101. In a wire, the stress at a certain point is \(\sigma\) and the corresponding strain is \(\epsilon\) in the linear elastic region. The elastic energy density is
ⓐ. \(\sigma+\epsilon\)
ⓑ. \(\frac{1}{2}\sigma\epsilon\)
ⓒ. \(\frac{\sigma}{\epsilon}\)
ⓓ. \(\frac{1}{2}\frac{\sigma}{\epsilon}\)
Correct Answer: \(\frac{1}{2}\sigma\epsilon\)
Explanation: Elastic energy density is the energy stored per unit volume. On a stress-strain graph, it is represented by the area under the graph. In the linear elastic region, the graph is a triangle with height \(\sigma\) and base \(\epsilon\). The triangular area is \(\frac{1}{2}\times\sigma\times\epsilon\). The ratio \(\frac{\sigma}{\epsilon}\) gives Young’s modulus, not energy density.
102. A material has Young’s modulus \(2.0\times10^{11}\,\text{Pa}\) and is stretched to a strain of \(1.0\times10^{-3}\) within the elastic limit. The elastic energy density is
ⓐ. \(2.0\times10^8\,\text{J m}^{-3}\)
ⓑ. \(1.0\times10^{-5}\,\text{J m}^{-3}\)
ⓒ. \(4.0\times10^5\,\text{J m}^{-3}\)
ⓓ. \(1.0\times10^5\,\text{J m}^{-3}\)
Correct Answer: \(1.0\times10^5\,\text{J m}^{-3}\)
Explanation: \( \textbf{Known data:} \) Young’s modulus \(Y=2.0\times10^{11}\,\text{Pa}\).
Strain \(\epsilon=1.0\times10^{-3}\).
\( \textbf{Required quantity:} \) Elastic energy density \(u\).
In the linear elastic region:
\[u=\frac{1}{2}Y\epsilon^2\]
This form is useful because \(Y\) and strain are directly given.
Substitute:
\[u=\frac{1}{2}(2.0\times10^{11})(1.0\times10^{-3})^2\]
Square the strain:
\[(1.0\times10^{-3})^2=1.0\times10^{-6}\]
Now calculate:
\[u=1.0\times10^{11}\times10^{-6}\,\text{J m}^{-3}\]
\[u=1.0\times10^5\,\text{J m}^{-3}\]
Energy density has unit \( \text{J m}^{-3} \), the same dimensional form as \( \text{Pa} \).
\( \textbf{Final answer:} \) \(1.0\times10^5\,\text{J m}^{-3}\).
103. A material is stretched in its linear elastic region. If the stress is \(\sigma\) and Young’s modulus is \(Y\), the elastic energy density can also be written as
ⓐ. \(\frac{2Y}{\sigma^2}\)
ⓑ. \(\frac{\sigma}{2Y^2}\)
ⓒ. \(\frac{\sigma^2}{2Y}\)
ⓓ. \(2\sigma Y\)
Correct Answer: \(\frac{\sigma^2}{2Y}\)
Explanation: In the linear elastic region, elastic energy density is the area under the stress-strain graph. The basic expression is \(u=\frac{1}{2}\sigma\epsilon\). For a material obeying Hooke’s law, \(\sigma=Y\epsilon\), so \(\epsilon=\frac{\sigma}{Y}\). Substituting this into the energy-density expression gives \(u=\frac{1}{2}\sigma\left(\frac{\sigma}{Y}\right)\). Therefore \(u=\frac{\sigma^2}{2Y}\). This form is useful when stress and Young’s modulus are known but strain is not directly given.
104. A wire is stretched within the elastic limit. The elastic energy density is \(4.0\times10^5\,\text{J m}^{-3}\), and the volume of the stretched part is \(2.5\times10^{-6}\,\text{m}^3\). The total elastic energy stored is
ⓐ. \(1.6\times10^{11}\,\text{J}\)
ⓑ. \(4.0\times10^5\,\text{J}\)
ⓒ. \(2.5\times10^{-6}\,\text{J}\)
ⓓ. \(1.0\,\text{J}\)
Correct Answer: \(1.0\,\text{J}\)
Explanation: \( \textbf{Known data:} \) Energy density \(u=4.0\times10^5\,\text{J m}^{-3}\).
