This set covers all five skills in the module: separating a hypothesis from its plan, spotting confounders, comparing repeats with an independent method, stating what data can support, and proposing safe improvements. The questions run from easier to harder, and the data are invented.
Work through the questions in order with a pencil. Then open each answer and compare. If you are unsure of the method, read the matching lesson first, starting with separating a hypothesis from its measurement plan.
Questions
1. A student writes: “Plants grown in red light will grow taller than plants grown in blue light.” Is this a hypothesis or part of a plan?
Show answer
It is a hypothesis. It makes a claim that links light colour and plant height, and data could contradict it. The choice of lamp, the number of plants and the measuring ruler belong in the plan.
2. A student investigates whether the height of a ramp affects how far a trolley rolls along the floor. Name the independent variable, the dependent variable (with a unit) and two control variables.
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Independent: ramp height (cm). Dependent: distance rolled (m). Controls: trolley mass, floor surface, release point, ramp surface (any two).
3. A student compares two paper towels. Brand X uses a 10 cm × 10 cm sheet soaked for 10 s. Brand Y uses a 15 cm × 15 cm sheet soaked for 30 s. Find the area of each sheet and name two confounding variables.
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Area X = 10 × 10 = 100 cm². Area Y = 15 × 15 = 225 cm². Confounders: sheet size and soaking time. Both could change the mass of water absorbed. Use the same size and time for both brands.
4. The time for water to cool from 60 °C to 50 °C was measured three times with a lid: 410 s, 425 s and 415 s. Without a lid the mean time was 305 s. Calculate the mean time with a lid, the range, and the difference between the means.
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Mean with lid = (410 + 425 + 415) ÷ 3 = 1250 ÷ 3 = 416.7 s, so 417 s (3 s.f.). Range = 425 − 410 = 15 s. Difference = 417 − 305 = 112 s, which is much larger than the 15 s range, so the lid has a clear effect in this setup.
5. A pendulum of length 0.50 m takes 28.4 s (mean) for 20 oscillations. Calculate the period, then g using g = 4π²L ÷ T². Give g to 2 significant figures.
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T = 28.4 ÷ 20 = 1.42 s. T² = 2.0164. 4π² = 39.478. g = 39.478 × 0.50 ÷ 2.0164 = 19.739 ÷ 2.0164 = 9.789, so 9.8 m/s².
6. Why was the pendulum timed for 20 oscillations rather than one? Use a human reaction time of 0.2 s to support your answer.
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A reaction time of 0.2 s is 0.2 ÷ 1.42 = 14% of a single period but only 0.2 ÷ 28.4 = 0.70% of 20 periods. Timing many oscillations makes the timing error a much smaller fraction of the measured time.
7. A balance reads 1.5 g too high because it was not set to zero. A student weighs a sample five times and gets values that agree within 0.1 g. Does this show the mass is accurate? What would reveal the problem?
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No. The agreement shows precision only. Every reading has the same offset. Checking against a known mass or a second balance would reveal it, which is an independent check.
8. Pondweed bubble counts (invented) were: 10 cm from the lamp, 42 bubbles per minute; 20 cm, 21; 30 cm, 14. Calculate distance × bubble count for each, and say which conclusion is supported: (a) the count fell as distance increased between 10 and 30 cm; (b) the count at 5 cm would be 84; (c) the bubbles prove that oxygen is made in the leaf.
Show answer
10 × 42 = 420, 20 × 21 = 420, 30 × 14 = 420. The product is constant, so the count is inversely proportional to distance in this range. (a) is supported. (b) is extrapolation outside 10 to 30 cm. (c) is an explanation about gas identity that counting bubbles does not test.
9. A ruler reads to ±0.1 cm. Calculate the percentage uncertainty when measuring a length of 2.0 cm and of 20.0 cm. What does this suggest about choosing a measurement?
Show answer
0.1 ÷ 2.0 = 5%. 0.1 ÷ 20.0 = 0.5%. A larger measured quantity gives a smaller percentage uncertainty, so measure a longer length where possible, for example a stack of identical items.
10. Four seedling heights are 12.1 cm, 12.4 cm, 15.9 cm and 12.2 cm. Identify the anomalous value, calculate the mean without it, and say what to do next.
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15.9 cm stands apart from the others. Mean without it = (12.1 + 12.4 + 12.2) ÷ 3 = 36.7 ÷ 3 = 12.2 cm. Do not simply delete it. Check the record for a mistake, and repeat the measurement if possible.
11. Water temperature fell by 3 °C during each run. A student suggests “use a digital stopwatch”. Is this connected to the limitation? Suggest a better improvement.
Show answer
Not connected. The stopwatch does not stop the temperature falling. A better improvement is an insulated container or a supervised warm-water bath, and recording the temperature at the start and end of each run.
12. A student wants a bigger temperature range and suggests using a very hot heater under a flask of liquid. Rewrite this as a safe improvement.
Show answer
Use warm water in a supervised water bath within the temperature range the teacher has approved, and add one more value inside that range. The aim of a wider range does not justify a higher hazard.
If you got these wrong
| What went wrong | Go to |
|---|---|
| Mixing up the claim and the method (Q1, Q2) | Hypothesis and measurement plan |
| Missing a difference between groups (Q3) | Confounding variables |
| Treating close repeats as proof of accuracy (Q4, Q5, Q6, Q7, Q9) | Repeat and independent method |
| Claiming more than the data show (Q8, Q10) | What a dataset can and cannot establish |
| Improvements that miss the limitation or ignore safety (Q11, Q12) | Safe improvement for a real limitation |
After the set, record each error with a short note in the mistake log and retest queue, and use the scientific investigation critic on one of your own practicals. Return to the module overview to plan another round.
If the same two or three slips keep returning, online one-to-one Co-ordinated Sciences tuition lets a teacher watch your reasoning on fresh questions and adjust the explanation to how you think.