AS Level Physics chapter 6: Deformation of solids MCQs with answers
By Team Megalecture · Updated 4 Nov 2026
50 exam-style questions on Deformation of solids, each with the answer and a short explanation. Ten of them are below to try. The chapter test draws ten fresh questions each time and is marked the moment you finish, with your score by topic.
Try each one, then open the answer. These ten stay the same; the test draws new ones every time.
1. A toy launcher uses a spring of spring constant 800 N m−1 compressed by 5.0 cm. It fires a 20 g ball vertically upwards. Ignoring air resistance and the energy kept by the spring, how high does the ball rise? (g = 9.81 m s−2)
A
5.1 × 10−3 m
B
5.1 m
C
10 m
D
50 m
Show answer
Answer: B. EP = ½kx2 = ½ × 800 × 0.0502 = 1.0 J = mgh, so h = 1.0 ÷ (0.020 × 9.81) = 5.1 m. Leaving out the ½ gives 10 m; leaving out g gives 50 m.
2. A spring obeys Hooke's law and has a spring constant of 80 N m−1. It is cut into two equal halves. What is the spring constant of one half?
A
40 N m−1
B
80 N m−1
C
160 N m−1
D
320 N m−1
Show answer
Answer: C. A given force stretches every part of the spring equally, so half the spring gives half the extension for the same force: k = F/x doubles to 160 N m−1.
3. On a force–extension graph, one small square represents 0.5 N on the force axis and 2 mm on the extension axis. The area under the curve up to a certain extension is 36 small squares. How much work was done in stretching the sample to that extension?
A
3.6 × 10−3 J
B
1.8 × 10−2 J
C
3.6 × 10−2 J
D
36 J
Show answer
Answer: C. Each square = 0.5 × 2 × 10−3 = 1.0 × 10−3 J, so work = 36 × 1.0 × 10−3 = 3.6 × 10−2 J. Leaving mm unconverted gives 36 J; halving the area wrongly gives 1.8 × 10−2 J.
4. A graph of extension (y-axis) against force (x-axis) for a spring is a straight line through the origin with gradient 2.5 mm N−1. What is the spring constant?
A
2.5 × 10−3 N m−1
B
0.40 N m−1
C
400 N m−1
D
4.0 × 105 N m−1
Show answer
Answer: C. The gradient is x/F = 2.5 × 10−3 m N−1, and k = F/x is its reciprocal: 1 ÷ 2.5 × 10−3 = 400 N m−1. Taking the reciprocal without converting mm gives 0.40.
5. A wire of volume 2.0 × 10−6 m3 is under a stress of 2.0 × 108 Pa and a strain of 1.0 × 10−3, within the limit of proportionality. How much elastic potential energy is stored in the wire?
A
0.20 J
B
0.40 J
C
1.0 × 105 J
D
2.0 × 105 J
Show answer
Answer: A. Energy = ½Fx = ½(stress × A)(strain × L) = ½ × stress × strain × volume = ½ × 2.0 × 108 × 1.0 × 10−3 × 2.0 × 10−6 = 0.20 J. 1.0 × 105 is the energy per unit volume in J m−3.
6. A spring has a natural length of 12.0 cm. With a 4.0 N load hanging from it, its length is 15.0 cm. What is the extension with a 6.0 N load? The limit of proportionality is not reached.
A
3.0 cm
B
4.5 cm
C
16.5 cm
D
22.5 cm
Show answer
Answer: B. Extension with 4.0 N = 15.0 − 12.0 = 3.0 cm, so extension ∝ load gives 3.0 × 6.0/4.0 = 4.5 cm (new length 16.5 cm). Scaling the length instead of the extension gives 22.5 cm.
7. In an experiment to find the Young modulus of a metal, the wire used is long and thin. What is the main reason?
A
to produce an extension large enough to measure with a small percentage uncertainty
B
to keep the stress small
C
to reduce the weight of the wire
D
to keep the wire below its elastic limit
Show answer
Answer: A. Metals extend very little. A long wire gives a larger extension for a given strain, and a thin wire gives a larger stress for a given load, so the extension can be measured more precisely.
8. A steel wire of length 2.0 m and cross-sectional area 1.0 × 10−6 m2 is stretched by 1.0 mm, within its limit of proportionality. The Young modulus of steel is 2.0 × 1011 Pa. How much elastic potential energy is stored?
A
0.025 J
B
0.050 J
C
0.10 J
D
50 J
Show answer
Answer: B. F = EAx/L = 2.0 × 1011 × 1.0 × 10−6 × 1.0 × 10−3 ÷ 2.0 = 100 N. Energy = ½Fx = ½ × 100 × 1.0 × 10−3 = 0.050 J. Leaving out the ½ gives 0.10 J.
9. On a force–extension graph for a wire loaded then unloaded within its elastic region, the loading and unloading lines coincide. What does this show?
A
the wire has deformed plastically
B
all the energy stored on loading is returned on unloading
C
the wire has broken
D
the material obeys Hooke's law only on unloading
Show answer
Answer: B. Coinciding lines mean no energy is lost: elastic behaviour. A loop between the lines (hysteresis, as in rubber) shows energy dissipated.
10. A rubber band is stretched and then released slowly. It returns to its original length, but the unloading curve on its force–extension graph lies below the loading curve. Which statement is correct?
A
the deformation is plastic
B
the band stores more energy on unloading than on loading
C
the area between the two curves is zero
D
the deformation is elastic, and the area between the curves is energy transferred to thermal energy
Show answer
Answer: D. It returns to its original length, so it is elastic. The work done on loading (area under the loading curve) exceeds the work returned (area under the unloading curve); the difference heats the rubber.
50, each with the answer and a short explanation, across 2 topics: Stress and strain, Elastic and plastic behaviour. Every chapter test draws ten of them, so you can take it many times.
Is the AS Level Physics chapter 6 test free?
Yes. The test, the answers and the explanations are free and need no sign-up. Sign in only if you want your scores kept across devices.
How is the chapter test marked?
The moment you press Finish: your score, the right answer and an explanation for every question, and the ones you got wrong saved in My mistakes to retry.
