Free Deformation of solids questions for AS Level Physics (9702), part of Mechanics & materials. Each question has the answer and the marking points an examiner would look for.
Generate a full Deformation of Solids worksheet →All Mechanics & materials questionsA metal wire is loaded beyond its elastic limit and then unloaded.
1 The Young modulus can be found from the gradient of the stress–strain graph in the linear region.
2 The limit of proportionality always occurs after the elastic limit.
3 Work done during loading is greater than the energy released during unloading.
Which statements are correct?
Answer: B
A wire of length 2.5 m and diameter 1.0 mm has a Young modulus of 7.0 × 1010 Pa. A tension of 100 N is applied.
What is the extension of the wire?
Answer: B
What does the area under a force–extension graph represent?
Answer: B
A wire of length 2.00 m and diameter 0.60 mm is suspended vertically and a load of 80 N is hung from it. The Young modulus of the metal is 2.0 × 1011 Pa.
Define the Young modulus.
Calculate the stress in the wire.
Calculate the extension of the wire.
Use your answer to (c) to calculate the elastic potential energy stored in the wire.
Describe the difference between elastic and plastic deformation.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | stress / strain stress divided by strain (in the region where Hooke’s law applies) B1 | 1 |
| (b) | 2.83 × 108 Pa A = π(0.30 × 10−3)2 C1 stress = 2.83 × 108 Pa A1 | 2 |
| (c) | 2.83 × 10−3 m strain = stress/E C1 extension = 2.83 × 10−3 m A1 | 2 |
| (d) | 0.113 J E = ½Fx C1 0.113 J A1 | 2 |
| (e) | elastic: returns to original shape when load removed; plastic: permanent extension elastic: the material returns to its original length when the load is removed B1 plastic: permanent deformation remains B1 | 2 |
A wire of length 2.8 m and diameter 0.51 mm extends by 1 mm when a tensile force of 112 N is applied. Calculate the Young modulus of the material.
| Part | Answer and marking points | Marks |
|---|---|---|
| 1540000000000 Pa A = πd2/4 C1 stress = F/A C1 strain = e/L C1 1540000000000 A1 | 4 |
A wire of length 1.5 m and diameter 0.5 mm is made of a metal with Young modulus 2.0 × 1011 Pa. A load of 80 N is hung from it.
Calculate the stress in the wire.
Calculate the extension of the wire.
Calculate the elastic potential energy stored.
| Part | Answer and marking points | Marks |
|---|---|---|
| (a) | 4.07 × 108 Pa A = πd2/4 C1 4.07 × 108 Pa A1 | 2 |
| (b) | 3.06 × 10−3 m strain = stress/E, x = strain × L C1 3.06 × 10−3 m A1 | 2 |
| (c) | 0.122 J ½Fx C1 0.122 J A1 | 2 |
Riyaz Shaikh has mentored IGCSE, IB and A Level students for 15+ years, one-on-one, online and in Mumbai. The hardest questions on this site are the ones he uses to find a student's weak points.
Meet Riyaz →Original Cambridge-style practice questions written for mentorriyaz.in. These are not Cambridge past paper questions, and this site is not affiliated with or endorsed by Cambridge University Press & Assessment. Updated 2026-11-04. Paper generator · mentorriyaz.in