401. A rod of length \(1.0\,\text{m}\) is rigidly fixed between two supports so that its thermal expansion is completely prevented. Its Young's modulus is
\[
Y=2.0\times10^{11}\,\text{Pa},
\]
and its linear-expansion coefficient is
\[
\alpha=1.2\times10^{-5}\,\text{K}^{-1}.
\]
The temperature is raised by \(50\,\text{K}\), and the material remains within the linear elastic range. What thermal stress develops?
ⓐ. \(60\,\text{MPa}\) tensile
ⓑ. \(120\,\text{MPa}\) compressive
ⓒ. \(120\,\text{MPa}\) tensile
ⓓ. \(240\,\text{MPa}\) compressive
Correct Answer: \(120\,\text{MPa}\) compressive
Explanation: If the rod were free, its thermal strain would be
\[
\epsilon_{\text{th}}=\alpha\Delta T
\]
Substitution gives
\[
\epsilon_{\text{th}}
=
(1.2\times10^{-5})(50)
=
6.0\times10^{-4}
\]
The rigid supports prevent this extension.
Therefore an equal-magnitude mechanical strain is imposed in the opposite sense.
\(\textbf{Convert blocked strain into elastic stress:}\)
Within the linear elastic range,
\[
\sigma=Y\epsilon
\]
Hence,
\[
\sigma
=
(2.0\times10^{11})(6.0\times10^{-4})
\]
Thus,
\[
\sigma=1.2\times10^{8}\,\text{Pa}
\]
Therefore,
\[
\sigma=120\,\text{MPa}
\]
Heating makes the rod try to expand, so prevention of that expansion produces compressive stress.
\(\textbf{Final answer:}\) \(120\,\text{MPa}\) compressive.
402. A fully constrained metal rod has
\[
Y=2.0\times10^{11}\,\text{Pa},
\qquad
\alpha=1.0\times10^{-5}\,\text{K}^{-1}.
\]
Its allowable tensile stress magnitude is \(1.0\times10^{8}\,\text{Pa}\). Starting from a stress-free state, by how much may its temperature be lowered before this tensile-stress limit is reached?
ⓐ. \(25\,\text{K}\)
ⓑ. \(50\,\text{K}\)
ⓒ. \(100\,\text{K}\)
ⓓ. \(200\,\text{K}\)
Correct Answer: \(50\,\text{K}\)
Explanation: Cooling makes a free rod tend to contract.
Because the rod is fully constrained, this contraction is prevented.
The resulting stress is tensile.
For complete prevention of thermal strain,
\[
|\sigma|=Y\alpha|\Delta T|
\]
\(\textbf{Apply the allowable-stress condition:}\)
\[
1.0\times10^{8}
=
(2.0\times10^{11})(1.0\times10^{-5})|\Delta T|
\]
The product in the denominator is
\[
(2.0\times10^{11})(1.0\times10^{-5})
=
2.0\times10^{6}
\]
Hence,
\[
|\Delta T|
=
\frac{1.0\times10^{8}}
{2.0\times10^{6}}
=
50\,\text{K}
\]
A larger temperature decrease would exceed the stated tensile-stress limit in the idealised linear model.
\(\textbf{Final answer:}\) Maximum temperature decrease \(=50\,\text{K}\).
403. Two metal rods A and B lie collinearly between rigid outer walls, with a \(3.0\,\text{mm}\) gap between their free inner ends. Initially neither rod is stressed. Their data are
\[
L_A=1.0\,\text{m},\quad \alpha_A=1.0\times10^{-5}\,\text{K}^{-1},
\]
\[
L_B=2.0\,\text{m},\quad \alpha_B=2.0\times10^{-5}\,\text{K}^{-1}.
\]
Both rods are heated uniformly by the same amount while their inner ends remain free until contact. At what temperature rise do the rods just close the gap?
ⓐ. \(30\,\text{K}\)
ⓑ. \(50\,\text{K}\)
ⓒ. \(60\,\text{K}\)
ⓓ. \(100\,\text{K}\)
Correct Answer: \(60\,\text{K}\)
Explanation: Before the rods touch, each rod expands freely into the gap.
The gap closes when the sum of their free expansions equals
\[
3.0\,\text{mm}=3.0\times10^{-3}\,\text{m}
\]
\(\textbf{Expansion of rod A:}\)
\[
\Delta L_A=\alpha_AL_A\Delta T
\]
Thus,
\[
\Delta L_A
=
(1.0\times10^{-5})(1.0)\Delta T
\]
\(\textbf{Expansion of rod B:}\)
\[
\Delta L_B
=
(2.0\times10^{-5})(2.0)\Delta T
=
4.0\times10^{-5}\Delta T
\]
Therefore,
\[
\Delta L_A+\Delta L_B
=
5.0\times10^{-5}\Delta T
\]
At first contact,
\[
5.0\times10^{-5}\Delta T
=
3.0\times10^{-3}
\]
Hence,
\[
\Delta T=60\,\text{K}
\]
No thermal stress need be introduced before contact because both inner ends are free to move into the gap.
