301. A cylindrical wire is stretched, and its diameter decreases. If the diameter is used to find lateral strain, the correct expression and sign for stretching are
ⓐ. \(\frac{d}{\Delta d}\), positive
ⓑ. \(\frac{\Delta d}{d}\), negative
ⓒ. \(\frac{\Delta d}{d}\), positive
ⓓ. \(\frac{F}{A}\), negative
Correct Answer: \(\frac{\Delta d}{d}\), negative
Explanation: Lateral strain is the fractional change in a sideways dimension. For a cylindrical wire, diameter can be used as the lateral dimension, so lateral strain is \(\frac{\Delta d}{d}\). During stretching, the diameter decreases, so \(\Delta d\) is negative. Therefore the lateral strain is negative. This sign is why Poisson’s ratio is written as \(\nu=-\frac{\text{lateral strain}}{\text{longitudinal strain}}\) for ordinary materials.
302. A wire is stretched to stress \(\sigma\) in the linear elastic region. Another wire of the same material is stretched to stress \(2\sigma\). The elastic energy density in the second wire is
ⓐ. twice that in the first wire
ⓑ. half that in the first wire
ⓒ. four times that in the first wire
ⓓ. the same as that in the first wire
Correct Answer: four times that in the first wire
Explanation: In the linear elastic region, elastic energy density in terms of stress is:
\[u=\frac{\sigma^2}{2Y}\]
For the same material, \(Y\) is constant.
For the first wire:
\[u_1=\frac{\sigma^2}{2Y}\]
For the second wire, the stress is \(2\sigma\):
\[u_2=\frac{(2\sigma)^2}{2Y}\]
\[u_2=\frac{4\sigma^2}{2Y}\]
\[u_2=4u_1\]
Energy density grows with the square of stress, not directly in the same ratio as stress.
\( \textbf{Final answer:} \) four times that in the first wire.
303. A material-testing note says: “The initial stress-strain slope is small, the breaking stress is high, and the material shows large plastic strain before fracture.” The best description is that the material is
ⓐ. less stiff but strong and ductile
ⓑ. very stiff but brittle
ⓒ. incompressible but unable to carry tension
ⓓ. perfectly elastic at all stresses
Correct Answer: less stiff but strong and ductile
Explanation: A small initial slope on a stress-strain graph means a smaller Young’s modulus, so the material is less stiff in the elastic region. A high breaking stress means it can withstand a large stress before fracture. Large plastic strain before fracture is a sign of ductility. These three observations describe different properties and should not be merged into one word. A material can be less stiff yet still strong and ductile.
304. A solid of volume \(8.0\times10^{-4}\,\text{m}^3\) is compressed by pressure. Its bulk modulus is \(1.6\times10^{11}\,\text{Pa}\), and the pressure increase is \(4.0\times10^7\,\text{Pa}\). If the same material is instead given a pressure increase twice as large, the magnitude of volume decrease becomes
ⓐ. \(2.0\times10^{-7}\,\text{m}^3\)
ⓑ. \(8.0\times10^{-7}\,\text{m}^3\)
ⓒ. \(4.0\times10^{-7}\,\text{m}^3\)
ⓓ. \(1.0\times10^{-7}\,\text{m}^3\)
Correct Answer: \(4.0\times10^{-7}\,\text{m}^3\)
Explanation: \( \textbf{First pressure increase:} \) \(V=8.0\times10^{-4}\,\text{m}^3\), \(B=1.6\times10^{11}\,\text{Pa}\), and \(\Delta P=4.0\times10^7\,\text{Pa}\).
The magnitude of volume change is:
\[\left|\Delta V\right|=\frac{V\Delta P}{B}\]
For the first pressure increase:
\[\left|\Delta V_1\right|=\frac{(8.0\times10^{-4})(4.0\times10^7)}{1.6\times10^{11}}\,\text{m}^3\]
Numerator:
\[(8.0\times10^{-4})(4.0\times10^7)=3.2\times10^4\]
\[\left|\Delta V_1\right|=\frac{3.2\times10^4}{1.6\times10^{11}}\,\text{m}^3\]
\[\left|\Delta V_1\right|=2.0\times10^{-7}\,\text{m}^3\]
If pressure increase is doubled, volume decrease magnitude also doubles in the elastic range:
\[\left|\Delta V_2\right|=2\left|\Delta V_1\right|\]
\[\left|\Delta V_2\right|=4.0\times10^{-7}\,\text{m}^3\]
\( \textbf{Final answer:} \) \(4.0\times10^{-7}\,\text{m}^3\).
