Measurements and their Errors
6 questions, 57 marks, every one with its mark scheme.
Or open them one at a time, as you finish each question.
Question 01
9 marksWhich unit is NOT an SI base unit?
Tick (✓) one box.
- ampere
- kelvin
- mole
- volt
Mark scheme
- volt1
Which length is equal to 2.5 mm?
Tick (✓) one box.
- 2.5 × 105 nm
- 2.5 × 103 μm
- 2.5 × 10−2 cm
- 2.5 × 108 pm
Mark scheme
- 2.5 × 103 μm1
An electric kettle transfers 0.12 kW h of energy to the water in it.
Calculate this energy in J.
Mark scheme
- 0.12 × 1000 × 36001
- 4.3 × 105 J1
The air resistance F on a parachutist falling at speed v is given by
F = kv2
where k is a constant.
Determine the SI base units of k.
Mark scheme
- newton written as kg m s−21
- kg m−11
Do not accept: N s2 m−2
A spherical raindrop has a diameter of 2.0 mm. The density of water is 1000 kg m−3.
Calculate the mass of the raindrop and hence state the order of magnitude of its mass in kg.
volume of a sphere = 4/3πr3
Mark scheme
- mass = 4.2 × 10−6 kg1
- order of magnitude 10−6 (kg)1
Do not accept: 4 × 10−6 as the order of magnitude; 1.0 × 10−6 as the order of magnitude
A student plots a graph of the kinetic energy of a trolley, in J, on the y-axis against the mass of the trolley, in kg, on the x-axis.
State the SI unit of the gradient of the graph.
Mark scheme
- J kg−11
Do not accept: J / kg
Question 02
11 marksA student is asked to determine the diameter of a nylon fishing line. The diameter is about 0.4 mm.
Name the instrument the student should use.
Mark scheme
- micrometer screw gauge1
Do not accept: vernier callipers
Explain why this instrument is more suitable than a 30 cm ruler for this measurement.
Mark scheme
- the micrometer has a smaller resolution than a ruler1
- so its readings have a smaller percentage uncertainty1
Do not accept: resolution is better
Before taking any readings, the student closes the jaws of the instrument with nothing between them. The reading is 0.02 mm.
Name the type of error this shows and state how the student should correct the readings.
Mark scheme
- systematic error1
- subtract zero error from each reading1
Explain why the student should use the ratchet when closing the jaws on the line.
Mark scheme
- tightening the thimble directly can change the diameter1
- giving a reading smaller than true value1
Describe how the student should take readings of the diameter to reduce the effect of random error.
Mark scheme
Any 3 from:
- repeat measurements at different points along the line1
- repeat measurements in different directions1
- reject anomalous readings before calculating the mean1
- calculate a mean1
Do not accept: repeats eliminate random error
Which statement describes the accuracy of a measurement?
Tick (✓) one box.
- How close the measurement is to the true value
- How little repeated measurements scatter about their mean
- The same results are obtained when the original experimenter repeats the measurement
- The same results are obtained when a different person repeats the measurement
Mark scheme
- How close the measurement is to the true value1
Question 03
12 marksA student investigates water dripping from a burette tap. She uses a stopwatch to measure the time taken for 20 drops to fall. She repeats the measurement five times.
| Reading | 1 | 2 | 3 | 4 | 5 |
|---|---|---|---|---|---|
| Time for 20 drops / s | 15.4 | 15.8 | 15.5 | 17.9 | 15.7 |
The student also reads the level of the water on the burette scale before and after the drops are collected.
State where the student's eye should be when reading the scale, and give a reason.
Mark scheme
- eye level with the bottom of the meniscus1
- to avoid parallax error1
Do not accept: measure at eye level, without the bottom of the meniscus
State what the student should do with the reading of 17.9 s before calculating a mean.
Mark scheme
- reject anomalous reading before calculating the mean1
Calculate the mean time for 20 drops and the percentage uncertainty in this mean.
Give the percentage uncertainty to 2 significant figures.
Mark scheme
- mean = 15.6 s1
- uncertainty = half the range = 0.2 s1
- percentage uncertainty = 0.2 / 15.6 × 100 = 1.3 %1
Do not accept: 0.013
Determine the time interval between two drops, together with its absolute uncertainty.
Mark scheme
- 15.6 ÷ 20 = 0.780 s1
- ± 0.010 s1
Explain why timing 20 drops, rather than a single drop, gives a smaller percentage uncertainty in the time interval between drops.
Mark scheme
- absolute uncertainty is the same but value is larger1
- so the percentage uncertainty is smaller1
The student replaces the stopwatch with a light gate connected to a data logger.
Give two advantages of using a data logger.