\(\textbf{Final answer:}\)
\[
\boxed{60\,\text{K}}
\]
404. Consider a freshwater lake cooling from a temperature above \(4^\circ\text{C}\) during winter. Which sequence best represents the role of the anomalous expansion of water?
ⓐ. Surface water cools below \(4^\circ\text{C}\), becomes continuously denser, sinks to the bottom, and the lake therefore freezes first from the bottom
ⓑ. Water cooling toward \(4^\circ\text{C}\) becomes denser and sinks; water cooled below \(4^\circ\text{C}\) becomes less dense, remains near the surface, and the lake can freeze from the top
ⓒ. Water has minimum density at \(4^\circ\text{C}\), so \(4^\circ\text{C}\) water remains permanently at the surface
ⓓ. Water expands continuously on cooling from room temperature to \(0^\circ\text{C}\), so no density-driven circulation occurs
Correct Answer: Water cooling toward \(4^\circ\text{C}\) becomes denser and sinks; water cooled below \(4^\circ\text{C}\) becomes less dense, remains near the surface, and the lake can freeze from the top
Explanation: Water behaves anomalously near \(0^\circ\text{C}\) to \(4^\circ\text{C}\).
As water cools from temperatures above \(4^\circ\text{C}\) toward \(4^\circ\text{C}\), its density increases.
The denser surface water therefore tends to sink and is replaced by warmer water from below.
This circulation helps much of the lake approach approximately \(4^\circ\text{C}\).
Water has its maximum density near \(4^\circ\text{C}\).
When surface water cools below \(4^\circ\text{C}\), it begins to expand and becomes less dense.
That colder water therefore tends to remain near the surface instead of sinking.
Continued cooling can then produce ice at the top while deeper water remains liquid.
This is why the anomalous expansion of water is central to the familiar top-down freezing of lakes and ponds.
Hence option B gives the correct sequence.
405.
The lengths of two freely expanding rods A and B are plotted against temperature \(t\) in degree Celsius. Both plots are straight lines described by \(L_A=1.000+2.0\times10^{-5}t\) metre and \(L_B=1.002+1.0\times10^{-5}t\) metre over the stated temperature range.
At what temperature do the rods have equal lengths, and which rod has the larger coefficient of linear expansion?
ⓐ. \(100^\circ\text{C}\), rod A
ⓑ. \(200^\circ\text{C}\), rod A
ⓒ. \(200^\circ\text{C}\), rod B
ⓓ. \(400^\circ\text{C}\), rod A
Correct Answer: \(200^\circ\text{C}\), rod A
Explanation: The graph equations contain both the reference lengths and the temperature slopes.
For a relation
\[
L=L_0(1+\alpha t),
\]
the graph slope equals
\[
L_0\alpha
\]
\(\textbf{Compare the expansion coefficients:}\)
For A,
\[
\alpha_A=
\frac{2.0\times10^{-5}}{1.000}
=
2.0\times10^{-5}\,\text{K}^{-1}
\]
For B,
\[
\alpha_B=
\frac{1.0\times10^{-5}}{1.002}
\approx9.98\times10^{-6}\,\text{K}^{-1}
\]
Thus A has the larger expansion coefficient.
\(\textbf{Find the intersection of the two length-temperature lines:}\)
At equal lengths,
\[
1.000+2.0\times10^{-5}t
=
1.002+1.0\times10^{-5}t
\]
Therefore,
\[
1.0\times10^{-5}t=0.002
\]
Hence,
\[
t=200^\circ\text{C}
\]
Rod B begins longer, but A expands more rapidly and catches up.
\(\textbf{Final answer:}\) \(200^\circ\text{C}\), and rod A has the larger coefficient.
406. At \(20^\circ\text{C}\), a circular hole in a metal ring has diameter \(4.000\,\text{cm}\), while a metal shaft has diameter \(4.004\,\text{cm}\). The coefficients of linear expansion of the ring and shaft are respectively
\[
2.0\times10^{-5}\,\text{K}^{-1}
\quad\text{and}\quad
1.0\times10^{-5}\,\text{K}^{-1}.
\]
Both are heated together through the same temperature rise. Approximately at what temperature will the shaft just pass through the ring?
ⓐ. \(70^\circ\text{C}\)
ⓑ. \(100^\circ\text{C}\)
ⓒ. \(120^\circ\text{C}\)
ⓓ. \(220^\circ\text{C}\)
Correct Answer: \(120^\circ\text{C}\)
Explanation: Both the hole and shaft expand, so the shaft expansion must not be neglected.
