1. A solid is slightly deformed by an external force and completely regains its original size and shape after the force is removed. The property directly demonstrated by the solid is
ⓐ. plasticity
ⓑ. elasticity
ⓒ. rigidity
ⓓ. viscosity
Correct Answer: elasticity
Explanation: Elasticity is the property by virtue of which a body tends to regain its original size and shape when the deforming force is removed.
The temporary change produced while the force acts is called elastic deformation.
The decisive observation here is recovery after unloading, not the magnitude of the visible deformation.
A body need not be perfectly rigid in order to be elastic; an elastic body may deform appreciably or only by an imperceptibly small amount.
Plastic behaviour is different because a permanent deformation remains after the deforming force is removed.
Rigidity describes resistance to deformation and is not identical to elasticity.
Therefore, complete recovery of the original dimensions after removal of the force demonstrates elasticity.
2. Consider the following statements about deformation of a solid.
Statement I: An apparently rigid steel bar may undergo a small deformation when an external force is applied, even if the deformation is not easily visible.
Statement II: For a deformed body maintained in static equilibrium, an internal restoring force develops that opposes the applied deforming force and balances it in magnitude.
ⓐ. Statement I is correct, but Statement II is incorrect
ⓑ. Statement I is incorrect, but Statement II is correct
ⓒ. Both Statement I and Statement II are correct
ⓓ. Both Statement I and Statement II are incorrect
Correct Answer: Both Statement I and Statement II are correct
Explanation: Statement I is correct because real solid bodies are not perfectly rigid.
A sufficiently large external force can produce deformation even in a material such as steel.
The deformation may be so small that it is not readily noticed by eye, but absence of visible deformation does not imply zero deformation.
Statement II is also correct for the stated static-equilibrium condition.
When a deforming force acts, interatomic interactions within the solid give rise to an internal restoring response.
For equilibrium, the restoring force balances the applied force in magnitude and acts oppositely to the deforming action.
This restoring response is the basis for defining stress within the material.
The two statements therefore describe complementary features of real elastic solids: deformation can occur, while an internal response develops against that deformation.
Hence both statements are correct.
3. A helical spring and a lump of putty are separately deformed by moderate external forces. After the forces are removed, the spring returns to its original form while the putty retains a permanent change of shape. Which conclusion is correct?
ⓐ. The spring shows elastic deformation, whereas the putty shows plastic deformation
ⓑ. The spring shows plastic deformation, whereas the putty shows elastic deformation
ⓒ. Both show elastic deformation because both were deformed by external forces
ⓓ. Both show plastic deformation because neither was perfectly rigid during loading
Correct Answer: The spring shows elastic deformation, whereas the putty shows plastic deformation
Explanation: The classification is determined by what happens after the deforming force is removed.
The spring returns to its original form, so its deformation under the stated conditions is elastic.
The putty retains a permanent change of shape, so it has undergone plastic deformation.
The fact that both objects changed shape while the force acted does not distinguish elastic from plastic behaviour.
Elasticity does not mean that a body cannot deform; rather, it concerns its tendency to recover after unloading.
Similarly, plasticity is identified by a permanent set that remains after the external force is removed.
Perfect rigidity is therefore not the criterion being tested.
The observed unloading behaviour makes option A the correct conclusion.
4. Two materials P and Q are subjected to the same tensile stress \(8.0\times10^{7}\,\text{Pa}\) within their proportional regions. The resulting longitudinal strains are \(4.0\times10^{-4}\) for P and \(1.6\times10^{-3}\) for Q. Which statement is correct?
ⓐ. The elastic modulus of P is one-fourth that of Q, and P is more elastic
ⓑ. The elastic modulus of P is four times that of Q, and P is less elastic
ⓒ. The two materials have equal elastic moduli because the applied stress is the same
ⓓ. The elastic modulus of P is four times that of Q, and P is more elastic
Correct Answer: The elastic modulus of P is four times that of Q, and P is more elastic
Explanation: \(\textbf{Common loading condition:}\)
Both materials experience the same tensile stress:
\[
\sigma=8.0\times10^{7}\,\text{Pa}
\]
Within the proportional region, the relevant elastic modulus is
\[
M=\frac{\sigma}{\epsilon}
\]
\(\textbf{For material P:}\)
\[
M_P=\frac{8.0\times10^{7}}{4.0\times10^{-4}}
=2.0\times10^{11}\,\text{Pa}
\]
\(\textbf{For material Q:}\)
\[
M_Q=\frac{8.0\times10^{7}}{1.6\times10^{-3}}
=5.0\times10^{10}\,\text{Pa}
\]
Therefore,
\[
\frac{M_P}{M_Q}
=\frac{2.0\times10^{11}}{5.0\times10^{10}}
=4
\]
So P has four times the elastic modulus of Q.
For the same relevant stress, the material with the larger modulus develops the smaller strain.
In the modulus-based Physics sense used here, P is therefore more elastic than Q.