Mark scheme
- reduces impact of statistical error in reading and recording data manually1
- data can be collected at a high rate1
Question 04
11 marksA trolley moves along a level runway between two marks. A student measures the distance s between the marks with a metre rule and the time t for the trolley to travel between them.
| Quantity | Value | Absolute uncertainty |
|---|---|---|
| s | 1.200 m | ± 0.002 m |
| t | 0.85 s | ± 0.02 s |
The position of each mark can be located to within ± 1 mm.
Explain why s has an absolute uncertainty of ± 2 mm.
Mark scheme
- s is found from two readings (one at each mark)1
- uncertainties in each reading are added, so absolute uncertainty = 2 × uncertainty in each reading1
Do not accept: because the smallest division is 1 mm
The trolley moves at constant speed v = s / t.
Calculate the percentage uncertainty in v.
Mark scheme
- percentage uncertainty in s = 0.17 %1
- percentage uncertainty in t = 2.4 %1
- percentage uncertainty in v = 0.17 + 2.4 = 2.5 %1
Do not accept: the mean of two separate percentage uncertainties
The mass of the trolley is 0.250 kg ± 0.4 %.
Calculate the kinetic energy of the trolley and its absolute uncertainty.
Give your answer in the form value ± uncertainty.
Mark scheme
- Ek = ½ × 0.250 × (1.200 / 0.85)2 = 0.249 J1
- 2 × % uncertainty in v = 5.0 %1
- percentage uncertainty in Ek = 0.4 + 5.0 = 5.4 %1
- absolute uncertainty = 0.014 J, so Ek = 0.25 ± 0.01 J1
Suggest how the student could reduce the percentage uncertainty in s, and explain why this works.
Mark scheme
- increase distance between the marks1
- same absolute uncertainty is a smaller fraction of a larger value, decreasing the percentage uncertainty1
Do not accept: repeat and average to reduce the percentage uncertainty in a distance; a ruler with smaller divisions to reduce the percentage uncertainty in a distance
Question 05
8 marksA student plots a graph of a time y against a length x for six sets of readings. She adds error bars to every point and draws a straight line of best fit passing through all error bars.
The absolute uncertainty in each value of y is ± 0.05 s.
State the total length, in s, of each error bar on the graph.
Mark scheme
- 0.10 s1
Describe how the student should draw the line of steepest acceptable gradient.
Mark scheme
- ruled through bottom of first error bar1
- through top of last error bar, passing through all error bars1
The steepest acceptable line has a gradient of 2.46 s m−1. The shallowest acceptable line has a gradient of 2.18 s m−1.
Determine the best value of the gradient and its percentage uncertainty.
Mark scheme
- best gradient = (2.46 + 2.18) / 2 = 2.32 s m−11
- ΔG = (2.46 − 2.18) / 2 = 0.14 s m−11
- percentage uncertainty = 0.14 / 2.32 × 100 = 6.0 %1
In a second experiment every value of y has a percentage uncertainty of 2 %. The first value of y is 1.50 s and the last value is 6.00 s.
Compare the lengths of the error bars on these two points.
Mark scheme
- the error bar on the last point is 4 times longer than on the first1
The student uses the gradient to calculate a value for the acceleration due to gravity of 9.62 m s−2 with an absolute uncertainty of 0.58 m s−2.
Write the result with a number of significant figures consistent with its uncertainty.
Mark scheme
- 9.6 ± 0.6 m s−21
Question 06
6 marksA student is given a reel of thin metal wire. She needs a value for the cross-sectional area A of the wire and its percentage uncertainty.
Describe how she should measure the diameter of the wire, how she should minimise random errors and systematic errors, and how she should determine A and the percentage uncertainty in A.
Mark scheme
| Level | Marks | Description |
|---|---|---|
| 3 | 5-6 | All three areas (measurement and systematic error, random error, processing) are covered in detail. The method is clear and logically ordered, and includes both the calculation of A and the doubling of the percentage uncertainty in d. |
| 2 | 3-4 | Two areas are covered in detail, or all three areas are covered with some omissions. The method is mostly clear but may lack a step such as the zero-error correction or the factor of 2 in the uncertainty. |
| 1 | 1-2 | One area is covered, or a few relevant points from different areas are given without a coherent method. |
| 0 | No relevant content |
Indicative content
- Measurement: micrometer screw gauge
- close jaws using ratchet with nothing between them to find the zero error
- subtract zero error from each reading
- use of ratchet, so the diameter of the wire is not changed
- Random error: repeat measurements at different points along the wire
- repeat measurements in different directions
- reject anomalous readings before calculating the mean
- calculate a mean diameter d
- Processing: uncertainty in d = half the range
- percentage uncertainty in d = ½ range / mean × 100
- A = πd2/4
- percentage uncertainty in A = 2 × % uncertainty in d