305. A wire of radius \(r\), length \(L\), and Young’s modulus \(Y\) is stretched by force \(F\). A second wire of the same material is to have the same extension under the same force but has length \(4L\). Its radius should be
ⓐ. \(4r\)
ⓑ. \(\frac{r}{2}\)
ⓒ. \(r\)
ⓓ. \(2r\)
Correct Answer: \(2r\)
Explanation: \( \textbf{Condition:} \) Same material and same force mean \(Y\) and \(F\) are unchanged.
Extension of a wire is:
\[\Delta L=\frac{FL}{AY}\]
For the extension to remain the same:
\[\frac{L}{A}=\frac{L'}{A'}\]
The second wire has:
\[L'=4L\]
Therefore:
\[\frac{L}{A}=\frac{4L}{A'}\]
Cancel \(L\):
\[\frac{1}{A}=\frac{4}{A'}\]
\[A'=4A\]
For a circular cross-section:
\[A=\pi r^2\]
If area becomes \(4A\), radius must become \(2r\).
\( \textbf{Final answer:} \) \(2r\).
306. The following claims are made after a full elastic test of a wire.
I. The slope of the stress-strain graph gives Young’s modulus.
II. The area under the force-extension graph gives work done.
III. The area under the stress-strain graph gives elastic energy density.
IV. The slope of the force-extension graph is always equal to Young’s modulus.
The suitable set is
ⓐ. I, II, and IV only
ⓑ. I, III, and IV only
ⓒ. II, III, and IV only
ⓓ. I, II, and III only
Correct Answer: I, II, and III only
Explanation: Statement I is correct because the stress-strain slope in the linear region is \(\frac{\text{stress}}{\text{strain}}=Y\). Statement II is correct because area under a force-extension graph gives work done in stretching. Statement III is correct because area under a stress-strain graph gives energy stored per unit volume. Statement IV is not correct because the force-extension slope is \(\frac{F}{\Delta L}=\frac{AY}{L}\), which depends on the wire dimensions. A graph’s slope or area must be interpreted using the quantities placed on its axes.
307. A cable must carry \(1.2\times10^4\,\text{N}\). Its length is \(5.0\,\text{m}\), \(Y=2.0\times10^{11}\,\text{Pa}\), breaking stress is \(6.0\times10^8\,\text{Pa}\), and the factor of safety is \(3\). If the extension must not exceed \(2.0\,\text{mm}\), the minimum cross-sectional area required is
ⓐ. \(1.5\times10^{-4}\,\text{m}^2\)
ⓑ. \(6.0\times10^{-5}\,\text{m}^2\)
ⓒ. \(3.0\times10^{-5}\,\text{m}^2\)
ⓓ. \(2.4\times10^{-4}\,\text{m}^2\)
Correct Answer: \(1.5\times10^{-4}\,\text{m}^2\)
Explanation: \( \textbf{Strength condition:} \) Safe working stress is:
\[\sigma_w=\frac{\sigma_b}{n}\]
\[\sigma_w=\frac{6.0\times10^8}{3}=2.0\times10^8\,\text{Pa}\]
Area required by strength is:
\[A_1=\frac{F}{\sigma_w}\]
\[A_1=\frac{1.2\times10^4}{2.0\times10^8}\,\text{m}^2\]
\[A_1=6.0\times10^{-5}\,\text{m}^2\]
\( \textbf{Extension condition:} \) Maximum extension is:
\[\Delta L_{\max}=2.0\,\text{mm}=2.0\times10^{-3}\,\text{m}\]
From:
\[\Delta L=\frac{FL}{AY}\]
\[A_2=\frac{FL}{Y\Delta L_{\max}}\]
\[A_2=\frac{(1.2\times10^4)(5.0)}{(2.0\times10^{11})(2.0\times10^{-3})}\,\text{m}^2\]
\[A_2=\frac{6.0\times10^4}{4.0\times10^8}\,\text{m}^2\]
\[A_2=1.5\times10^{-4}\,\text{m}^2\]
The larger area is needed because the cable must satisfy both the strength limit and the extension limit.