\(\textbf{Final answer:}\) P has four times the elastic modulus of Q and is more elastic.
5. Assertion: Steel may be regarded as more elastic than rubber in the modulus-based Physics sense even though rubber can undergo a much larger extension.
Reason: Under the same relevant stress, the material with the larger elastic modulus undergoes the smaller strain.
ⓐ. Assertion is true, but Reason is false
ⓑ. Assertion is false, but Reason is true
ⓒ. Both Assertion and Reason are true, and Reason explains Assertion
ⓓ. Both Assertion and Reason are true, but Reason does not explain Assertion
Correct Answer: Both Assertion and Reason are true, and Reason explains Assertion
Explanation: The Assertion is correct because greater stretchability and greater elasticity are not synonymous in this context.
Rubber can sustain a very large strain and may still recover its original dimensions after unloading.
However, when materials are compared through their elastic modulus, a larger modulus corresponds to a smaller strain for the same relevant stress.
Steel has a much larger elastic modulus than rubber for the corresponding deformation.
The Reason is therefore also correct.
From
\[
M=\frac{\text{stress}}{\text{strain}},
\]
a larger \(M\) at fixed stress means a smaller strain.
This is precisely why a material that stretches less, such as steel, can be called more elastic in the modulus-based comparison.
The Reason directly explains the apparently counter-intuitive Assertion.
The common misconception is to identify “more elastic” simply with “can be stretched farther.”
6. Material P has a linear tensile stress-strain response up to a point where \(\sigma=1.2\times10^{8}\,\text{Pa}\) and \(\epsilon=6.0\times10^{-4}\). Material Q also behaves linearly and has twice the stress-strain slope of P. Both are loaded from zero stress to the same final stress \(1.2\times10^{8}\,\text{Pa}\). Which statement is correct?
ⓐ. \(u_P:u_Q=2:1\), and Q has the larger elastic modulus
ⓑ. \(u_P:u_Q=1:2\), and Q has the larger elastic modulus
ⓒ. \(u_P:u_Q=2:1\), and P has the larger elastic modulus
ⓓ. \(u_P:u_Q=1:1\), because the final stresses are equal
Correct Answer: \(u_P:u_Q=2:1\), and Q has the larger elastic modulus
Explanation: \(\textbf{Slope comparison:}\)
In the linear region,
\[
M=\frac{\sigma}{\epsilon},
\]
so the stress-strain slope represents the relevant elastic modulus.
Q has twice the slope of P, hence
\[
M_Q=2M_P
\]
\(\textbf{Strain of Q at the same stress:}\)
Since \(\epsilon=\sigma/M\),
\[
\epsilon_Q=\frac{\epsilon_P}{2}
=\frac{6.0\times10^{-4}}{2}
=3.0\times10^{-4}
\]
\(\textbf{Elastic energy density:}\)
For stress increasing linearly from zero,
\[
u=\frac12\sigma\epsilon
\]
For P,
\[
u_P=\frac12(1.2\times10^{8})(6.0\times10^{-4})
=3.6\times10^{4}\,\text{J m}^{-3}
\]
For Q,
\[
u_Q=\frac12(1.2\times10^{8})(3.0\times10^{-4})
=1.8\times10^{4}\,\text{J m}^{-3}
\]
Thus,
\[
u_P:u_Q=2:1
\]
Geometrically, these energy densities are the triangular areas under the respective linear stress-strain graphs up to the common final stress.
Q is stiffer because its slope, and hence its modulus, is larger.
\(\textbf{Final answer:}\) \(u_P:u_Q=2:1\), with Q having the larger elastic modulus.
7. Match each loading situation with the corresponding stress-strain pair.
| Loading situation | Stress-strain pair |
|---|
| P. Equal and opposite forces act normally on the end faces of a cylinder | 1. Hydraulic stress — volume strain |
| Q. Tangential forces produce relative displacement between opposite faces | 2. Shearing stress — shearing strain |
| R. Uniform fluid pressure acts normally over the entire surface of a solid | 3. Longitudinal stress — longitudinal strain |
ⓐ. P-1, Q-2, R-3
ⓑ. P-3, Q-2, R-1
ⓒ. P-3, Q-1, R-2
ⓓ. P-2, Q-3, R-1
Correct Answer: P-3, Q-2, R-1
Explanation: For P, the applied forces are normal to the end faces and act along the length of the cylinder.
This produces tensile or compressive longitudinal stress, depending on whether the body is pulled or compressed.
The associated deformation is a change in length, so the corresponding strain is longitudinal strain.
Hence P matches 3.
For Q, the forces are tangential to the relevant surfaces and produce relative sliding displacement of opposite faces.
The restoring force per unit area is therefore shearing stress, while the deformation is measured by shearing strain.
Hence Q matches 2.
For R, pressure acts normally at every point over the surface and produces a uniform compression.
The corresponding internal stress is hydraulic stress, and the resulting fractional change is volume strain.
Hence R matches 1.
The important discrimination is the direction and distribution of the applied force relative to the surface.