Let the common temperature rise be \(\Delta T\).
\(\textbf{Expanded hole diameter:}\)
\[
d_r=4.000(1+2.0\times10^{-5}\Delta T)
\]
\(\textbf{Expanded shaft diameter:}\)
\[
d_s=4.004(1+1.0\times10^{-5}\Delta T)
\]
For a just fit,
\[
d_r=d_s
\]
Thus,
\[
4.000+8.000\times10^{-5}\Delta T
=
4.004+4.004\times10^{-5}\Delta T
\]
Therefore,
\[
3.996\times10^{-5}\Delta T
=
0.004
\]
Hence,
\[
\Delta T\approx100\,\text{K}
\]
The initial temperature is
\[
20^\circ\text{C}
\]
Therefore,
\[
T\approx20+100=120^\circ\text{C}
\]
The larger expansion coefficient of the ring allows its hole eventually to catch up with the expanding shaft.
\(\textbf{Final answer:}\) Approximately \(120^\circ\text{C}\).
407. A metal rod of length \(1.0\,\text{m}\), cross-sectional area \(2.0\times10^{-4}\,\text{m}^2\), coefficient of linear expansion \(2.0\times10^{-5}\,\text{K}^{-1}\), and Young's modulus \(1.0\times10^{11}\,\text{Pa}\) is fixed at one end. Initially there is a \(1.0\,\text{mm}\) gap between its other end and a rigid wall. The rod is heated uniformly through \(100\,\text{K}\). Assume linear elasticity after contact. What compressive stress and force develop at the final temperature?
ⓐ. \(5.0\times10^{7}\,\text{Pa}\) and \(1.0\times10^{4}\,\text{N}\)
ⓑ. \(1.0\times10^{8}\,\text{Pa}\) and \(2.0\times10^{4}\,\text{N}\)
ⓒ. \(2.0\times10^{8}\,\text{Pa}\) and \(2.0\times10^{4}\,\text{N}\)
ⓓ. No stress develops because the rod initially had a gap
Correct Answer: \(1.0\times10^{8}\,\text{Pa}\) and \(2.0\times10^{4}\,\text{N}\)
Explanation: The rod first expands freely until it closes the gap.
Only the additional thermal expansion after contact is mechanically prevented.
\(\textbf{Free expansion for the full temperature rise:}\)
\[
\Delta L_{\text{free}}
=
\alpha L\Delta T
\]
Thus,
\[
\Delta L_{\text{free}}
=
(2.0\times10^{-5})(1.0)(100)
=
2.0\times10^{-3}\,\text{m}
\]
Hence the free expansion would be
\[
2.0\,\text{mm}
\]
The available gap absorbs
\[
1.0\,\text{mm}
\]
Therefore the prevented part of the extension is
\[
1.0\,\text{mm}
=
1.0\times10^{-3}\,\text{m}
\]
\(\textbf{Mechanical strain created by the wall constraint:}\)
\[
|\epsilon|
=
\frac{1.0\times10^{-3}}{1.0}
=
1.0\times10^{-3}
\]
Hence,
\[
|\sigma|=Y|\epsilon|
\]
Thus,
\[
|\sigma|
=
(1.0\times10^{11})(1.0\times10^{-3})
=
1.0\times10^{8}\,\text{Pa}
\]
The stress is compressive because heating is being restrained.
\(\textbf{Force on the wall:}\)
\[
F=\sigma A
\]
Therefore,
\[
F=
(1.0\times10^{8})(2.0\times10^{-4})
=
2.0\times10^{4}\,\text{N}
\]
\(\textbf{Final answer:}\) Compressive stress \(=1.0\times10^{8}\,\text{Pa}\), force \(=2.0\times10^{4}\,\text{N}\).
408. A rod of length \(2.0\,\text{m}\) and cross-sectional area \(1.0\times10^{-4}\,\text{m}^2\) is completely prevented from expanding. Its Young's modulus and coefficient of linear expansion are
\[
Y=2.0\times10^{11}\,\text{Pa},
\qquad
\alpha=1.0\times10^{-5}\,\text{K}^{-1}.
\]
It is heated through \(50\,\text{K}\) while remaining in the linear elastic range. What are the magnitude of the force exerted on each support and the elastic strain energy stored in the rod?
