ 1
true temperature = (50.5 − 1.5) ÷ 0.96 = 51.0 °C (allow 50–52 °C read from the graph)
✓ 1
Part (d)
rise in readings = 24.0 °C; true rise = 24.0 ÷ 0.96 = 25.0 °C
✓ 1
Part (e)
the thermometer was only checked between 0 °C and 100 °C; outside this range the line is an extrapolation and the error may not follow the same straight line
✓ 1
Common error: Assuming a zero error cancels in a temperature rise and so the rise needs no correction. The zero error does cancel, but a scale error does not: the rise must still be divided by the gradient.
A ball rolls along a gently sloping track between two marks. The time between the marks is measured seven times with a hand-held stopwatch, and then three times with two light gates connected to a timer. Figure 1 shows the readings as a dot plot, with the values listed underneath.
(a)
Identify the anomalous stopwatch reading and suggest how it could have arisen.
(2)
(b)
Calculate the mean of the remaining stopwatch readings.
(1)
(c)
Estimate the uncertainty in this mean.
(1)
(d)
Discuss whether the student needs to take more repeat readings with the light gates.
(1)
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Answer
Mark
Notes
Part (a)
1.71 s: it lies far from the other six, which are all within 1.38–1.45 s
✓ 1
the stopwatch was stopped late, e.g. the finish was missed or the button was not pressed firmly
half the range of the six readings = (1.45 − 1.38) ÷ 2 = 0.035 s (± 0.04 s)
✓ 1
Part (d)
no: the three light-gate readings agree to within 0.003 s, which is tiny compared with the time measured (about 0.1%), so further repeats would not change the mean noticeably; repeats matter for the stopwatch, where reaction time causes a large random scatter
✓ 1
Exam tip: Repeat readings are worth taking when the readings scatter. Look at the spread first: if repeats agree to the last digit shown, more of them add little.
A student uses an oscilloscope to check the frequency scale of a signal generator. Figure 1 shows the trace when the signal generator dial is set to 480 Hz, with the oscilloscope settings beside the screen.
(a)
Use Figure 1 to determine the period of the signal.
(2)
(b)
Calculate the frequency of the signal.
(1)
(c)
Explain why the student measures the length of two complete cycles rather than one.
(1)
(d)
The length of 8.0 divisions is uncertain by ± 0.2 div. Calculate the percentage uncertainty in the frequency.
(1)
(e)
Comment on the frequency scale of the signal generator.
(1)
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Answer
Mark
Notes
Part (a)
two cycles occupy 8.0 divisions, so one cycle = 4.0 div
✓ 1
T = 4.0 × 0.5 ms = 2.0 ms
✓ 1
Part (b)
f = 1 ÷ 2.0 × 10−3 = 5.0 × 102 Hz (500 Hz)
✓ 1
Part (c)
the uncertainty in judging where the trace crosses the gridlines is the same, but it is a smaller fraction of a longer length, so the percentage uncertainty is smaller
✓ 1
Part (d)
0.2 ÷ 8.0 × 100 = 2.5%
✓ 1
Part (e)
the dial reads 20 Hz (4%) below the measured 500 Hz; this is larger than the 2.5% uncertainty, so the dial is not accurately calibrated — use the oscilloscope value or make a calibration graph
✓ 1
Exam tip: Use the full width of the screen: measure across as many complete cycles as fit, between points where the trace crosses a gridline, then divide by the number of cycles.
The speed v of a long, low wave pulse in shallow water of depth d is predicted to be v = √(gd). A student has a rectangular plastic tray 0.600 m long, a supply of water and any standard laboratory equipment.
(a)
Plan an experiment to test this relationship and to determine a value for g. Your plan should include the measurements, the instruments, the control of variables, a safety point and how the results are analysed.
(6)
(b)
At a depth of 12 mm, the pulse travels four lengths of the tray in 7.0 s. Calculate the speed of the pulse and hence a value for g.