Therefore, the complete matching is P-3, Q-2, R-1.
8. A uniform wire of cross-sectional area \(A\) is in equilibrium under equal and opposite pulls of magnitude \(F\) at its two ends. A second wire is subjected to equal and opposite pulls of magnitude \(2F\) and has cross-sectional area \(A/2\). What is the ratio of the tensile stress in the second wire to that in the first?
ⓐ. \(1\)
ⓑ. \(2\)
ⓒ. \(4\)
ⓓ. \(8\)
Correct Answer: \(4\)
Explanation: \(\textbf{First wire:}\)
Although equal pulls act at both ends, the tension carried by any cross-section of the wire is \(F\), not \(2F\).
Therefore,
\[
\sigma_1=\frac{F}{A}
\]
Adding the two opposite external force magnitudes would incorrectly give \(2F/A\); that would double-count the tension.
\(\textbf{Second wire:}\)
Its tension is \(2F\), while its cross-sectional area is \(A/2\).
Hence,
\[
\sigma_2=\frac{2F}{A/2}
\]
Therefore,
\[
\sigma_2=\frac{4F}{A}
\]
\(\textbf{Comparison:}\)
\[
\frac{\sigma_2}{\sigma_1}
=
\frac{4F/A}{F/A}
=4
\]
The stress increases because the tensile force is doubled while the area carrying that force is simultaneously halved.
These two changes multiply rather than cancel.
\(\textbf{Final answer:}\) The second wire has four times the tensile stress of the first.
9. Use the arrangement described below.
A solid block of height \(0.50\,\text{m}\) has its lower face fixed. A tangential force \(2.0\times10^{4}\,\text{N}\) acts uniformly over its upper face of area \(2.0\times10^{-2}\,\text{m}^2\). The upper face is displaced sideways by \(1.0\,\text{mm}\). The deformation is small.
Which set of values is correct for the shearing strain \(\gamma\), shear modulus \(G\), and angular deformation \(\theta\)?
ⓐ. \(\gamma=5.0\times10^{-4}\), \(G=2.0\times10^{9}\,\text{Pa}\), \(\theta\approx5.0\times10^{-4}\,\text{rad}\)
ⓑ. \(\gamma=2.0\times10^{-3}\,\text{m}\), \(G=5.0\times10^{8}\,\text{Pa}\), \(\theta\approx2.0\times10^{-3}\,\text{rad}\)
ⓒ. \(\gamma=2.0\times10^{-3}\), \(G=2.0\times10^{9}\,\text{Pa}\), \(\theta\approx4.0\times10^{-3}\,\text{rad}\)
ⓓ. \(\gamma=2.0\times10^{-3}\), \(G=5.0\times10^{8}\,\text{Pa}\), \(\theta\approx2.0\times10^{-3}\,\text{rad}\)
Correct Answer: \(\gamma=2.0\times10^{-3}\), \(G=5.0\times10^{8}\,\text{Pa}\), \(\theta\approx2.0\times10^{-3}\,\text{rad}\)
Explanation: \(\textbf{Convert the lateral displacement:}\)
\[
\Delta x=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}
\]
\(\textbf{Shearing strain:}\)
For a block of height \(L\),
\[
\gamma=\frac{\Delta x}{L}
\]
Thus,
\[
\gamma=\frac{1.0\times10^{-3}}{0.50}
=2.0\times10^{-3}
\]
Strain is a ratio of lengths and therefore has no SI unit.
\(\textbf{Shearing stress:}\)
\[
\tau=\frac{F}{A}
=\frac{2.0\times10^{4}}{2.0\times10^{-2}}
=1.0\times10^{6}\,\text{Pa}
\]
\(\textbf{Shear modulus:}\)
\[
G=\frac{\tau}{\gamma}
=\frac{1.0\times10^{6}}{2.0\times10^{-3}}
=5.0\times10^{8}\,\text{Pa}
\]
\(\textbf{Angular deformation:}\)
For a small deformation,
\[
\gamma=\tan\theta\approx\theta
\]
Hence,
\[
\theta\approx2.0\times10^{-3}\,\text{rad}
\]
The displacement itself is not the strain; it must be divided by the height over which the shear occurs.
\(\textbf{Final answer:}\) \(\gamma=2.0\times10^{-3}\), \(G=5.0\times10^{8}\,\text{Pa}\), and \(\theta\approx2.0\times10^{-3}\,\text{rad}\).
10. A cylindrical wire of radius \(1.0\,\text{mm}\) and length \(2.0\,\text{m}\) is stretched by a tensile force of \(314\,\text{N}\). Its extension is \(1.0\,\text{mm}\), and the response is within the proportional region. Take \(\pi=3.14\). Which set correctly gives the tensile stress, longitudinal strain, and slope of the stress-strain graph?