ⓐ. \(5.0\times10^{3}\,\text{N}\) and \(2.5\,\text{J}\)
ⓑ. \(2.0\times10^{4}\,\text{N}\) and \(5.0\,\text{J}\)
ⓒ. \(1.0\times10^{4}\,\text{N}\) and \(10\,\text{J}\)
ⓓ. \(1.0\times10^{4}\,\text{N}\) and \(5.0\,\text{J}\)
Correct Answer: \(1.0\times10^{4}\,\text{N}\) and \(5.0\,\text{J}\)
Explanation: The thermal strain that would occur freely is
\[
\epsilon_{\text{th}}=\alpha\Delta T
\]
Thus,
\[
\epsilon_{\text{th}}
=
(1.0\times10^{-5})(50)
=
5.0\times10^{-4}
\]
Because expansion is completely prevented, an equal-magnitude elastic strain is produced in compression.
\(\textbf{Thermal stress:}\)
\[
\sigma=Y\epsilon
\]
Therefore,
\[
\sigma
=
(2.0\times10^{11})(5.0\times10^{-4})
=
1.0\times10^{8}\,\text{Pa}
\]
\(\textbf{Support force:}\)
\[
F=\sigma A
\]
Thus,
\[
F=
(1.0\times10^{8})(1.0\times10^{-4})
=
1.0\times10^{4}\,\text{N}
\]
\(\textbf{Elastic energy density:}\)
Within the linear elastic range,
\[
u=\frac12\sigma\epsilon
\]
Hence,
\[
u=
\frac12(1.0\times10^{8})(5.0\times10^{-4})
=
2.5\times10^{4}\,\text{J m}^{-3}
\]
The rod volume is
\[
V=AL=(1.0\times10^{-4})(2.0)
=2.0\times10^{-4}\,\text{m}^3
\]
Therefore,
\[
U=uV
=
(2.5\times10^{4})(2.0\times10^{-4})
=
5.0\,\text{J}
\]
\(\textbf{Final answer:}\) Force \(=1.0\times10^{4}\,\text{N}\), elastic energy \(=5.0\,\text{J}\).
409. Rods A and B are joined end to end and are free to expand. Their data are
\[
L_A=1.0\,\text{m},\quad \alpha_A=1.0\times10^{-5}\,\text{K}^{-1},
\]
\[
L_B=2.0\,\text{m},\quad \alpha_B=2.0\times10^{-5}\,\text{K}^{-1}.
\]
The combination is heated uniformly through \(100\,\text{K}\). What are the increase in total length and the effective coefficient of linear expansion of the whole \(3.0\,\text{m}\) combination?
ⓐ. \(5.0\,\text{mm}\) and \(1.67\times10^{-5}\,\text{K}^{-1}\)
ⓑ. \(3.0\,\text{mm}\) and \(1.0\times10^{-5}\,\text{K}^{-1}\)
ⓒ. \(5.0\,\text{mm}\) and \(2.50\times10^{-5}\,\text{K}^{-1}\)
ⓓ. \(6.0\,\text{mm}\) and \(2.0\times10^{-5}\,\text{K}^{-1}\)
Correct Answer: \(5.0\,\text{mm}\) and \(1.67\times10^{-5}\,\text{K}^{-1}\)
Explanation: The two rods are free, so their thermal expansions simply add.
\(\textbf{Expansion of A:}\)
\[
\Delta L_A
=
\alpha_AL_A\Delta T
\]
Thus,
\[
\Delta L_A
=
(1.0\times10^{-5})(1.0)(100)
=
1.0\times10^{-3}\,\text{m}
\]
Hence,
\[
\Delta L_A=1.0\,\text{mm}
\]
\(\textbf{Expansion of B:}\)
\[
\Delta L_B
=
(2.0\times10^{-5})(2.0)(100)
=
4.0\times10^{-3}\,\text{m}
\]
Thus,
\[
\Delta L_B=4.0\,\text{mm}
\]
Therefore,
\[
\Delta L_{\text{total}}
=
5.0\,\text{mm}
\]
Define the effective coefficient by
\[
\Delta L_{\text{total}}
=
\alpha_{\text{eff}}L_{\text{total}}\Delta T
\]
Hence,
\[
\alpha_{\text{eff}}
=
\frac{5.0\times10^{-3}}
{(3.0)(100)}
\]
Therefore,
\[
\alpha_{\text{eff}}
\approx1.67\times10^{-5}\,\text{K}^{-1}
\]
The effective coefficient is length-weighted rather than the simple arithmetic mean of the two coefficients.
\(\textbf{Final answer:}\)
\[
\boxed{5.0\,\text{mm},\quad 1.67\times10^{-5}\,\text{K}^{-1}}
\]
410. At \(20^\circ\text{C}\), a circular metal plug fits exactly into a circular hole in a different metal plate. The plug and hole both have diameter \(4.000\,\text{cm}\). Their linear-expansion coefficients are
\[
\alpha_{\text{plug}}=2.0\times10^{-5}\,\text{K}^{-1},
\qquad
\alpha_{\text{plate}}=1.0\times10^{-5}\,\text{K}^{-1}.