(3)
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Answer
Mark
Notes
Part (a)
Indicative content, one mark each:
—
vary the depth d over at least five values (e.g. 5 mm to 30 mm); measure the depth with a ruler/vernier depth gauge at several places in the tray and average
✓ 1
make a single pulse by quickly lifting and lowering one end of the tray (or dipping a ruler), the same way each time
✓ 1
time the pulse over several lengths of the tray (there and back) with a stopwatch, or film it with a video camera against a scale; v = total distance ÷ time
✓ 1
repeat each timing and take the mean; keep the tray level (check with a spirit level) and the water at the same temperature
✓ 1
plot v2 against d: a straight line through the origin confirms the relationship, and the gradient = g
✓ 1
safety: mop up spills at once so nobody slips, and keep electrical equipment away from the water
✓ 1
Part (b)
v = 4 × 0.600 ÷ 7.0 = 0.343 m s−1
✓ 1
g = v2 ÷ d = 0.3432 ÷ 0.012
✓ 1
= 9.8 m s−2 (9.80)
✓ 1
Exam tip: In an extended plan, cover the full checklist: what is varied and measured (with instruments), technique, control variables, repeats, safety and the graph. Each point should be specific to this experiment.
A student uses a light meter to investigate how the intensity I of the light from a small lamp depends on the distance r from the lamp. For a point source, I = P/(4πr2), where P is the power emitted as light. The meter reading, in lux, is proportional to the intensity. With the lamp switched off the meter reads 6 lux. With the lamp on, the meter reads 218 lux at r = 0.200 m.
(a)
State the graph the student should plot to test the relationship, and what its gradient represents.
(2)
(b)
Explain what the 6 lux reading represents and how the student should allow for it.
(2)
(c)
Calculate the reading the meter should show at r = 0.400 m if the relationship holds.
(2)
(d)
Suggest why the relationship may not hold when the meter is very close to the lamp.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
plot the meter reading (corrected for background) against 1/r2
✓ 1
a straight line through the origin is expected; the gradient is proportional to P/4π (it equals P/4π if the readings are converted to intensity in W m−2)
✓ 1
Part (b)
background light from the room (a zero error on every reading)
✓ 1
subtract 6 lux from every reading, or darken the room so the background reading is zero
✓ 1
Part (c)
light from the lamp at 0.200 m = 218 − 6 = 212 lux; at double the distance it is ¼ of this = 53 lux
✓ 1
meter reading = 53 + 6 = 59 lux
✓ 1
Part (d)
close up, the lamp is no longer small compared with r, so it does not behave as a point source (and the exact distance to the filament is uncertain)
✓ 1
Model sketch
Figure: expected graph: straight line through the origin
Common error: Applying the inverse-square law to the meter readings without first removing the background. Only the light from the lamp follows 1/r2; the background stays the same at every distance.
A student measures the extension of a spring when a load is added. He clamps a metre rule vertically beside the spring and reads the position of a pointer fixed to the bottom of the spring, before and after adding the load. Three problems are noticed: (1) the zero end of the metre rule is worn, so every reading is 2 mm larger than it should be; (2) he reads the pointer with his eye above it; (3) after the load is added the spring is still bouncing gently when he takes the reading.
Figure: problem (2): the eye is above the level of the pointer; not to scale
(a)
State and explain the effect of problem (1) on the extension.
(2)
(b)
Problem (2) causes a systematic error. Explain why, and state how to avoid it.
(2)
(c)
Problem (3) causes a random error. Explain why, and state how to reduce it.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
no effect on the extension
✓ 1
the extension is the difference between two readings, and both are 2 mm too large, so the error cancels
✓ 1
Part (b)
the eye is in the same wrong position for every reading, so parallax shifts every reading in the same direction
✓ 1
place the eye level with the pointer, looking perpendicular to the rule
✓ 1
Part (c)
the pointer is at a different point of its bounce each time it is read, so readings scatter either side of the rest position
✓ 1
wait for the spring to come to rest (or steady it by hand) before reading, and repeat and average
✓ 1
Exam tip: A constant error in every reading cancels when you take a difference — but it does not cancel when you use a single reading, such as a length measured from a worn end.