Volume \(V=2.5\times10^{-6}\,\text{m}^3\).
\( \textbf{Required quantity:} \) Total elastic energy \(U\).
Energy density means energy per unit volume:
\[u=\frac{U}{V}\]
Rearranging:
\[U=uV\]
Substitute the values:
\[U=(4.0\times10^5)(2.5\times10^{-6})\,\text{J}\]
Multiply the numerical factors:
\[4.0\times2.5=10.0\]
Combine powers of ten:
\[10^5\times10^{-6}=10^{-1}\]
\[U=10.0\times10^{-1}\,\text{J}=1.0\,\text{J}\]
\( \textbf{Final answer:} \) \(1.0\,\text{J}\).
105. Two rods are under the same tensile stress in the linear elastic region. Rod \(P\) has Young’s modulus \(Y\), while rod \(Q\) has Young’s modulus \(2Y\). The elastic energy density in \(Q\), compared with that in \(P\), is
ⓐ. half as large
ⓑ. twice as large
ⓒ. four times as large
ⓓ. the same
Correct Answer: half as large
Explanation: \( \textbf{Useful relation:} \) For a given stress in the linear elastic region,
\[u=\frac{\sigma^2}{2Y}\]
The stress is the same for both rods, so \(\sigma^2\) is unchanged.
For rod \(P\):
\[u_P=\frac{\sigma^2}{2Y}\]
For rod \(Q\), the modulus is \(2Y\):
\[u_Q=\frac{\sigma^2}{2(2Y)}\]
\[u_Q=\frac{\sigma^2}{4Y}\]
Compare with \(u_P=\frac{\sigma^2}{2Y}\):
\[u_Q=\frac{1}{2}u_P\]
A stiffer material stores less elastic energy density under the same stress because it develops less strain.
\( \textbf{Final answer:} \) half as large.
106. A stress-strain graph is linear from the origin up to a working point. At the working point, the stress is \(8.0\times10^7\,\text{Pa}\) and the strain is \(2.0\times10^{-4}\). The elastic energy density is
ⓐ. \(1.6\times10^4\,\text{J m}^{-3}\)
ⓑ. \(8.0\times10^3\,\text{J m}^{-3}\)
ⓒ. \(4.0\times10^{11}\,\text{J m}^{-3}\)
ⓓ. \(4.0\times10^{-12}\,\text{J m}^{-3}\)
Correct Answer: \(8.0\times10^3\,\text{J m}^{-3}\)
Explanation: \( \textbf{Graph meaning:} \) Energy density is the area under the stress-strain graph.
Since the graph is a straight line from the origin, the area is triangular.
Stress at the point:
\[\sigma=8.0\times10^7\,\text{Pa}\]
Strain at the point:
\[\epsilon=2.0\times10^{-4}\]
The energy density is:
\[u=\frac{1}{2}\sigma\epsilon\]
Substitute:
\[u=\frac{1}{2}(8.0\times10^7)(2.0\times10^{-4})\]
First multiply:
\[(8.0\times10^7)(2.0\times10^{-4})=16.0\times10^3\]
Now take half:
\[u=8.0\times10^3\,\text{J m}^{-3}\]
The factor \(\frac{1}{2}\) comes from the triangular graph area, not from unit conversion.
\( \textbf{Final answer:} \) \(8.0\times10^3\,\text{J m}^{-3}\).
107. Bulk modulus describes a material’s resistance to
ⓐ. change in volume under hydraulic stress
ⓑ. change in colour under heating
ⓒ. change in electric charge under voltage
ⓓ. change in mass under gravity
Correct Answer: change in volume under hydraulic stress
Explanation: Bulk modulus is the elastic modulus connected with volume change. It compares hydraulic stress, or pressure change, with volumetric strain. A material with a large bulk modulus undergoes only a small fractional volume change for a given pressure change. This makes it different from Young’s modulus, which deals with length change, and shear modulus, which deals with shape change. The word “bulk” points to the whole volume of the body rather than one length or one surface layer.