ⓐ. \(1.0\times10^{8}\,\text{Pa},\ 5.0\times10^{-4},\ 5.0\times10^{10}\,\text{Pa}\)
ⓑ. \(1.0\times10^{8}\,\text{Pa},\ 5.0\times10^{-4},\ 2.0\times10^{11}\,\text{Pa}\)
ⓒ. \(1.0\times10^{8}\,\text{Pa},\ 2.0\times10^{-3},\ 5.0\times10^{10}\,\text{Pa}\)
ⓓ. \(1.0\times10^{11}\,\text{Pa},\ 5.0\times10^{-4},\ 2.0\times10^{14}\,\text{Pa}\)
Correct Answer: \(1.0\times10^{8}\,\text{Pa},\ 5.0\times10^{-4},\ 2.0\times10^{11}\,\text{Pa}\)
Explanation: \(\textbf{Cross-sectional area:}\)
The radius is
\[
r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}
\]
Therefore,
\[
A=\pi r^2
=3.14(1.0\times10^{-3})^2
=3.14\times10^{-6}\,\text{m}^2
\]
\(\textbf{Tensile stress:}\)
\[
\sigma=\frac{F}{A}
=\frac{314}{3.14\times10^{-6}}
=1.0\times10^{8}\,\text{Pa}
\]
\(\textbf{Longitudinal strain:}\)
\[
\epsilon=\frac{\Delta L}{L}
=\frac{1.0\times10^{-3}}{2.0}
=5.0\times10^{-4}
\]
\(\textbf{Stress-strain slope:}\)
Because the point lies in the proportional region,
\[
\text{slope}=\frac{\sigma}{\epsilon}
\]
Hence,
\[
\text{slope}
=\frac{1.0\times10^{8}}{5.0\times10^{-4}}
=2.0\times10^{11}\,\text{Pa}
\]
For tensile deformation this slope is the Young's modulus of the material in the linear region.
The radius must be squared when calculating area, while strain remains dimensionless.
\(\textbf{Final answer:}\) \(1.0\times10^{8}\,\text{Pa},\ 5.0\times10^{-4},\ 2.0\times10^{11}\,\text{Pa}\).
11. An experiment gives the following loading data for a material.
| Stress \(\sigma\) in MPa | Strain \(\epsilon\) |
|---|
| 50 | \(2.0\times10^{-4}\) |
| 100 | \(4.0\times10^{-4}\) |
| 150 | \(6.0\times10^{-4}\) |
| 200 | \(1.0\times10^{-3}\) |
Which conclusion is best supported by these data alone?
ⓐ. Hooke's law holds through \(150\,\text{MPa}\); at \(200\,\text{MPa}\) it fails, while permanent deformation remains undetermined
ⓑ. Hooke's law is obeyed through \(200\,\text{MPa}\), and unloading from \(200\,\text{MPa}\) must give complete recovery
ⓒ. Hooke's law fails from \(100\,\text{MPa}\) onward, so the material is plastic above \(100\,\text{MPa}\)
ⓓ. Hooke's law is obeyed through \(150\,\text{MPa}\), and the \(200\,\text{MPa}\) point must lie beyond the elastic limit
Correct Answer: Hooke's law holds through \(150\,\text{MPa}\); at \(200\,\text{MPa}\) it fails, while permanent deformation remains undetermined
Explanation: \(\textbf{Check proportionality:}\)
Hooke's law requires stress to be directly proportional to strain.
For the first point,
\[
\frac{\sigma}{\epsilon}
=
\frac{50\times10^{6}}{2.0\times10^{-4}}
=2.5\times10^{11}\,\text{Pa}
\]
For \(100\,\text{MPa}\),
\[
\frac{\sigma}{\epsilon}
=
\frac{100\times10^{6}}{4.0\times10^{-4}}
=2.5\times10^{11}\,\text{Pa}
\]
For \(150\,\text{MPa}\),
\[
\frac{\sigma}{\epsilon}
=
\frac{150\times10^{6}}{6.0\times10^{-4}}
=2.5\times10^{11}\,\text{Pa}
\]
Thus the first three points are consistent with a single straight-line stress-strain relation.
At \(200\,\text{MPa}\),
\[
\frac{\sigma}{\epsilon}
=
\frac{200\times10^{6}}{1.0\times10^{-3}}
=2.0\times10^{11}\,\text{Pa},
\]
so proportionality has been lost.
However, loss of Hooke-law proportionality does not by itself prove permanent deformation.
A material can pass beyond its proportional limit yet remain below its elastic limit, so unloading can still restore its original dimensions.
The table contains only loading data and gives no unloading observation.
\(\textbf{Final conclusion:}\) Hooke's law is supported up to \(150\,\text{MPa}\), while plasticity at \(200\,\text{MPa}\) cannot be inferred from these data alone.
12. Use the stress-strain graph description below.
For a metal, the curve is linear from O to A. From A to B it is nonlinear, but unloading still restores the original dimensions. B is the yield point or elastic limit. Point C lies beyond B in the plastic region. D is the maximum-stress point on the curve, and fracture occurs later at E.