\( \textbf{Final answer:} \) \(1.5\times10^{-4}\,\text{m}^2\).
308. A stress-strain curve of a metal is described as follows: the initial part is straight, unloading from a later curved part still gives full recovery, unloading from a still later part leaves a permanent set, and the highest point of stress occurs before final fracture. The correct interpretation is
ⓐ. Hookean, elastic non-linear, plastic, then ultimate-strength behaviour
ⓑ. plastic, Hookean, zero-stress, then recovery after fracture
ⓒ. hydraulic, shear, Poisson-ratio, then volume-strain behaviour
ⓓ. breaking-point, proportional-limit, elastic-limit, then recovery behaviour
Correct Answer: Hookean, elastic non-linear, plastic, then ultimate-strength behaviour
Explanation: The initial straight part of the graph represents Hooke’s law behaviour because stress and strain are proportional there. A later curved part can still be elastic if unloading from it gives full recovery. Once unloading leaves a permanent set, the material has entered plastic behaviour. The highest stress on the curve is the ultimate tensile strength. Final fracture can occur after this highest stress point, especially in a ductile material. The graph must be read using both shape and unloading behaviour, not by one label alone.
309. A solid block is compressed uniformly and also tested in shear. In the compression test, \(\Delta P=9.0\times10^6\,\text{Pa}\), \(V=3.0\times10^{-4}\,\text{m}^3\), and \(\Delta V=-1.5\times10^{-8}\,\text{m}^3\). In the shear test, \(\tau=2.4\times10^6\,\text{Pa}\) and \(\gamma=1.2\times10^{-4}\). The values of \(B\) and \(G\) are respectively
ⓐ. \(2.0\times10^{10}\,\text{Pa}\) and \(1.8\times10^{11}\,\text{Pa}\)
ⓑ. \(6.0\times10^{-15}\,\text{Pa}\) and \(2.9\times10^2\,\text{Pa}\)
ⓒ. \(1.8\times10^{10}\,\text{Pa}\) and \(2.0\times10^{11}\,\text{Pa}\)
ⓓ. \(1.8\times10^{11}\,\text{Pa}\) and \(2.0\times10^{10}\,\text{Pa}\)
Correct Answer: \(1.8\times10^{11}\,\text{Pa}\) and \(2.0\times10^{10}\,\text{Pa}\)
Explanation: \( \textbf{Bulk modulus calculation:} \)
\[B=-\frac{\Delta P}{\Delta V/V}\]
First find the volumetric strain:
\[\frac{\Delta V}{V}=\frac{-1.5\times10^{-8}}{3.0\times10^{-4}}\]
\[\frac{\Delta V}{V}=-0.50\times10^{-4}=-5.0\times10^{-5}\]
Now substitute:
\[B=-\frac{9.0\times10^6}{-5.0\times10^{-5}}\,\text{Pa}\]
\[B=1.8\times10^{11}\,\text{Pa}\]
\( \textbf{Shear modulus calculation:} \)
\[G=\frac{\tau}{\gamma}\]
\[G=\frac{2.4\times10^6}{1.2\times10^{-4}}\,\text{Pa}\]
\[G=2.0\times10^{10}\,\text{Pa}\]
The pressure-volume data determine \(B\), while the tangential stress and shearing strain determine \(G\).
\( \textbf{Final answer:} \) \(1.8\times10^{11}\,\text{Pa}\) and \(2.0\times10^{10}\,\text{Pa}\).
