AQA A-Level Physics Paper 3 (3A), June 2025: Question 4

📐 Step-by-Step Calculation

1. Read the vernier scale at Q:
   Main scale reading = 24.5 mm (or 24.7 mm depending on alignment). Looking closely at Figure 11:
   The vernier zero is past 24.5 mm (at ~24.7 mm).
   Vernier coincident mark = 3 tenths, so reading x = 24.73 mm .
2. Count the number of turns between P and Q:
   Counting coils from point P to point Q in the enlarged view gives n = 13 turns .
3. Calculate thickness (t):
   t = (24.73 - 9.26) / 13 = 15.47 / 13 = 1.19 mm
   This demonstrates approximately 1.2 mm .

✅ Mark Scheme Breakdown

• Mark 1: Correct method expression:
  (24.73 - 9.26) / n where n is an integer OR (x - 9.26) / 13 where x is their vernier reading (must be to 2 d.p.). Allow 15.47 / n .
• Mark 2: Correct final value of t = 1.19 mm (contingent on showing working; accept >3 sf that rounds to 1.19).

❌ Common Errors

• Misreading the vernier scale tenths: reading 24.23 mm or forgetting to add the vernier coincidence to the main scale.
• Miscounting the turns: count spaces or ridges carefully between P and Q (exactly 13 turns).
• Failing to show a value more precise than "1.2 mm" in a "Show that" question. You must show at least 3 s.f. (e.g. 1.19 mm ) to prove you didn't work backwards.

🧠 Exam Technique

In "Show that approximately [value]" questions, examiners require you to quote your unrounded calculation to at least one more significant figure than given in the prompt, alongside full algebraic or numerical working.

📐 Step-by-Step Calculation

1. Determine effective coil diameter:
   Taking the centre of the wire:
   D_eff = D_tube + t = 18.0 + 1.19 = 19.19 mm
   (Mark scheme also allows taking the outer diameter: 18.0 + 2t = 20.38 mm , or simple tube diameter 18.0 mm ).
2. Calculate circumference per turn:
   C = π × D_eff = π × 19.19 = 60.29 mm
3. Total length (L) for 13 turns:
   L = n × C = 13 × 60.29 ≈ 784 mm
   (If using 18.0 mm: L ≈ 735 mm ; if using 18 + 2t : L ≈ 832 mm ).

✅ Mark Scheme Breakdown

• Mark 1: Effective diameter: 18.0 + (2 × t) or 18.0 + t (accept t = 1.2 mm ).
• Mark 2: Circumference formula: π × diameter .
• Mark 3: Total length = n × circumference (using their n from 04.1). Must be given to 3 s.f. if a whole number of mm.

🧠 Exam Technique: Significant Figures

The mark scheme explicitly specifies that the final answer must be given to 3 significant figures (e.g. 784 mm ). Giving 783.7 mm loses the final mark!

💡 Key Knowledge: Coil Geometry

When wire wraps around a cylinder of diameter D , the axis of the wire lies on a cylinder of diameter D + t . Hence, one single turn has length π(D + t) .

✅ Correct Identification & Justification

Position: B

Explanation: Position B marks the equilibrium position (centre of oscillation). At this point, the test tube has maximum speed / maximum kinetic energy, meaning the transit time past the marker is at a minimum, which minimises human reaction-time error in starting/stopping the stopwatch.

🧠 Marking Guidance Notes

• Mark 1: Identifies B (must reject A and D; C only accepted under strict condition of an added mark on the tube at equilibrium).
• Mark 2: State that the tube is moving fastest / has maximum kinetic energy / transit time is least.
• Note: Simply stating "it is the equilibrium position" or "it is the centre of oscillation" is neutral and gains NO explanation mark. You must refer to speed / velocity / kinetic energy.

📐 Step-by-Step Calculation

1. Establish the relationship:
   T ∝ 1 / √ρ ==> T × √ρ = constant ==> T² × ρ = constant
2. Equate oil and milk systems:
   T_oil² × ρ_oil = T_milk² × ρ_milk
3. Rearrange for density of milk:
   ρ_milk = ρ_oil × (T_oil / T_milk)²
4. Substitute values:
   ρ_milk = (8.49 × 10²) × (0.587 / 0.534)²
   ρ_milk = 849 × (1.09925)² = 849 × 1.20835 ≈ 1025.9 kg m⁻³
5. Round to 3 significant figures:
   ρ_milk = 1.03 × 10³ kg m⁻³ (or 1030 kg m⁻³ ; allow 4 sf: 1026 kg m⁻³).

✅ Mark Scheme Breakdown

• Mark 1: Valid algebraic expression or intermediate evaluation:
  e.g., ρ_milk = ρ_oil × (T_oil² / T_milk²) OR calculates T²ρ = 0.587² × 849 = 293 OR T√ρ = 0.587 × √849 = 17.1 .
• Mark 2: Final answer 1.03 × 10³ kg m⁻³ (must be 3 s.f., though range 1025 to 1030 condoned).

💡 Understanding the Physics

To increase the number of oscillations before motion ceases, we need more energy in the system initially or less energy loss per cycle (reduced drag).

Crucially, the prompt states: without changing T. Therefore, you CANNOT change the liquid, change the test tube diameter, or change the mass of the tube.

✅ Acceptable Modifications & Rationales (Max 3 Marks)

Award marks for pairs of [Modification + Matching Rationale]:

• Option A:
  Modification: Increase depth of liquid / add more oil to beaker.
  Rationale: Allows a larger initial displacement / larger amplitude / provides more initial energy.
• Option B:
  Modification: Use a wider diameter beaker (increase distance between tube and beaker walls).
  Rationale: Reduces viscous drag / fluid friction / velocity gradient between tube and walls.

❌ Disallowed / Neutral Answers

• "Use a narrower beaker": Incorrect — this increases drag / viscous shear, increasing damping.
• "Use a less viscous liquid / change liquid": The question specifically specifies "for a particular liquid".
• "Add more mass to the tube": Changes T (violates question constraint).
• Vague statements: "Use a bigger beaker" is neutral without specifying wider/larger diameter. Simply stating "to reduce damping" is neutral for the rationale mark.

🧠 Exam Technique: State and Explain

Always link your proposed physical action directly to an underlying physical mechanism (e.g. initial potential energy or fluid shear/drag resistance). Make sure your suggestion does not inadvertently alter variables held constant by the question stem.