A toy company wants to know the stiffness of the elastic cord used in a foam-rocket launcher. A student hangs a 0.300 m length of the cord from a clamp and adds loads from 2.0 N to 10.0 N in steps of 2.0 N, measuring the extension each time. His graph of load F against extension x is a straight line through the origin, passing through the points (0.050 m, 2.4 N) and (0.200 m, 9.6 N).
Figure: the student's graph of load against extension
(a)
Describe how the student could measure each extension accurately.
(2)
(b)
Determine the stiffness of the cord from the graph.
(2)
(c)
State one variable, other than the load, that should be kept constant, and explain why.
(1)
(d)
In the launcher the cord is stretched by 0.35 m. Discuss whether the company can use the student's value of stiffness to find the force at this extension.
(2)
Show mark scheme
Answer
Mark
Notes
Part (a)
fix a pointer / horizontal pin to the bottom of the cord and read it against a vertical metre rule, checked vertical with a set square
✓ 1
eye level with the pointer; extension = loaded reading − unloaded reading
✓ 1
Part (b)
gradient = (9.6 − 2.4) ÷ (0.200 − 0.050)
✓ 1
= 48 N m−1
✓ 1
Part (c)
the original length of the cord (use the same 0.300 m piece), because a longer cord extends more for the same load / the temperature, because the stiffness of rubber depends on temperature
✓ 1
Part (d)
0.35 m is outside the range tested (the largest load, 10.0 N, gives only 10.0 ÷ 48 = 0.21 m), so the result would be an extrapolation
✓ 1
the cord may stop obeying Hooke's law at larger extensions, so the force should be measured over the full working range before relying on it
✓ 1
Exam tip: When asked to comment on a result in context, compare the conditions of the real use with the conditions actually tested. A result is only trustworthy inside the range of the data.
A student investigates a simple pendulum, for which T = 2π√(L/g), where L is the distance from the point of suspension to the centre of the bob. She measures a length ℓ from the point of suspension to the top of the bob, and the period T. The graph shows her results, plotted as T2 against ℓ, with her best-fit line.
(a)
Explain why the student plots T2 against ℓ rather than T against ℓ.
(1)
(b)
Use the graph to determine a value for g.
(3)
(c)
The line does not pass through the origin. Use the graph to explain why, and determine the diameter of the bob.
(2)
(d)
State and explain whether measuring to the top of the bob affects the value of g found from the graph.
(1)
Show mark scheme
Answer
Mark
Notes
Part (a)
T2 = (4π2/g)L, so T2 against length gives a straight line, whose gradient can be measured; T against ℓ is a curve
the line meets the ℓ axis at −0.025 m (allow −0.020 to −0.030 m): the true length L = ℓ + r, so T2 = 0 when ℓ = −r
✓ 1
r = 0.025 m, so the diameter of the bob = 0.050 m (50 mm)
✓ 1
Part (d)
no: the same length r is missing from every reading (a systematic error), which shifts the line sideways but does not change its gradient
✓ 1
Exam tip: When a graph that should pass through the origin has an intercept, ask what constant amount has been left out of every reading. The gradient is unaffected, and the intercept often tells you the size of the missing amount.
Figure 1 shows apparatus for investigating stationary waves on a wire. The frequency f of the fundamental mode of vibration between the bridges is predicted to be f = (1/2ℓ)√(T/μ), where T is the tension and μ is the mass per unit length of the wire.
(a)
Plan an experiment, using this apparatus, to test how f depends on T. Your plan should include the measurements and how they are made, the variables to be controlled, a safety precaution, and how the results are used.