\]
Both are heated freely through \(100\,\text{K}\). Which statement is correct?
ⓐ. The hole becomes \(0.004\,\text{cm}\) larger than the plug
ⓑ. The plug becomes \(0.004\,\text{cm}\) larger in diameter than the hole
ⓒ. They remain an exact fit because both dimensions expand
ⓓ. The hole contracts by \(0.004\,\text{cm}\)
Correct Answer: The plug becomes \(0.004\,\text{cm}\) larger in diameter than the hole
Explanation: A hole expands according to the thermal-expansion coefficient of the plate.
\(\textbf{Expansion of the plug:}\)
\[
\Delta d_p
=
\alpha_pd\Delta T
\]
Thus,
\[
\Delta d_p
=
(2.0\times10^{-5})(4.000)(100)
=
0.008\,\text{cm}
\]
\(\textbf{Expansion of the hole:}\)
\[
\Delta d_h
=
(1.0\times10^{-5})(4.000)(100)
=
0.004\,\text{cm}
\]
Therefore the final plug diameter exceeds the final hole diameter by
\[
0.008-0.004
=
0.004\,\text{cm}
\]
The common misconception is to treat the hole as contracting when the surrounding plate expands.
Instead, the hole itself expands, but here it expands less than the plug because its material has the smaller coefficient.
\(\textbf{Final answer:}\) The plug exceeds the hole diameter by \(0.004\,\text{cm}\).
411. An isotropic solid is heated through \(100\,\text{K}\). Its mass remains constant, and its density decreases by approximately \(0.60\%\). Assuming small expansion, what is its coefficient of linear expansion?
ⓐ. \(0.5\times10^{-5}\,\text{K}^{-1}\)
ⓑ. \(1.0\times10^{-5}\,\text{K}^{-1}\)
ⓒ. \(2.0\times10^{-5}\,\text{K}^{-1}\)
ⓓ. \(6.0\times10^{-5}\,\text{K}^{-1}\)
Correct Answer: \(2.0\times10^{-5}\,\text{K}^{-1}\)
Explanation: For unchanged mass,
\[
\rho=\frac{m}{V}
\]
For a small increase in volume, the fractional density decrease is approximately equal to the fractional volume increase:
\[
\left|\frac{\Delta\rho}{\rho}\right|
\approx
\frac{\Delta V}{V}
\]
For an isotropic solid,
\[
\frac{\Delta V}{V}
\approx
3\alpha\Delta T
\]
The observed density decrease is
\[
0.60\%=6.0\times10^{-3}
\]
Therefore,
\[
6.0\times10^{-3}
=
3\alpha(100)
\]
Hence,
\[
\alpha
=
\frac{6.0\times10^{-3}}{300}
\]
Thus,
\[
\alpha=2.0\times10^{-5}\,\text{K}^{-1}
\]
This is a reverse-inference use of density change rather than a direct length measurement.
\(\textbf{Final answer:}\)
\[
\boxed{\alpha=2.0\times10^{-5}\,\text{K}^{-1}}
\]
412.
A graph of the volume of a fixed mass of water against temperature from \(0^\circ\text{C}\) to \(10^\circ\text{C}\) decreases from \(0^\circ\text{C}\) to a minimum at \(4^\circ\text{C}\), then increases from \(4^\circ\text{C}\) to \(10^\circ\text{C}\).
Which conclusion necessarily follows from this graph?
ⓐ. Water has minimum density at \(4^\circ\text{C}\)
ⓑ. Water at \(4^\circ\text{C}\) is denser than the same mass of water at both \(0^\circ\text{C}\) and \(10^\circ\text{C}\)
ⓒ. Water at \(0^\circ\text{C}\) must be denser than water at \(10^\circ\text{C}\)
ⓓ. Water expands continuously throughout the interval \(0^\circ\text{C}\) to \(10^\circ\text{C}\)
Correct Answer: Water at \(4^\circ\text{C}\) is denser than the same mass of water at both \(0^\circ\text{C}\) and \(10^\circ\text{C}\)
Explanation: The graph refers to a fixed mass of water.
Density is
\[
\rho=\frac{m}{V}.
\]
For fixed \(m\), density is greatest when volume is smallest.
The graph states that the minimum volume occurs at
\[
4^\circ\text{C}.
\]
Therefore water has its maximum density near \(4^\circ\text{C}\).
Its volume at both \(0^\circ\text{C}\) and \(10^\circ\text{C}\) is greater than the minimum value at \(4^\circ\text{C}\).
Hence its density at each of those temperatures is lower than at \(4^\circ\text{C}\).
The graph as described does not specify whether the \(0^\circ\text{C}\) volume is greater or smaller than the \(10^\circ\text{C}\) volume, so their mutual density ordering cannot be inferred.