310. A table gives observations from three different materials tested under suitable elastic conditions.
| Observation | Best inference |
|---|
| P. Small strain under a given tensile stress | 1. Large \(Y\) |
| Q. Small fractional volume change under a given pressure rise | 2. Large \(B\) |
| R. Small angular distortion under a given tangential stress | 3. Large \(G\) |
| S. Small working stress compared with breaking stress | 4. Large factor of safety |
The correct matching is
ⓐ. \(P-2\), \(Q-1\), \(R-4\), \(S-3\)
ⓑ. \(P-3\), \(Q-4\), \(R-1\), \(S-2\)
ⓒ. \(P-4\), \(Q-3\), \(R-2\), \(S-1\)
ⓓ. \(P-1\), \(Q-2\), \(R-3\), \(S-4\)
Correct Answer: \(P-1\), \(Q-2\), \(R-3\), \(S-4\)
Explanation: A small tensile strain under a given tensile stress means the material has a large Young’s modulus \(Y\). A small fractional volume change under pressure indicates a large bulk modulus \(B\). A small angular distortion under tangential stress indicates a large shear modulus \(G\). A working stress far below the breaking stress means the factor of safety is large. The same unit \( \text{Pa} \) for the moduli does not make their physical roles interchangeable.
311. A circular wire of radius \(r\) and length \(L\) stores elastic energy \(U\) when stretched by a force \(F\). A second wire of the same material has radius \(2r\), length \(2L\), and is stretched by the same force. The stored energy in the second wire is
ⓐ. \(\frac{U}{2}\)
ⓑ. \(2U\)
ⓒ. \(\frac{U}{4}\)
ⓓ. \(4U\)
Correct Answer: \(\frac{U}{2}\)
Explanation: \( \textbf{Energy relation:} \) For a wire stretched within the elastic range:
\[U=\frac{F^2L}{2AY}\]
The force \(F\) and Young’s modulus \(Y\) are unchanged.
For the first wire:
\[U\propto\frac{L}{A}\]
If the radius changes from \(r\) to \(2r\), the area becomes:
\[A'=\pi(2r)^2=4A\]
The second wire has:
\[L'=2L\]
Therefore:
\[\frac{U'}{U}=\frac{L'/A'}{L/A}\]
\[\frac{U'}{U}=\frac{(2L)/(4A)}{L/A}\]
\[\frac{U'}{U}=\frac{1}{2}\]
The doubled length increases stored energy, but the four times larger area reduces it more strongly.
\( \textbf{Final answer:} \) \(\frac{U}{2}\).
312. A rod of length \(L\) is heated by \(\Delta T\). Case \(P\): it is free to expand. Case \(Q\): it is completely prevented from expanding. The most accurate comparison is
ⓐ. \(P\) expands freely; \(Q\) develops stress
ⓑ. \(P\) develops larger stress because it expands visibly
ⓒ. \(Q\) has no stress because its length is unchanged
ⓓ. both cases have the same stress for the same temperature rise
Correct Answer: \(P\) expands freely; \(Q\) develops stress
Explanation: A freely heated rod can expand by its natural thermal amount. Since the expansion is allowed, very little thermal stress is produced by the expansion alone. If the rod is completely prevented from expanding, the natural thermal strain is not allowed to appear as a length change. The supports then exert forces on the rod and stress develops. Visible expansion and internal stress are not the same condition; the largest stress can occur when the actual length change is prevented.
313. A wire is stretched to stress \(\sigma\) in the linear elastic range. Its Young’s modulus is \(Y\) and its Poisson’s ratio is \(\nu\). The approximate lateral strain is
ⓐ. \(+\nu\frac{\sigma}{Y}\)
ⓑ. \(-\frac{Y}{\nu\sigma}\)
ⓒ. \(\frac{\sigma Y}{\nu}\)
ⓓ. \(-\nu\frac{\sigma}{Y}\)
Correct Answer: \(-\nu\frac{\sigma}{Y}\)
Explanation: In the linear elastic range, the longitudinal strain is:
\[\epsilon_L=\frac{\sigma}{Y}\]
Poisson’s ratio is:
\[\nu=-\frac{\epsilon_{\text{lat}}}{\epsilon_L}\]
Rearranging:
\[\epsilon_{\text{lat}}=-\nu\epsilon_L\]
Substitute \(\epsilon_L=\frac{\sigma}{Y}\):
\[\epsilon_{\text{lat}}=-\nu\frac{\sigma}{Y}\]
The negative sign shows lateral contraction when the wire is stretched. The expression is dimensionless because \(\frac{\sigma}{Y}\) is a ratio of two quantities measured in \( \text{Pa} \).