If the specimen is loaded from O to C and then completely unloaded, which statement is correct?
ⓐ. It returns exactly to its original dimensions, and D represents the fracture point
ⓑ. It returns exactly to its original dimensions, and D represents the ultimate tensile strength
ⓒ. It retains a permanent set, and D represents the ultimate tensile strength
ⓓ. It retains a permanent set, and B represents the ultimate tensile strength
Correct Answer: It retains a permanent set, and D represents the ultimate tensile strength
Explanation: Point B marks the end of the elastic region in the stated graph description.
Point C lies beyond B, so the specimen has entered the plastic-deformation region.
If the load is removed from C, the stress can fall to zero without the strain returning to zero.
The remaining non-zero strain corresponds to a permanent set.
Therefore, loading to C does not lead to complete recovery of the original dimensions.
Point D has a different physical meaning.
D is the point at which the tensile stress reaches its maximum value on the typical engineering stress-strain curve.
That stress is the ultimate tensile strength of the material.
Fracture occurs later at E, so ultimate tensile strength and fracture point need not coincide.
A common mistake is to assume that any point before fracture is still elastic; the elastic-to-plastic transition has already occurred at B.
Thus the correct combination is permanent set after unloading from C and ultimate tensile strength at D.
13. Use the graph information below.
In their linear tensile regions, material P has stress-strain slope \(2.0\times10^{11}\,\text{Pa}\) and material Q has slope \(1.0\times10^{11}\,\text{Pa}\). A tensile stress \(1.5\times10^{8}\,\text{Pa}\) lies inside the linear region for both materials. Each specimen is loaded gradually from zero stress to this value.
What are the ratios \(\epsilon_Q/\epsilon_P\) and \(u_Q/u_P\), where \(u\) is the elastic energy stored per unit volume?
ⓐ. \(1/2,\ 1/2\)
ⓑ. \(1/2,\ 2\)
ⓒ. \(2,\ 1/2\)
ⓓ. \(2,\ 2\)
Correct Answer: \(2,\ 2\)
Explanation: \(\textbf{Interpret the graph slopes:}\)
In the linear tensile region,
\[
M=\frac{\sigma}{\epsilon},
\]
so the given slopes are the elastic moduli.
Thus,
\[
M_P=2.0\times10^{11}\,\text{Pa},
\qquad
M_Q=1.0\times10^{11}\,\text{Pa}
\]
\(\textbf{Strain of P:}\)
\[
\epsilon_P=\frac{\sigma}{M_P}
=\frac{1.5\times10^{8}}{2.0\times10^{11}}
=7.5\times10^{-4}
\]
\(\textbf{Strain of Q:}\)
\[
\epsilon_Q=\frac{\sigma}{M_Q}
=\frac{1.5\times10^{8}}{1.0\times10^{11}}
=1.5\times10^{-3}
\]
Therefore,
\[
\frac{\epsilon_Q}{\epsilon_P}=2
\]
\(\textbf{Energy-density comparison:}\)
For linear loading from zero stress,
\[
u=\frac12\sigma\epsilon
\]
Both materials reach the same final stress, so at this common stress \(u\) is directly proportional to the corresponding strain.
Hence,
\[
\frac{u_Q}{u_P}
=
\frac{\epsilon_Q}{\epsilon_P}
=2
\]
Numerically,
\[
u_P=\frac12(1.5\times10^{8})(7.5\times10^{-4})
=5.625\times10^{4}\,\text{J m}^{-3},
\]
while
\[
u_Q=1.125\times10^{5}\,\text{J m}^{-3}
\]
The less stiff material Q undergoes the larger strain and therefore encloses the larger triangular area under the graph at the same final stress.
\(\textbf{Final answer:}\)
\[
\boxed{\epsilon_Q/\epsilon_P=2\quad\text{and}\quad u_Q/u_P=2}
\]
14. A typical tensile stress-strain curve for a metal has a linear region O-A, a nonlinear but elastic region A-B, a plastic region beyond B, an ultimate tensile-strength point D, and a fracture point E. Consider the following statements.
Statement I: A specimen unloaded from a point between A and B can regain its original dimensions even though stress is no longer proportional to strain there.
Statement II: A specimen unloaded from a point C lying beyond B in the plastic region can retain a permanent set.
Statement III: D and E need not be the same point.
Statement IV: If D and E are very close, the material is characteristically ductile.
Statement V: Beyond D, additional strain can occur even though the engineering stress decreases before fracture.
ⓐ. I, III and IV only
ⓑ. I, II and IV only
ⓒ. II, IV and V only
ⓓ. I, II, III and V only
Correct Answer: I, II, III and V only
Explanation: Statement I is correct.
The region A-B is explicitly described as nonlinear but still elastic, so Hooke's law no longer applies there while complete recovery on unloading can still occur.
Statement II is also correct because C lies beyond the elastic limit B.
Unloading from the plastic region can leave non-zero strain at zero stress, which is the permanent set.