Water also contracts, rather than expands, when heated from \(0^\circ\text{C}\) to \(4^\circ\text{C}\).
Hence option B is the necessary conclusion.
413. A lake is initially nearly uniform at \(8^\circ\text{C}\). During cooling, its surface water first reaches \(6^\circ\text{C}\), and much later, after the lake has approached \(4^\circ\text{C}\), the surface cools further to \(2^\circ\text{C}\). Which sequence is physically consistent with the anomalous expansion of water?
ⓐ. The \(6^\circ\text{C}\) surface layer tends to remain above \(8^\circ\text{C}\) water, and the \(2^\circ\text{C}\) layer sinks below \(4^\circ\text{C}\) water
ⓑ. The \(6^\circ\text{C}\) layer tends to sink through \(8^\circ\text{C}\) water, whereas the \(2^\circ\text{C}\) layer tends to remain above \(4^\circ\text{C}\) water
ⓒ. Both the \(6^\circ\text{C}\) and \(2^\circ\text{C}\) layers sink because colder water is always denser
ⓓ. Neither layer moves because water density is independent of temperature
Correct Answer: The \(6^\circ\text{C}\) layer tends to sink through \(8^\circ\text{C}\) water, whereas the \(2^\circ\text{C}\) layer tends to remain above \(4^\circ\text{C}\) water
Explanation: Above \(4^\circ\text{C}\), cooling water increases its density.
Therefore surface water cooling from \(8^\circ\text{C}\) to \(6^\circ\text{C}\) becomes denser than the warmer water beneath it.
That denser surface layer tends to sink, promoting circulation.
This process helps the lake approach approximately
\[
4^\circ\text{C}.
\]
Water has maximum density near \(4^\circ\text{C}\).
When the surface cools further from \(4^\circ\text{C}\) toward \(2^\circ\text{C}\), it behaves anomalously and becomes less dense.
The \(2^\circ\text{C}\) water therefore tends to remain above the denser \(4^\circ\text{C}\) water.
Continued cooling can consequently lead to freezing first at the surface.
Thus the direction of buoyant motion reverses on crossing the temperature of maximum density.
414. Two completely constrained rods A and B are made of the same material and undergo the same temperature rise. Rod A has length \(L\) and area \(A\). Rod B has length \(2L\) and area \(A/2\). Assume linear elasticity. If the compressive stresses are \(\sigma_A,\sigma_B\), the support forces are \(F_A,F_B\), and the elastic energies stored are \(U_A,U_B\), which relation is correct?
ⓐ. \(\sigma_B=\sigma_A,\ F_B=F_A/2,\ U_B=U_A\)
ⓑ. \(\sigma_B=2\sigma_A,\ F_B=F_A,\ U_B=2U_A\)
ⓒ. \(\sigma_B=\sigma_A/2,\ F_B=F_A/4,\ U_B=U_A/2\)
ⓓ. \(\sigma_B=\sigma_A,\ F_B=2F_A,\ U_B=2U_A\)
Correct Answer: \(\sigma_B=\sigma_A,\ F_B=F_A/2,\ U_B=U_A\)
Explanation: For complete prevention of thermal expansion,
\[
\sigma=Y\alpha\Delta T
\]
Both rods have the same material and the same temperature rise.
Therefore,
\[
\sigma_A=\sigma_B
\]
\(\textbf{Compare support forces:}\)
\[
F=\sigma A
\]
Rod B has half the area, so
\[
F_B=\sigma\frac A2=\frac{F_A}{2}
\]
\(\textbf{Compare stored elastic energies:}\)
The elastic energy density is
\[
u=\frac12\sigma\epsilon
\]
Since both rods have the same stress and blocked strain, they have the same energy density.
Their volumes are
\[
V_A=AL
\]
and
\[
V_B=\left(\frac A2\right)(2L)=AL
\]
Thus,
\[
V_A=V_B
\]
Equal energy density and equal volume imply
\[
U_A=U_B
\]
The thermal stress itself is independent of rod length and area in this fully constrained idealisation.
\(\textbf{Final answer:}\)
\[
\boxed{\sigma_B=\sigma_A,\ F_B=F_A/2,\ U_B=U_A}
\]
415.
For a fully constrained rod, a graph of compressive stress magnitude \(\sigma\) against temperature rise \(\Delta T\) is a straight line through the origin with slope \(2.4\,\text{MPa K}^{-1}\).
The rod has Young's modulus \(2.0\times10^{11}\,\text{Pa}\) and cross-sectional area \(3.0\times10^{-4}\,\text{m}^2\). What are its coefficient of linear expansion and the force on each support after a \(40\,\text{K}\) rise?