108. For a solid compressed uniformly by an increase in pressure \(\Delta P\), the bulk modulus is commonly written as \(B=-\frac{\Delta P}{\Delta V/V}\). The negative sign is used because
ⓐ. pressure is always a negative physical quantity
ⓑ. pressure increase reduces volume
ⓒ. volume has no SI unit
ⓓ. bulk modulus must be negative for solids
Correct Answer: pressure increase reduces volume
Explanation: When external pressure on a solid is increased, the volume usually decreases. With the usual sign convention, a decrease in volume means \(\Delta V\) is negative. The pressure increase \(\Delta P\) is positive, so \(\frac{\Delta P}{\Delta V/V}\) would be negative without the extra sign. The negative sign is included so that \(B\) becomes a positive material constant for ordinary compression. The sign records the opposite directions of pressure increase and volume change.
109. A solid of volume \(V\) is compressed by pressure so that its volume changes by \(\Delta V\). The ratio \(\frac{\Delta V}{V}\) in the bulk modulus formula represents
ⓐ. longitudinal stress
ⓑ. Young’s modulus
ⓒ. volumetric strain
ⓓ. shearing stress
Correct Answer: volumetric strain
Explanation: Volumetric strain is the fractional change in volume of a body. It is written as \(\frac{\Delta V}{V}\), where \(V\) is the original volume and \(\Delta V\) is the change in volume. In bulk modulus, this volumetric strain is compared with hydraulic stress or pressure change. Longitudinal stress involves force per area in stretching or compression along a length. Shearing stress involves tangential force per area and is not measured by \(\frac{\Delta V}{V}\).
110. The SI unit and dimensional formula of bulk modulus are respectively
ⓐ. \( \text{m} \) and \([L]\)
ⓑ. \( \text{Pa} \) and \([ML^{-1}T^{-2}]\)
ⓒ. dimensionless and \([M^0L^0T^0]\)
ⓓ. \( \text{J} \) and \([ML^2T^{-2}]\)
Correct Answer: \( \text{Pa} \) and \([ML^{-1}T^{-2}]\)
Explanation: Bulk modulus is hydraulic stress divided by volumetric strain. Hydraulic stress has the same unit as pressure, which is \( \text{Pa} \) or \( \text{N m}^{-2} \). Volumetric strain is dimensionless because it is a ratio of two volumes. Therefore bulk modulus has the same unit and dimensions as stress. Its dimensional formula is \([ML^{-1}T^{-2}]\), just like pressure, Young’s modulus, and shear modulus.
111. A liquid-like pressure is applied equally on all faces of a small solid cube. The cube’s volume decreases slightly but its shape remains almost similar. This situation is the most suitable setup for measuring
ⓐ. bulk modulus
ⓑ. Young’s modulus only
ⓒ. shear modulus only
ⓓ. Poisson’s ratio only
Correct Answer: bulk modulus
Explanation: Equal pressure on all faces produces hydraulic stress. The main deformation is a change in volume rather than a change in one length or a sideways distortion. Bulk modulus relates this pressure change to the fractional volume change. Young’s modulus would be more suitable for stretching or compressing a wire along one direction. Shear modulus would be used when tangential forces change shape with little volume change.
112. A material has a large bulk modulus. Under the same pressure increase, compared with a material of smaller bulk modulus, it will show
ⓐ. larger fractional volume change
ⓑ. zero stress in every case
ⓒ. larger strain because it is stiffer
ⓓ. smaller fractional volume change
Correct Answer: smaller fractional volume change
Explanation: Bulk modulus measures resistance to volume change. From the magnitude relation \(B=\frac{\Delta P}{|\Delta V|/V}\), a larger \(B\) means a smaller value of \(\frac{|\Delta V|}{V}\) for the same pressure change. Such a material is less compressible. This does not mean the material experiences zero stress; it means the deformation produced by that stress is smaller. The idea is parallel to stiffness in stretching, but here the deformation is volumetric.
113. In a compression test, pressure on a solid is increased by \(2.0\times10^7\,\text{Pa}\). The fractional decrease in volume is \(4.0\times10^{-5}\). The bulk modulus is
ⓐ. \(5.0\times10^{11}\,\text{Pa}\)
ⓑ. \(8.0\times10^2\,\text{Pa}\)
ⓒ. \(5.0\times10^{-12}\,\text{Pa}\)
ⓓ. \(2.0\times10^7\,\text{Pa}\)
Correct Answer: \(5.0\times10^{11}\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) Pressure increase \(\Delta P=2.0\times10^7\,\text{Pa}\).