Statement III is correct because D denotes ultimate tensile strength whereas E denotes fracture.
A specimen can therefore continue to deform after reaching its maximum engineering stress before it finally breaks.
Statement IV is incorrect.
If the ultimate-strength point and fracture point are close, there is little post-maximum deformation and the material is characterised as brittle rather than ductile.
A larger separation is associated with greater ductile deformation.
Statement V is correct for the typical curve described: after D, additional strain can occur while the engineering stress falls until fracture at E.
The key is to keep proportionality, elastic recovery, ultimate strength and fracture as distinct ideas rather than treating them as one boundary.
Hence I, II, III and V only are correct.
15. The following idealised data refer to three materials in a tensile test.
| Material | Elastic modulus | Yield strength | Ultimate tensile strength | Fracture strain |
|---|
| P | \(200\,\text{GPa}\) | \(220\,\text{MPa}\) | \(350\,\text{MPa}\) | \(0.04\) |
| Q | \(120\,\text{GPa}\) | \(300\,\text{MPa}\) | \(500\,\text{MPa}\) | \(0.18\) |
| R | \(160\,\text{GPa}\) | \(260\,\text{MPa}\) | \(420\,\text{MPa}\) | \(0.10\) |
A tensile stress of \(80\,\text{MPa}\) lies in the linear region for all three materials. Which conclusion is correct?
ⓐ. Q develops the least strain; P has the greatest tensile strength and greatest ductility
ⓑ. P develops the least strain; P has the greatest tensile strength, while Q has the greatest ductility
ⓒ. P develops the least strain; Q has the greatest ultimate tensile strength and the greatest fracture strain
ⓓ. R develops the least strain; Q has both the largest elastic modulus and the largest ultimate tensile strength
Correct Answer: P develops the least strain; Q has the greatest ultimate tensile strength and the greatest fracture strain
Explanation: \(\textbf{Compare stiffness first:}\)
Within the linear region,
\[
\epsilon=\frac{\sigma}{M}.
\]
All three specimens experience the same stress, so the material with the largest elastic modulus develops the least strain.
For P,
\[
\epsilon_P=\frac{80\times10^{6}}{200\times10^{9}}
=4.0\times10^{-4}.
\]
For Q,
\[
\epsilon_Q=\frac{80\times10^{6}}{120\times10^{9}}
\approx6.67\times10^{-4}.
\]
For R,
\[
\epsilon_R=\frac{80\times10^{6}}{160\times10^{9}}
=5.0\times10^{-4}.
\]
Therefore P has the smallest strain and is the stiffest of the three.
\(\textbf{Compare tensile strength:}\)
Ultimate tensile strength is a strength measure and must not be confused with elastic modulus.
The largest tabulated ultimate tensile strength is
\[
500\,\text{MPa},
\]
which belongs to Q.
\(\textbf{Compare ductility:}\)
The largest listed fracture strain is
\[
0.18,
\]
again belonging to Q, indicating the greatest deformation before fracture among the three data sets.
Thus P is stiffest under the common elastic stress, while Q has the greatest ultimate tensile strength and the greatest fracture strain.
This illustrates why stiffness, strength and ductility are separate material properties.
\(\textbf{Final answer:}\) Option C.
16. Two materials are tested by loading them to several stresses and then completely unloading them after each reading.
| Material | Stress | Measured strain | Observation after unloading |
|---|
| X | \(20\,\text{MPa}\) | \(1.0\times10^{-4}\) | Original dimensions recovered |
| X | \(40\,\text{MPa}\) | \(2.0\times10^{-4}\) | Original dimensions recovered |
| X | \(60\,\text{MPa}\) | \(3.0\times10^{-4}\) | Original dimensions recovered |
| Y | \(2\,\text{MPa}\) | \(0.02\) | Original dimensions recovered |
| Y | \(4\,\text{MPa}\) | \(0.07\) | Original dimensions recovered |
| Y | \(6\,\text{MPa}\) | \(0.16\) | Original dimensions recovered |
Which conclusion is justified?
ⓐ. X is Hookean over the tested range, while Y is elastic but non-Hookean over the tested range
ⓑ. X is plastic whereas Y is elastic because Y undergoes the larger strains
ⓒ. Both X and Y obey Hooke's law because both recover completely after unloading
ⓓ. Y is more Hookean than X because its strain increases more rapidly with stress
Correct Answer: X is Hookean over the tested range, while Y is elastic but non-Hookean over the tested range
Explanation: Complete recovery after unloading is the evidence needed to classify the tested deformation as elastic.
Both X and Y recover their original dimensions after every listed loading condition, so both exhibit elastic behaviour over their respective tested ranges.
Hooke's law requires an additional condition: stress must be directly proportional to strain.
\(\textbf{Check material X:}\)
\[
\frac{20\times10^{6}}{1.0\times10^{-4}}
=
2.0\times10^{11}\,\text{Pa},
\]
\[
\frac{40\times10^{6}}{2.0\times10^{-4}}
=
2.0\times10^{11}\,\text{Pa},
\]
and
\[
\frac{60\times10^{6}}{3.0\times10^{-4}}
=
2.0\times10^{11}\,\text{Pa}.