ⓐ. \(1.2\times10^{-5}\,\text{K}^{-1}\) and \(28.8\,\text{kN}\)
ⓑ. \(2.4\times10^{-5}\,\text{K}^{-1}\) and \(14.4\,\text{kN}\)
ⓒ. \(1.2\times10^{-5}\,\text{K}^{-1}\) and \(9.6\,\text{kN}\)
ⓓ. \(4.8\times10^{-6}\,\text{K}^{-1}\) and \(28.8\,\text{kN}\)
Correct Answer: \(1.2\times10^{-5}\,\text{K}^{-1}\) and \(28.8\,\text{kN}\)
Explanation: For a completely constrained rod,
\[
\sigma=Y\alpha\Delta T
\]
Therefore the slope of the \(\sigma\)-versus-\(\Delta T\) graph is
\[
Y\alpha
\]
\(\textbf{Infer the expansion coefficient from the graph slope:}\)
\[
Y\alpha
=
2.4\,\text{MPa K}^{-1}
=
2.4\times10^6\,\text{Pa K}^{-1}
\]
Thus,
\[
\alpha
=
\frac{2.4\times10^6}
{2.0\times10^{11}}
\]
Hence,
\[
\alpha=1.2\times10^{-5}\,\text{K}^{-1}
\]
\(\textbf{Stress after a 40\,\text{K} rise:}\)
Using the graph slope,
\[
\sigma
=
(2.4\,\text{MPa K}^{-1})(40\,\text{K})
=
96\,\text{MPa}
\]
Thus,
\[
\sigma=9.6\times10^7\,\text{Pa}
\]
\(\textbf{Support force:}\)
\[
F=\sigma A
\]
Therefore,
\[
F
=
(9.6\times10^7)(3.0\times10^{-4})
=
2.88\times10^4\,\text{N}
\]
Hence,
\[
F=28.8\,\text{kN}
\]
\(\textbf{Final answer:}\)
\[
\boxed{\alpha=1.2\times10^{-5}\,\text{K}^{-1},\quad F=28.8\,\text{kN}}
\]
416. Successive steel rails, each \(20\,\text{m}\) long, are laid with an expansion gap of \(15\,\text{mm}\) at a temperature of \(10^\circ\text{C}\). The coefficient of linear expansion of steel is \(1.2\times10^{-5}\,\text{K}^{-1}\). Neglect any restraint. What gap remains when the temperature reaches \(60^\circ\text{C}\)?
ⓐ. \(1\,\text{mm}\)
ⓑ. \(12\,\text{mm}\)
ⓒ. \(6\,\text{mm}\)
ⓓ. \(3\,\text{mm}\)
Correct Answer: \(3\,\text{mm}\)
Explanation: The temperature rise is
\[
\Delta T=60-10=50\,\text{K}
\]
Each rail attempts to expand by
\[
\Delta L=\alpha L\Delta T
\]
Substitution gives
\[
\Delta L
=
(1.2\times10^{-5})(20)(50)
\]
Hence,
\[
\Delta L
=
1.2\times10^{-2}\,\text{m}
\]
Therefore,
\[
\Delta L=12\,\text{mm}
\]
In a long sequence of identical freely expanding rails, each joint gap must accommodate approximately the expansion associated with one rail length.
Thus the remaining gap is
\[
15-12=3\,\text{mm}
\]
The gap prevents significant compressive thermal stress as long as it remains open.
\(\textbf{Final answer:}\)
\[
\boxed{3\,\text{mm}}
\]
417. A metal rod is exactly \(1.0000\,\text{m}\) long at \(20^\circ\text{C}\) and has coefficient of linear expansion
\[
\alpha_r=2.0\times10^{-5}\,\text{K}^{-1}.
\]
It is measured at \(120^\circ\text{C}\) using a metal scale that was correct at \(20^\circ\text{C}\) and has coefficient
\[
\alpha_s=1.0\times10^{-5}\,\text{K}^{-1}.
\]
Assuming both rod and scale are at \(120^\circ\text{C}\), what reading does the expanded scale give approximately?
ⓐ. \(0.9990\,\text{m}\)
ⓑ. \(1.0000\,\text{m}\)
ⓒ. \(1.0010\,\text{m}\)
ⓓ. \(1.0020\,\text{m}\)
Correct Answer: \(1.0010\,\text{m}\)
Explanation: The rod and the scale both expand, but by different fractional amounts.