314. A tensile test gives the following data for a uniform wire: \(F=250\,\text{N}\), \(L=2.0\,\text{m}\), \(r=0.50\,\text{mm}\), and extension \(0.40\,\text{mm}\). Taking \(\pi=3.14\), Young’s modulus is closest to
ⓐ. \(1.6\times10^{11}\,\text{Pa}\)
ⓑ. \(8.0\times10^{10}\,\text{Pa}\)
ⓒ. \(1.6\times10^{12}\,\text{Pa}\)
ⓓ. \(3.2\times10^{12}\,\text{Pa}\)
Correct Answer: \(1.6\times10^{12}\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) \(F=250\,\text{N}\), \(L=2.0\,\text{m}\), \(r=0.50\,\text{mm}=5.0\times10^{-4}\,\text{m}\), and \(\Delta L=0.40\,\text{mm}=4.0\times10^{-4}\,\text{m}\).
Area of the wire is:
\[A=\pi r^2\]
\[A=3.14(5.0\times10^{-4})^2\,\text{m}^2\]
\[A=3.14(25\times10^{-8})\,\text{m}^2\]
\[A=7.85\times10^{-7}\,\text{m}^2\]
Young’s modulus is:
\[Y=\frac{FL}{A\Delta L}\]
Substitute:
\[Y=\frac{(250)(2.0)}{(7.85\times10^{-7})(4.0\times10^{-4})}\,\text{Pa}\]
Denominator:
\[(7.85\times10^{-7})(4.0\times10^{-4})=3.14\times10^{-10}\]
Therefore:
\[Y=\frac{500}{3.14\times10^{-10}}\,\text{Pa}\]
\[Y\approx1.6\times10^{12}\,\text{Pa}\]
Converting both radius and extension into metres before substitution is necessary.
\( \textbf{Final answer:} \) \(1.6\times10^{12}\,\text{Pa}\).
315. A student says that a material with high breaking stress must always have a high Young’s modulus. The best response is that
ⓐ. breaking stress and Young’s modulus are the same quantity
ⓑ. strength and stiffness are different properties
ⓒ. Young’s modulus is meaningful only after breaking
ⓓ. high breaking stress means strain has dimensions
Correct Answer: strength and stiffness are different properties
Explanation: Breaking stress tells the stress at which the material fails. Young’s modulus tells how much strain is produced by a given stress in the elastic region. A material may be stiff but not very strong, or less stiff but able to undergo larger stress before breaking. These properties are found from different parts or meanings of the stress-strain behaviour. Material selection often needs both stiffness and strength rather than assuming one from the other.
316. A beam is bent downward by a load at its middle. A designer increases the depth of the beam while keeping much of the material near the top and bottom faces. The main reason this improves resistance to bending is that
ⓐ. outer layers are farther from the neutral layer
ⓑ. the neutral layer becomes the only stressed layer
ⓒ. Young’s modulus of the material becomes zero
ⓓ. bending changes into pure volume compression
Correct Answer: outer layers are farther from the neutral layer
Explanation: In a bent beam, layers on one side of the neutral layer are stretched and layers on the other side are compressed. The farther a layer is from the neutral layer, the larger its change in length for the same curvature. Placing material near the top and bottom faces therefore helps resist bending effectively. This is the idea behind shapes such as \(I\)-beams. The neutral layer itself has little change in length, so material near it is not as effective for bending resistance.
317. A wire of volume \(2.0\times10^{-6}\,\text{m}^3\) is stretched in the linear elastic region. The stress is \(1.0\times10^8\,\text{Pa}\), and Young’s modulus is \(2.0\times10^{11}\,\text{Pa}\). The total elastic energy stored is
ⓐ. \(1.0\times10^{-1}\,\text{J}\)
ⓑ. \(5.0\times10^{-2}\,\text{J}\)
ⓒ. \(2.5\times10^4\,\text{J}\)
ⓓ. \(5.0\times10^4\,\text{J}\)
Correct Answer: \(5.0\times10^{-2}\,\text{J}\)
Explanation: \( \textbf{Known data:} \) Volume \(V=2.0\times10^{-6}\,\text{m}^3\).