\]
The ratio \(\sigma/\epsilon\) is constant, so X is Hookean over the tested range.
\(\textbf{Check material Y:}\)
For Y, the corresponding ratios are
\[
1.0\times10^{8}\,\text{Pa},\qquad
5.71\times10^{7}\,\text{Pa},\qquad
3.75\times10^{7}\,\text{Pa},
\]
approximately.
They are clearly not constant, so stress is not proportional to strain for Y over the tested range.
Yet Y still recovers completely after unloading, so its behaviour is elastic even though it is non-Hookean.
This is the essential distinction illustrated by elastomer-like materials: a large elastic range does not require a linear stress-strain relation throughout that range.
\(\textbf{Final answer:}\) X is Hookean in the tested range, whereas Y is elastic but non-Hookean.
17. A wire of length \(L\), cross-sectional area \(A\) and Young's modulus \(Y\) is stretched by a force \(F\), producing extension \(\Delta L_1\). It is replaced by another wire of the same material and the same volume, but of length \(2L\). The second wire is stretched by a force \(F/2\), and its extension is \(\Delta L_2\). Which relation is correct?
ⓐ. \(\sigma_2/\sigma_1=1/2,\ \Delta L_2/\Delta L_1=1\)
ⓑ. \(\sigma_2/\sigma_1=2,\ \Delta L_2/\Delta L_1=4\)
ⓒ. \(\sigma_2/\sigma_1=1,\ \Delta L_2/\Delta L_1=2\)
ⓓ. \(\sigma_2/\sigma_1=1,\ \Delta L_2/\Delta L_1=4\)
Correct Answer: \(\sigma_2/\sigma_1=1,\ \Delta L_2/\Delta L_1=2\)
Explanation: \(\textbf{Geometry imposed by equal volume:}\)
For the first wire,
\[
V=AL
\]
For the second wire,
\[
V=A_2(2L)
\]
Since the volumes are equal,
\[
A_2=\frac{A}{2}
\]
\(\textbf{Compare the tensile stresses:}\)
For the first wire,
\[
\sigma_1=\frac{F}{A}
\]
For the second wire,
\[
\sigma_2=\frac{F/2}{A/2}=\frac{F}{A}
\]
Therefore,
\[
\frac{\sigma_2}{\sigma_1}=1
\]
Because both wires are of the same material and remain in the proportional region, equal stress also produces equal longitudinal strain.
Thus,
\[
\epsilon_2=\epsilon_1
\]
Since \(\epsilon=\Delta L/L\),
\[
\frac{\Delta L_2}{2L}=\frac{\Delta L_1}{L}
\]
Hence,
\[
\Delta L_2=2\Delta L_1
\]
The same result follows directly from
\[
\Delta L=\frac{FL}{AY}
\]
The reduced force is exactly compensated in stress by the reduced area, while the doubled length doubles the absolute extension.
\(\textbf{Final answer:}\) \(\sigma_2/\sigma_1=1\) and \(\Delta L_2/\Delta L_1=2\).
18. Two uniform wires P and Q are joined end-to-end and stretched by a tensile force of \(200\,\text{N}\). Each wire has cross-sectional area \(2.0\times10^{-6}\,\text{m}^2\). Wire P has length \(1.0\,\text{m}\) and Young's modulus \(2.0\times10^{11}\,\text{Pa}\). Wire Q has length \(0.50\,\text{m}\). If the total extension of the combination is \(1.0\,\text{mm}\), the Young's modulus of Q is
ⓐ. \(5.0\times10^{10}\,\text{Pa}\)
ⓑ. \(4.0\times10^{11}\,\text{Pa}\)
ⓒ. \(2.0\times10^{11}\,\text{Pa}\)
ⓓ. \(1.0\times10^{11}\,\text{Pa}\)
Correct Answer: \(1.0\times10^{11}\,\text{Pa}\)
Explanation: The two wires are connected end-to-end, so the same tensile force acts through both wires when the junction is in equilibrium.
\(\textbf{Extension contributed by P:}\)
For a wire in the proportional region,
\[
\Delta L=\frac{FL}{AY}
\]
Therefore,
\[
\Delta L_P=
\frac{(200)(1.0)}
{(2.0\times10^{-6})(2.0\times10^{11})}
\]
Thus,
\[
\Delta L_P=5.0\times10^{-4}\,\text{m}
=0.50\,\text{mm}
\]
\(\textbf{Use the measured total extension:}\)
The total extension is the sum of the extensions of the two series segments:
\[
\Delta L_{\text{total}}=\Delta L_P+\Delta L_Q
\]
Hence,
\[
1.0\,\text{mm}=0.50\,\text{mm}+\Delta L_Q
\]
So,
\[
\Delta L_Q=0.50\,\text{mm}
=5.0\times10^{-4}\,\text{m}
\]
\(\textbf{Infer the modulus of Q:}\)
Rearranging the extension relation,
\[
Y_Q=\frac{FL_Q}{A\Delta L_Q}
\]
Substitution gives
\[
Y_Q=
\frac{(200)(0.50)}
{(2.0\times10^{-6})(5.0\times10^{-4})}
\]
Therefore,
\[
Y_Q=1.0\times10^{11}\,\text{Pa}
\]
The moduli are not added directly; the physically additive quantity for these series segments is their elongation.
