AQA AS Level Physics Paper 2, June 2024: Question 2
8 marks · Medium difficulty · Short Answer
Determine the force applied in a Brinell hardness test using graph data and explain advantages of measuring indentation diameter over depth.
Practise this questionQuestion
Question text
02 The Brinell test determines the hardness of the surface of a material.
Figure 7 shows a steel sphere on the surface of a material being tested.
Figure 7
In the test, a load F is applied to a steel sphere of diameter D and an indentation of
depth h is produced in the material. Figure 8 shows one test.
Figure 8
The Brinell hardness number B is given by
F
B =
πg Dh
where F is in N, g is in N kg−1 and D and h are in mm.
The unit of B is kg mm−2.
Using the same steel sphere, the value of h was measured for five materials.
B was calculated for each material.
For each material:
• F was the same 11
• D = 10.0 mm.
Figure 9 is a plot of B against h.
Figure 9
02.1 Determine the value of F that was used to produce Figure 9.
[1 mark]
F = N
02.2 Brass was not one of the five materials tested.
When brass was tested using these values of F and D, the value of h = 1.60 mm.
Determine, using Figure 9, B for brass.
[2 marks]
B for brass = kg mm−2
02.3 B for lead is about 5 kg mm−2.
Show that this result cannot be obtained with the steel sphere and the value of F used
to produce Figure 9.
Go on to suggest how the test can be modified to determine B for lead.
[2 marks]
The Brinell hardness number can be determined by measuring the diameter d of the
circular indentation rather than h.
Figure 10 shows d.
Figure 10
For the indentation created in brass, d = 7.33 mm.
02.4 Suggest a suitable instrument that could have been used to measure this value of d.
[1 mark]
02.5 For the indentation created in brass, h = 1.60 mm.
Explain one advantage of finding B by measuring d rather than h.
[2 marks]
END OF SECTION A
Section B
Answer all questions in this section.
Mark scheme
Show the mark scheme
Question Answers Additional Comments/Guidance Mark AO
02.1 value in range 2.9 × 104 to 3.0 × 104 (N) Use of data from any point (plotted or using their 1 AO2
line or using their B for brass) is acceptable
02.2 smooth curve through at least 4 saltires 1a 1a Reject thick or discontinuous lines 2 1 × AO1
1a can be awarded if no credit gained in 1b or 1 × AO2
2b
correct read off at 1.60 mm, leading to answer in range 58 to 2a 2 or 3 sf values only
64 (kg mm−2)
2a
OR
1b Condone use of D and h in metres if also seen
𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐹𝐹
use of 𝐵𝐵 = 1b (and penalised) in 02.1
𝜋𝜋×𝑔𝑔×10×1.6
B 2b 2 or 3 sf values only
consistent calculation of 2b
𝑡𝑡ℎ𝑒𝑒𝑒𝑒𝑒𝑒 𝐹𝐹
2b Their B should be
02.3 𝐹𝐹 2 92 × AO3
uses 𝐵𝐵 = to: 1 Expect h = 19 mm
𝜋𝜋𝑔𝑔𝜋𝜋ℎ
evaluate h, and compare to radius/diameter of steel sphere 1 Condone ‘steel ball will be completely pushed
into the lead’ for comparison
OR
evaluate (minimum value of) B based on radius/diameter of 1 Reject references to graph scale e.g. ‘h scale
steel sphere, and compare to 5 (kg mm–2) only goes up to 3.5 mm on graph’
reduce F
OR 2 Condone ‘use a steel sphere with D > 19 mm’ or
increase D 2 ‘use a bigger sphere’.
02.4 travelling microscope 1 AO1
OR
micrometer / screw gauge
OR
digital vernier calliper
02.5 1 mark for an advantage AND 1 mark for a relevant 2 2 × AO3
explanation. No credit for an explanation without
d is (always) larger (than h) 1a the relevant advantage.
so percentage / % uncertainty is smaller 1b h
Allow reverse arguments throughout e.g. ‘ is
(always) smaller than d ‘
OR
2a Allow ‘can take multiple readings of d’ or ‘h can
d only be measured once’
can be measured in different directions 2a
2b Allow ‘can identify anomalous readings’ or ‘can
so can obtain an average 2b
reduce the effect of random error’
OR
idea that readings for d are clearer to judge (than for h) 3a 3a Allow ‘difficult to see where the centre of
indentation is for h’ or wtte.
so measurement is closer to true value / more accurate 3b
3a Allow ‘easier to define d’. Reject ‘easier to
measure d ’.
3b Allow idea that parallax error can be reduced.
Total 8
How to answer it
Materials Testing: The Brinell Hardness Test
What this question tests
This question assesses your ability to apply a non-standard algebraic formula involving mechanics and materials ( B = F / (pi * g * D * h) ), interpret graphical data, evaluate experimental limitations using physical constraints, select appropriate measuring instruments for macroscopic dimensions, and critically compare experimental techniques regarding uncertainty and precision.
Determine the value of F
✅ Correct Answer
Value in the range 2.9 x 10⁴ to 3.0 x 10⁴ N
📐 Calculation Method
Rearrange the given formula for force: F = B * pi * g * D * h .
Take coordinates from any plotted point on Figure 9 (e.g., at h = 1.0 mm , B = 100 kg mm⁻² , taking g = 9.81 N kg⁻¹ and D = 10.0 mm ).
❌ Common Errors
Forgetting to convert units or misreading coordinate values from the grid lines on Figure 9.
Determine B for Brass
✅ Correct Answer
A value in the range 58 to 64 kg mm⁻²
💡 Key Knowledge
You can either interpolate directly from a smooth trend line drawn through the data points at h = 1.60 mm , or calculate B directly using the constant force F found in 02.1.
🧠 Exam Technique
Ensure any drawn curve is smooth and passes through at least 4 of the plotted crosses without using thick, disjointed sketching lines.
Evaluating Lead Testing Limitations
✅ Correct Answer
Part 1 (The problem): For lead, h evaluates to approximately 19 mm (using B = 5 ), which is greater than or comparable to the steel sphere diameter D = 10 mm (the ball would sink entirely into the lead).
Part 2 (The modification): Reduce the applied load F or increase the sphere diameter D .
❌ Common Errors
Stating vaguely that "the graph scale only goes up to 3.5 mm". The physical limitation is that the indentation depth exceeds the sphere size, making the formula invalid.
Selecting a Measuring Instrument
✅ Correct Answer
Any one of:
- Travelling microscope
- Micrometer / screw gauge
- Digital vernier calliper
❌ Common Errors
Naming standard rulers or metre rules, which lack the sub-millimetre resolution required to measure indentation diameters accurately.
Advantages of Measuring Diameter over Depth
✅ Correct Answer
Advantage 1: The indentation diameter d is always larger than the depth h .
Advantage 2 (The link): Therefore, the percentage uncertainty in measuring d is smaller than that for h .
💡 Alternative Valid Points
- d can be measured in multiple directions to find an average, reducing random error.
- The edges of the circular indentation are clearer to judge visually than depth, reducing parallax and systematic errors.
Topics
Physics · Practical skills · 3.4 Mechanics and materials · Data analysis · Uncertainty and evaluation
Question and mark scheme from the AQA AS Level Physics examination, Paper 2, June 2024. QuestionVault is an independent revision resource; questions remain the copyright of the awarding body.