AQA A-Level Physics: Estimation of Physical Quantities — Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Estimation of Physical Quantities study guide

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Section A

1. Explain what it means for two quantities to be described as of the same order of magnitude. [2]

2. A calculation gives a result of 3.4 × 10⁵. State the order of magnitude of this result. [1]

Section B

3. Estimate, to the nearest order of magnitude, the diameter of an atom in metres, stating the piece of general physics knowledge you used. [2]

4. Estimate, to the nearest order of magnitude, the number of seconds in one day. Show your working. [2]

5. Estimate, to the nearest order of magnitude, the volume of air in an average classroom in m³, stating the dimensions you assumed. [3]

6. A person walks at an estimated speed of 1 m s⁻¹ for an estimated 30 minutes. Estimate, to the nearest order of magnitude, the distance walked, showing how the two estimates are combined. [3]

7. Estimate the number of water molecules in a 250 ml glass of water, to the nearest order of magnitude. You may use: molar mass of water ≈ 18 g mol⁻¹, the Avogadro constant ≈ 6 × 10²³ mol⁻¹, and density of water ≈ 1000 kg m⁻³. [4]

8. A student calculates the kinetic energy of a car travelling on a motorway and obtains 4.5 × 10¹² J. Using order-of-magnitude estimates for a plausible mass and speed, explain why this answer must be wrong. [4]

9. A student estimates that a stadium measuring roughly 50 m by 50 m, with each spectator needing about 0.5 m², can hold 5 × 10² spectators. Identify the error in this estimate and give a more realistic order-of-magnitude answer. [3]

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Answers

1. Two quantities are of the same order of magnitude if they are within a factor of about 3 of the same power of ten [1] — that is, judging closeness by ratio rather than by numerical difference, the power of ten closest to each quantity’s value is the same power [1].

2. 10⁶ [1]. (Order of magnitude is judged by ratio, on a logarithmic scale: the crossover between 10⁵ and 10⁶ is at √10 × 10⁵ ≈ 3.16 × 10⁵, and 3.4 × 10⁵ lies above it.)

3. ≈ 10⁻¹⁰ m [1]. This uses the general physics knowledge that atoms are of the order of a tenth of a nanometre across, far smaller than anything visible even under an optical microscope [1].

4. 24 hours × 60 minutes × 60 seconds = 86 400 s [1], which rounds to the nearest order of magnitude as 10⁵ s [1] — a well-known Fermi estimate: “a day is about 10⁵ seconds.”

5. Reasonable assumed dimensions: roughly 10 m × 8 m × 3 m high [1]. Volume ≈ 10 × 8 × 3 = 240 m³ [1], which is order of magnitude 10² m³ [1]. (Any dimensions of the right rough scale, correctly multiplied and rounded, gain full credit.)

6. Speed = 1 m s⁻¹ is order 10⁰ m s⁻¹; time = 30 minutes = 1800 s is order 10³ s [1] [1]. Distance = speed × time ≈ 1 × 1800 = 1800 m, which is order of magnitude 10³ m [1] — combining two order-of-magnitude estimates by multiplying them together, exactly as the derived-estimate technique requires.

7. Mass of water: 250 ml = 2.5 × 10⁻⁴ m³, so mass = volume × density = 2.5 × 10⁻⁴ × 1000 = 0.25 kg = 250 g [1]. Moles = 250 ÷ 18 ≈ 13.9 mol [1]. Number of molecules = 13.9 × 6 × 10²³ ≈ 8.3 × 10²⁴ [1], which rounds to the nearest order of magnitude as 10²⁵ molecules [1] — a classic derived estimate, since it combines a volume, a density, a molar mass and the Avogadro constant, none of which alone gives the answer.

8. A reasonable estimate for a car’s mass is ≈ 10³ kg, and a reasonable motorway speed is ≈ 30 m s⁻¹ [1] [1]. Kinetic energy = ½mv² ≈ ½ × 10³ × 30² ≈ 4.5 × 10⁵ J, order of magnitude 10⁶ J [1]. The student’s answer of 4.5 × 10¹² J is 10⁷ times this, about seven orders of magnitude too large to be a plausible car’s kinetic energy, so it must contain an error [1] — exactly the kind of sanity check order-of-magnitude reasoning is used for.

9. The correct calculation is area ÷ area-per-spectator = (50 × 50) ÷ 0.5 = 2500 ÷ 0.5 = 5000 [1], which is 5 × 10³ and so of order of magnitude 10⁴ (5 lies above the crossover at √10 ≈ 3.16) [1]. The student’s answer of 5 × 10² (order 10³) is a factor of ten, one order of magnitude, too small — most likely from a slip of a factor of ten somewhere in the arithmetic [1]. A more realistic estimate is therefore about 5 × 10³ spectators, of order 10⁴.

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Where marks are usually lost

• Giving an estimate as a long string of exact-looking decimal digits rather than as a single order of magnitude (a power of ten).
• Forgetting to convert all quantities to consistent units before combining several estimates, especially mixing minutes/hours with seconds.
• Treating “order of magnitude” as meaning “round to one significant figure” rather than “the nearest power of ten”, judged by ratio (a leading figure above √10 ≈ 3.16 rounds up to the next power of ten).
• Not stating the assumption or piece of general knowledge behind an estimate, which is where the method marks are awarded.
• Failing to use a quick order-of-magnitude check to catch a numerical answer that is wrong by several powers of ten.