AS Physics: Deformation of Solids — Practice Questions

These are original questions written for Marlbridge, in the style and at the standard of the examination. They are not reproduced past-paper questions — examination boards hold copyright in their own papers. Use these alongside the official past papers available free from your board.

Related: Deformation of Solids revision notes

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Section A

1. State Hooke’s law in full. [2]

2. Distinguish between elastic and plastic deformation. [2]

3. Distinguish between the terms strong, stiff and tough. [3]

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Section B

4. A wire of original length 2.50 m and diameter 0.42 mm is stretched by a load of 45 N, producing an extension of 3.6 mm.

(a) Calculate the cross-sectional area of the wire. [2]

(b) Calculate the stress. [2]

(c) Calculate the strain. [2]

(d) Calculate the Young modulus. [2]

(e) A second wire of the same material but twice the diameter is used. State, with a reason, the value of its Young modulus. [2]

5. A student determines the Young modulus of a metal using a long thin wire.

(a) Explain why a long wire is used. [2]

(b) Explain why a thin wire is used. [2]

(c) The student measures the diameter at three points along the wire and takes a mean. Explain why. [2]

(d) State which graph should be plotted and what its gradient represents. [2]

6. A spring obeys Hooke’s law with spring constant 25 N m⁻¹.

(a) Calculate the energy stored when it is extended by 0.12 m. [2]

(b) A rubber band is stretched and released. Its loading and unloading curves differ. State what the area between them represents. [2]

7. A wire of original length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² extends by 1.5 mm under a load of 60 N.

(a) Calculate the stress, strain and Young modulus. [3]

(b) Distinguish between the limit of proportionality and the elastic limit. [2]

8. A student sets up the standard experiment to determine the Young modulus of a wire.

(a) Describe how a reference wire, mounted alongside the test wire, improves the experiment. [2]

(b) Explain why the force–extension graph must remain a straight line for Eₚ = ½Fx to be valid, and what must be done instead if it is not. [2]

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Answers

1. The extension is directly proportional to the applied force [1], up to the limit of proportionality [1]. The qualifier is part of the law.

2. Elastic — the material returns to its original shape when the load is removed [1]. Plastic — a permanent extension remains after the load is removed [1].

3. Strong — high breaking stress [1]. Stiff — high Young modulus [1]. Tough — absorbs a large amount of energy before fracture [1].

4. (a) r = 0.21 mm = 2.1 × 10⁻⁴ m [1] A = πr² = π × (2.1 × 10⁻⁴)² = 1.385 × 10⁻⁷ m² [1]. Using the diameter instead of the radius is the standard error.

(b) σ = F ÷ A = 45 ÷ (1.385 × 10⁻⁷) [1] = 3.25 × 10⁸ Pa [1].

(c) ε = x ÷ L = 3.6 × 10⁻³ ÷ 2.50 [1] = 1.44 × 10⁻³ [1].

(d) E = σ ÷ ε = (3.25 × 10⁸) ÷ (1.44 × 10⁻³) [1] = 2.26 × 10¹¹ Pa [1].

(e) The same, 2.26 × 10¹¹ Pa [1], because the Young modulus is a property of the material, not of the specimen’s dimensions [1].

5. (a) A long wire gives a larger, more easily measurable extension [1], reducing the percentage uncertainty in the extension measurement [1].

(b) A thin wire has a small cross-sectional area, so a modest load produces a large stress [1] and hence a measurable extension without needing dangerous masses [1].

(c) The wire may not be perfectly uniform along its length [1]; taking a mean reduces the effect of this and of random error in the measurement [1].

(d) Plot stress against strain [1]; the gradient of the straight-line region is the Young modulus [1].

6. (a) E = ½kx² = 0.5 × 25 × 0.12² [1] = 0.18 J [1].

(b) The energy dissipated during the loading–unloading cycle [1], transferred mostly to thermal energy, which is why a repeatedly stretched rubber band warms up [1].

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7. (a) stress = F/A = 60 ÷ (1.0 × 10⁻⁶) = 6.0 × 10⁷ Pa [1]; strain = x/L = (1.5 × 10⁻³) ÷ 2.0 = 7.5 × 10⁻⁴ [1]; E = stress/strain = 8.0 × 10¹⁰ Pa [1].

(b) The limit of proportionality is where extension stops being proportional to force [1]; the elastic limit is the point beyond which deformation becomes at least partly plastic — for many materials this lies close to, but not exactly at, the limit of proportionality [1].

8. (a) The unstretched reference wire is measured against the test wire to cancel out effects such as thermal expansion that would otherwise be mistaken for extension due to the load [1], since both wires expand or contract by the same amount with any temperature change [1].

(b) Eₚ = ½Fx = ½kx² only holds while F = kx, i.e. within the limit of proportionality, so the area under the graph is a triangle [1]. Beyond that point the graph curves, so the work done stretching the wire must instead be found from the actual area under the loading force–extension curve [1]. That area gives the recoverable elastic energy only if the wire is still behaving elastically (returns fully to its original length when unloaded); if the wire has been stretched past its elastic limit, some of that work is not recovered on unloading, and the true recoverable energy is instead the (smaller) area under the separate unloading curve.

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Where marks are usually lost

• Omitting “up to the limit of proportionality” from Hooke’s law.
• Using the diameter instead of the radius when finding area.
• Treating the Young modulus as a property of the object.
• Confusing strong, stiff and tough.
• Using ½Fx on a non-linear force–extension graph.