Fractional decrease in volume:
\[\frac{|\Delta V|}{V}=4.0\times10^{-5}\]
\( \textbf{Required quantity:} \) Bulk modulus \(B\).
For magnitude calculation:
\[B=\frac{\Delta P}{|\Delta V|/V}\]
Substitute:
\[B=\frac{2.0\times10^7}{4.0\times10^{-5}}\,\text{Pa}\]
Divide numerical factors:
\[\frac{2.0}{4.0}=0.50\]
Divide powers of ten:
\[10^7/10^{-5}=10^{12}\]
Therefore:
\[B=0.50\times10^{12}\,\text{Pa}=5.0\times10^{11}\,\text{Pa}\]
The fractional volume change has no unit, so \(B\) keeps the unit \( \text{Pa} \).
\( \textbf{Final answer:} \) \(5.0\times10^{11}\,\text{Pa}\).
114. A report gives \(\Delta P=+6.0\times10^6\,\text{Pa}\), \(V=3.0\times10^{-3}\,\text{m}^3\), and \(\Delta V=-1.5\times10^{-8}\,\text{m}^3\). The bulk modulus is
ⓐ. \(-1.2\times10^{12}\,\text{Pa}\)
ⓑ. \(3.0\times10^{-15}\,\text{Pa}\)
ⓒ. \(1.2\times10^{12}\,\text{Pa}\)
ⓓ. \(4.0\times10^6\,\text{Pa}\)
Correct Answer: \(1.2\times10^{12}\,\text{Pa}\)
Explanation: \( \textbf{Known data:} \) \(\Delta P=+6.0\times10^6\,\text{Pa}\).
Original volume:
\[V=3.0\times10^{-3}\,\text{m}^3\]
Change in volume:
\[\Delta V=-1.5\times10^{-8}\,\text{m}^3\]
\( \textbf{Required quantity:} \) Bulk modulus \(B\).
Use:
\[B=-\frac{\Delta P}{\Delta V/V}\]
First calculate volumetric strain:
\[\frac{\Delta V}{V}=\frac{-1.5\times10^{-8}}{3.0\times10^{-3}}\]
\[\frac{\Delta V}{V}=-0.50\times10^{-5}=-5.0\times10^{-6}\]
Now substitute:
\[B=-\frac{6.0\times10^6}{-5.0\times10^{-6}}\,\text{Pa}\]
\[B=1.2\times10^{12}\,\text{Pa}\]
The two negative signs cancel because compression gives a negative volume change but a positive modulus.
\( \textbf{Final answer:} \) \(1.2\times10^{12}\,\text{Pa}\).
115. Consider the following statements about bulk modulus.
I. It relates pressure change to volumetric strain.
II. Its SI unit is \( \text{Pa} \).
III. A larger value usually means the material is less compressible.
IV. It is dimensionless because volumetric strain is dimensionless.
The suitable set is
ⓐ. I, II, and III only
ⓑ. I, II, and IV only
ⓒ. I, III, and IV only
ⓓ. II, III, and IV only
Correct Answer: I, II, and III only
Explanation: Statement I is correct because bulk modulus compares hydraulic stress with volumetric strain. Statement II is correct because pressure has unit \( \text{Pa} \), and volumetric strain has no unit. Statement III is correct because a large \(B\) means only a small fractional volume change occurs for a given pressure change. Statement IV is not correct because only the denominator is dimensionless; the numerator is pressure. Bulk modulus therefore has the same dimensions as stress, not zero dimensions.
116. Compressibility \(K_c\) is related to bulk modulus \(B\) by
ⓐ. \(K_c=\frac{1}{B}\)
ⓑ. \(K_c=B^2\)
ⓒ. \(K_c=2B\)
ⓓ. \(K_c=\frac{B}{2}\)
Correct Answer: \(K_c=\frac{1}{B}\)
Explanation: Compressibility measures how easily a material undergoes volume change under pressure. Bulk modulus measures resistance to volume change. Since these ideas are opposite in meaning, compressibility is the reciprocal of bulk modulus. Thus \(K_c=\frac{1}{B}\). A material with a high \(B\) has low compressibility, while a material with a low \(B\) is more compressible.