\(\textbf{Final answer:}\)
\[
\boxed{Y_Q=1.0\times10^{11}\,\text{Pa}}
\]
19. A light rigid horizontal plate is suspended by two vertical circular wires P and Q attached symmetrically at equal distances from its centre. The wires have equal initial lengths. A load is placed at the centre of the plate, and the plate remains horizontal after a small elastic extension of the wires. If \(Y_P=2Y_Q\), what must be the ratio of their diameters \(d_Q/d_P\)?
ⓐ. \(1/\sqrt2\)
ⓑ. \(1\)
ⓒ. \(2\)
ⓓ. \(\sqrt2\)
Correct Answer: \(\sqrt2\)
Explanation: Because the load acts at the centre and the suspension points are symmetrically placed, rotational equilibrium requires equal tensions in the two wires.
Thus,
\[
T_P=T_Q
\]
The plate remains horizontal, so the lower ends of the initially equal-length wires move downward by equal amounts.
Therefore,
\[
\Delta L_P=\Delta L_Q
\]
For either wire,
\[
\Delta L=\frac{TL}{AY}
\]
Since \(T\), \(L\), and \(\Delta L\) are equal for P and Q,
\[
A_PY_P=A_QY_Q
\]
Given
\[
Y_P=2Y_Q,
\]
we obtain
\[
A_Q=2A_P
\]
For a circular wire,
\[
A=\frac{\pi d^2}{4}
\]
Hence,
\[
\frac{A_Q}{A_P}
=
\frac{d_Q^2}{d_P^2}
=2
\]
Therefore,
\[
\frac{d_Q}{d_P}=\sqrt2
\]
The wire with the smaller Young's modulus must have the larger area if equal tension is to produce the same extension.
Simply assigning equal diameters would make Q extend more and the rigid plate would tilt.
\(\textbf{Final answer:}\)
\[
\boxed{d_Q/d_P=\sqrt2}
\]
20. A wire of length \(1.5\,\text{m}\) and diameter \(2.0\,\text{mm}\) has Young's modulus \(1.0\times10^{11}\,\text{Pa}\). For this idealised wire, assume the stress-strain relation remains linear up to a yield stress of \(2.0\times10^{8}\,\text{Pa}\). Take \(\pi=3.14\). What are the largest tensile load corresponding to the yield stress and the extension of the wire at that stress?
ⓐ. \(628\,\text{N}\) and \(3.0\,\text{mm}\)
ⓑ. \(314\,\text{N}\) and \(3.0\,\text{mm}\)
ⓒ. \(628\,\text{N}\) and \(1.5\,\text{mm}\)
ⓓ. \(1256\,\text{N}\) and \(6.0\,\text{mm}\)
Correct Answer: \(628\,\text{N}\) and \(3.0\,\text{mm}\)
Explanation: \(\textbf{Area carrying the tensile load:}\)
The radius is
\[
r=1.0\,\text{mm}=1.0\times10^{-3}\,\text{m}
\]
Hence,
\[
A=\pi r^2
=3.14(1.0\times10^{-3})^2
=3.14\times10^{-6}\,\text{m}^2
\]
\(\textbf{Load corresponding to yield stress:}\)
Tensile stress is
\[
\sigma=\frac{F}{A}
\]
Therefore,
\[
F_y=\sigma_yA
\]
Substituting,
\[
F_y=(2.0\times10^{8})(3.14\times10^{-6})
=628\,\text{N}
\]
\(\textbf{Strain at the stated yield stress:}\)
Under the given linear-response assumption,
\[
Y=\frac{\sigma}{\epsilon}
\]
Thus,
\[
\epsilon_y=\frac{\sigma_y}{Y}
=\frac{2.0\times10^{8}}{1.0\times10^{11}}
=2.0\times10^{-3}
\]
\(\textbf{Convert strain to extension:}\)
\[
\epsilon_y=\frac{\Delta L_y}{L}
\]
Therefore,
\[
\Delta L_y=(2.0\times10^{-3})(1.5)
=3.0\times10^{-3}\,\text{m}
\]
Hence,
\[
\Delta L_y=3.0\,\text{mm}
\]
Yield strength and Young's modulus describe different properties: one sets the stress scale for onset of yielding, while the other determines strain in the linear elastic region.
\(\textbf{Final answer:}\)
\[
\boxed{628\,\text{N}\quad\text{and}\quad 3.0\,\text{mm}}
\]