(6)
(b)
With the values shown in Figure 1, the fundamental frequency is 82.0 Hz. Calculate the mass per unit length of the wire. g = 9.81 N kg−1.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
Indicative content, one mark each:
—
vary the tension by changing the mass m on the hanger over at least five values; T = mg, with the masses checked on a balance
✓ 1
keep ℓ constant (measure the distance between the bridges with a metre rule) and use the same wire, so μ is constant
✓ 1
start at a low frequency and increase it slowly until the wire vibrates with one loop of largest amplitude between the bridges; read f from the signal generator
✓ 1
repeat each value approaching resonance from above and below and take the mean; check the generator's frequency scale with an oscilloscope
✓ 1
plot f2 against T: a straight line through the origin confirms the prediction (gradient = 1/(4ℓ2μ))
✓ 1
safety: place a box of sand or foam under the hanger in case the wire snaps, keep feet clear, and wear eye protection
✓ 1
Part (b)
T = 1.50 × 9.81 = 14.7 N
✓ 1
μ = T ÷ (2lf)2 = 14.7 ÷ (2 × 0.600 × 82.0)2
✓ 1
= 1.52 × 10−3 kg m−1
✓ 1
Exam tip: An extended plan earns credit point by point: what is varied and how, what is measured and with what, what is kept constant and how, repeats, the graph and what a straight line shows, and one specific safety precaution.
A student wants the thickness of kitchen aluminium foil, which is far too thin to measure reliably in a single layer. Instead she measures a rectangular sheet with a metre rule (resolution 1 mm), folds it and finds its mass on a top-pan balance (resolution 0.01 g), as shown in Figure 1. The density of aluminium is 2700 kg m−3.
(a)
Use the data in Figure 1 to calculate the thickness of the foil.
(2)
(b)
Explain, with a calculation, why a micrometer screw gauge used on a single layer would not give a reliable value.
(2)
(c)
Taking each uncertainty as half the resolution, calculate the percentage uncertainty in the mass reading and in the 150 mm side of the sheet. Use your answers to explain why the mass method is preferred.
(2)
(d)
Suggest why the value found may still differ from the thickness at a particular point on the foil.
each measurement has a percentage uncertainty well under 1%, far smaller than the 17% for a micrometer on a single layer, so the mass method gives a much more reliable thickness
✓ 1
Part (d)
the method gives the mean thickness over the whole sheet and assumes the foil is pure aluminium of the stated density; the foil may vary in thickness from place to place
✓ 1
Exam tip: When a quantity is too small to measure directly, look for a method that turns it into larger measurements, such as mass, area or a stack of many layers, and compare the percentage uncertainties to justify the choice.
A student uses a laser and a diffraction grating with 300 lines per mm to measure the wavelength of the laser light. Figure 1 is a plan view of the arrangement. A metre rule is fixed along the screen, and the positions of the bright spots are labelled with their readings.
(a)
State one hazard of using the laser and a precaution to reduce the risk.
(2)
(b)
Explain why the student measures the distance between the two second-order spots rather than from the central spot to one first-order spot.
(2)
(c)
Use the readings in Figure 1 to calculate the wavelength of the laser light.
(3)
Show mark scheme
Answer
Mark
Notes
Part (a)
the direct or reflected beam entering the eye can damage the retina
✓ 1
never look along the beam; remove watches and other shiny objects and keep the beam below eye level / use a low-power (class 2) laser and warning signs
✓ 1
Part (b)
the distance between the second-order spots is much larger, so the same reading uncertainty gives a smaller percentage uncertainty
✓ 1
using spots on both sides means the exact position of the centre is not needed, and halving the distance averages any small tilt of the screen
✓ 1
Part (c)
distance from the centre to second order = (83.9 − 16.1) ÷ 2 = 33.9 cm; tan θ = 0.339 ÷ 0.800, θ = 22.96°
✓ 1
d = 1 ÷ 300 000 = 3.333 × 10−6 m
✓ 1
λ = d sin θ ÷ 2 = 3.333 × 10−6 × sin 22.96° ÷ 2 = 6.50 × 10−7 m (650 nm)
✓ 1
Common error: Using the full distance between the two second-order spots as the distance to one of them. Halve it first, then find θ from tan θ = distance on screen ÷ distance to the screen.
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