10 questions🇬🇧English
🪢 Stress, Strain & the Young Modulus – A-level
Physics · A-level — practise Hooke's law, stress and strain, Young modulus calculations and material graphs.
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Questions
What is the SI unit of tensile stress?
N m
N m⁻¹
Pa (N m⁻²)
It has no unit
20s
How is tensile strain defined?
Extension ÷ original length
Extension ÷ final length
Force ÷ extension
Force ÷ cross-sectional area
20s
A spring obeys Hooke's law and extends by 4.0 cm when a 2.0 N force is applied. What is its spring constant?
8.0 N m⁻¹
5.0 N m⁻¹
0.50 N m⁻¹
50 N m⁻¹
30s
Which statement best describes the elastic limit of a material?
Beyond it, the material deforms permanently
It is the maximum stress the material can withstand
It is the stress at which the material fractures
Beyond it, extension is no longer proportional to force
30s
For a ductile metal such as copper, what happens at the yield point?
The material fractures suddenly
Stress becomes directly proportional to strain
Strain increases greatly with little increase in stress
The material returns to its original length
30s
A spring with spring constant 200 N m⁻¹ is stretched by 0.10 m within its limit of proportionality. How much elastic strain energy is stored?
20 J
10 J
1.0 J
2.0 J
30s
Which description matches the stress–strain graph of a brittle material such as glass?
A curve with no straight section and a hysteresis loop
A straight line up to fracture with no plastic region
A yield point followed by a large plastic region
Necking after the ultimate tensile stress, then fracture
30s
Rubber is stretched and then released. On its force–extension graph, what does the area between the loading and unloading curves represent?
Elastic strain energy stored at maximum stretch
The Young modulus of the rubber
Energy transferred to heat in the rubber
Work done by the rubber as it contracts
30s
A wire of length 2.0 m and cross-sectional area 1.0 × 10⁻⁶ m² extends by 1.0 mm under a 50 N load. What is its Young modulus?
1.0 × 10¹⁴ Pa
2.5 × 10¹⁰ Pa
1.0 × 10¹¹ Pa
1.0 × 10⁸ Pa
60s
In the required practical to find the Young modulus of a metal, why is a long, thin wire used?
To keep the stress below the limit of proportionality
To reduce the Young modulus of the metal being tested
To give a larger extension and lower percentage uncertainty
To stop the wire heating up as it is loaded
30s