117. The unit of compressibility is
ⓐ. \( \text{Pa} \)
ⓑ. \( \text{N} \)
ⓒ. \( \text{J m}^{-3} \)
ⓓ. \( \text{Pa}^{-1} \)
Correct Answer: \( \text{Pa}^{-1} \)
Explanation: Compressibility is the reciprocal of bulk modulus. Since bulk modulus has the unit \( \text{Pa} \), compressibility has the unit \( \text{Pa}^{-1} \). This is also clear from its meaning as fractional volume change per unit pressure change. Fractional volume change is dimensionless, so the remaining unit is inverse pressure. Energy density may share the dimensional form of pressure, but compressibility uses the reciprocal unit.
118. Material \(P\) has bulk modulus \(8.0\times10^{10}\,\text{Pa}\), while material \(Q\) has bulk modulus \(2.0\times10^{10}\,\text{Pa}\). The compressibility of \(Q\), compared with that of \(P\), is
ⓐ. four times as large
ⓑ. one-fourth as large
ⓒ. twice as large
ⓓ. the same
Correct Answer: four times as large
Explanation: \( \textbf{Relation needed:} \) Compressibility is:
\[K_c=\frac{1}{B}\]
For material \(P\):
\[K_{cP}=\frac{1}{8.0\times10^{10}}\]
For material \(Q\):
\[K_{cQ}=\frac{1}{2.0\times10^{10}}\]
Compare \(Q\) with \(P\):
\[\frac{K_{cQ}}{K_{cP}}=\frac{1/(2.0\times10^{10})}{1/(8.0\times10^{10})}\]
\[\frac{K_{cQ}}{K_{cP}}=\frac{8.0\times10^{10}}{2.0\times10^{10}}=4\]
The material with smaller bulk modulus is more compressible.
\( \textbf{Final answer:} \) four times as large.
119. A solid has bulk modulus \(1.5\times10^{11}\,\text{Pa}\). Its compressibility is closest to
ⓐ. \(1.5\times10^{11}\,\text{Pa}^{-1}\)
ⓑ. \(3.0\times10^{11}\,\text{Pa}^{-1}\)
ⓒ. \(7.5\times10^{10}\,\text{Pa}^{-1}\)
ⓓ. \(6.7\times10^{-12}\,\text{Pa}^{-1}\)
Correct Answer: \(6.7\times10^{-12}\,\text{Pa}^{-1}\)
Explanation: \( \textbf{Known data:} \) Bulk modulus:
\[B=1.5\times10^{11}\,\text{Pa}\]
\( \textbf{Required quantity:} \) Compressibility \(K_c\).
The relation is:
\[K_c=\frac{1}{B}\]
Substitute:
\[K_c=\frac{1}{1.5\times10^{11}}\,\text{Pa}^{-1}\]
Since \(\frac{1}{1.5}=0.667\):
\[K_c=0.667\times10^{-11}\,\text{Pa}^{-1}\]
\[K_c=6.67\times10^{-12}\,\text{Pa}^{-1}\]
To two significant figures, this is \(6.7\times10^{-12}\,\text{Pa}^{-1}\).
\( \textbf{Final answer:} \) \(6.7\times10^{-12}\,\text{Pa}^{-1}\).
120. For many ordinary comparisons, gases are more compressible than liquids and solids. In terms of bulk modulus, this means gases usually have
ⓐ. larger bulk modulus
ⓑ. smaller bulk modulus
ⓒ. infinite shear modulus
ⓓ. zero volumetric strain under pressure
Correct Answer: smaller bulk modulus
Explanation: Compressibility is the reciprocal of bulk modulus. A more compressible substance undergoes a larger fractional volume change for the same pressure change. Therefore it has a smaller bulk modulus. Gases usually show much larger volume changes under pressure than liquids and solids. This comparison concerns volume change and should not be mixed with shear rigidity or tensile stiffness.