Stress \(\sigma=1.0\times10^8\,\text{Pa}\).
Young’s modulus \(Y=2.0\times10^{11}\,\text{Pa}\).
Use the energy-density relation:
\[u=\frac{\sigma^2}{2Y}\]
Substitute:
\[u=\frac{(1.0\times10^8)^2}{2(2.0\times10^{11})}\,\text{J m}^{-3}\]
\[u=\frac{1.0\times10^{16}}{4.0\times10^{11}}\,\text{J m}^{-3}\]
\[u=2.5\times10^4\,\text{J m}^{-3}\]
Total energy is:
\[U=uV\]
\[U=(2.5\times10^4)(2.0\times10^{-6})\,\text{J}\]
\[U=5.0\times10^{-2}\,\text{J}\]
The energy density must be multiplied by volume to obtain the total stored energy.
\( \textbf{Final answer:} \) \(5.0\times10^{-2}\,\text{J}\).
318. A material is reported to have \(Y=1.2\times10^{11}\,\text{Pa}\) and \(G=4.8\times10^{10}\,\text{Pa}\). If the material is isotropic, its Poisson’s ratio is
ⓐ. \(0.50\)
ⓑ. \(0.20\)
ⓒ. \(0.25\)
ⓓ. \(1.25\)
Correct Answer: \(0.25\)
Explanation: \( \textbf{Known data:} \) \(Y=1.2\times10^{11}\,\text{Pa}\).
\(G=4.8\times10^{10}\,\text{Pa}\).
For an isotropic elastic material:
\[Y=2G(1+\nu)\]
Divide by \(2G\):
\[\frac{Y}{2G}=1+\nu\]
Rearrange:
\[\nu=\frac{Y}{2G}-1\]
Substitute:
\[\nu=\frac{1.2\times10^{11}}{2(4.8\times10^{10})}-1\]
\[\nu=\frac{1.2\times10^{11}}{9.6\times10^{10}}-1\]
\[\nu=1.25-1\]
\[\nu=0.25\]
The isotropic condition is needed before using this relation among elastic constants.
\( \textbf{Final answer:} \) \(0.25\).
319. A full test record for a cable gives these entries: working load, cross-sectional area, original length, extension, and breaking stress. The pair of quantities needed to check the factor of safety is
ⓐ. original length with extension only
ⓑ. Young’s modulus with original length only
ⓒ. extension with volume only
ⓓ. working stress and breaking stress
Correct Answer: working stress and breaking stress
Explanation: Factor of safety compares breaking stress with working stress. The working stress is found from the working load divided by cross-sectional area, \(\sigma_w=\frac{F}{A}\). The breaking stress is then compared with this working stress using \(n=\frac{\sigma_b}{\sigma_w}\). Original length and extension are useful for strain or Young’s modulus calculations, but they do not directly give the safety factor. Strength checking begins with stress, not with extension alone.
320. A final review table contains four claims.
I. A straight stress-strain graph through the origin implies constant modulus in that range.
II. A larger area under a loading-unloading hysteresis loop means more energy loss per cycle.
III. A larger bulk modulus means a larger fractional volume change for the same pressure change.
IV. A larger factor of safety means the working stress is farther below breaking stress.
The correct set is
ⓐ. I, II, and IV only
ⓑ. I and III only
ⓒ. II and III only
ⓓ. I, II, III, and IV
Correct Answer: I, II, and IV only
Explanation: Statement I is correct because the slope of a straight stress-strain graph is constant, giving a constant elastic modulus in that range. Statement II is correct because the area enclosed by a hysteresis loop represents energy lost during one loading-unloading cycle. Statement III is not correct because a larger bulk modulus means smaller fractional volume change for the same pressure change. Statement IV is correct because factor of safety is the ratio of breaking stress to working stress. These claims connect graph reading, energy loss, volume stiffness, and safe design in one chapter-level check.