\(\textbf{Actual rod length at 120^\circ\text{C}:}\)
The temperature rise is
\[
100\,\text{K}
\]
Thus,
\[
L_r
=
1.0000(1+\alpha_r\Delta T)
\]
Hence,
\[
L_r
=
1.0000(1+2.0\times10^{-3})
=
1.0020\,\text{m}
\]
\(\textbf{Actual length represented by one nominal metre of the hot scale:}\)
\[
L_{\text{scale metre}}
=
1.0000(1+\alpha_s\Delta T)
\]
Thus,
\[
L_{\text{scale metre}}
=
1.0010\,\text{m}
\]
Therefore the indicated reading is
\[
R=
\frac{1.0020}{1.0010}\,\text{m}
\]
Hence,
\[
R\approx1.0010\,\text{m}
\]
The scale itself has expanded, so it does not report the rod's actual hot length of \(1.0020\,\text{m}\) in its original calibration units.
\(\textbf{Final answer:}\) Approximately \(1.0010\,\text{m}\).
418. A solid cube has side \(10.0\,\text{cm}\) and coefficient of linear expansion \(2.0\times10^{-5}\,\text{K}^{-1}\). It contains a spherical cavity of volume \(100\,\text{cm}^3\). The solid is heated freely through \(100\,\text{K}\). Assuming small isotropic expansion, which pair correctly gives the approximate increases in the cube's outer volume and the cavity volume?
ⓐ. \(2.0\,\text{cm}^3\) and \(0.20\,\text{cm}^3\)
ⓑ. \(4.0\,\text{cm}^3\) and \(0.40\,\text{cm}^3\)
ⓒ. \(6.0\,\text{cm}^3\) and \(0.60\,\text{cm}^3\)
ⓓ. \(6.0\,\text{cm}^3\) and zero
Correct Answer: \(6.0\,\text{cm}^3\) and \(0.60\,\text{cm}^3\)
Explanation: For an isotropic solid,
\[
\beta\approx3\alpha
\]
Thus,
\[
\beta
=
3(2.0\times10^{-5})
=
6.0\times10^{-5}\,\text{K}^{-1}
\]
\(\textbf{Outer cube volume change:}\)
The initial outer volume is
\[
V_o=(10.0)^3=1000\,\text{cm}^3
\]
Therefore,
\[
\Delta V_o
=
\beta V_o\Delta T
\]
Thus,
\[
\Delta V_o
=
(6.0\times10^{-5})(1000)(100)
=
6.0\,\text{cm}^3
\]
\(\textbf{Expansion of the cavity:}\)
A cavity expands as though it were filled with the same material.
Therefore,
\[
\Delta V_c
=
\beta V_c\Delta T
\]
Hence,
\[
\Delta V_c
=
(6.0\times10^{-5})(100)(100)
=
0.60\,\text{cm}^3
\]
The cavity does not remain fixed merely because it contains no solid material.
\(\textbf{Final answer:}\) Outer-volume increase \(=6.0\,\text{cm}^3\), cavity-volume increase \(=0.60\,\text{cm}^3\).
419. A rod is fully constrained between rigid supports. At the initial temperature it already carries a tensile stress of \(40\,\text{MPa}\). Its Young's modulus is \(2.0\times10^{11}\,\text{Pa}\) and coefficient of linear expansion is \(1.0\times10^{-5}\,\text{K}^{-1}\). It is then heated through \(30\,\text{K}\), with linear superposition valid. What is its final stress?
ⓐ. \(20\,\text{MPa}\) tensile
ⓑ. \(20\,\text{MPa}\) compressive
ⓒ. \(100\,\text{MPa}\) tensile
ⓓ. \(100\,\text{MPa}\) compressive
Correct Answer: \(20\,\text{MPa}\) compressive
Explanation: Heating a fully constrained rod produces a compressive thermal stress.
\(\textbf{Magnitude of the added thermal stress:}\)
\[
\sigma_{\text{th}}=Y\alpha\Delta T
\]
Substitution gives
\[
\sigma_{\text{th}}
=
(2.0\times10^{11})
(1.0\times10^{-5})
(30)
\]
Therefore,
\[
\sigma_{\text{th}}
=
6.0\times10^{7}\,\text{Pa}
=
60\,\text{MPa}
\]
The initial stress is
\[
40\,\text{MPa}
\]
tensile.
The thermal contribution is
\[
60\,\text{MPa}
\]
compressive.
Taking tensile as positive,
\[
\sigma_f=+40-60=-20\,\text{MPa}
\]
The negative sign indicates compression.
The heating first cancels the initial tension and then produces a net compressive state.
\(\textbf{Final answer:}\) \(20\,\text{MPa}\) compressive.
420. Consider the following statements about thermal expansion.
Statement I: A circular hole in a uniformly heated isotropic plate expands rather than contracts.
Statement II: A freely expanding rod develops no thermal stress solely because its temperature changes, provided no mechanical restraint is present.
Statement III: For a fully constrained rod in the linear elastic range, the thermal-stress magnitude \(Y\alpha\Delta T\) is independent of the rod's length.
Statement IV: For an ordinary isotropic solid of fixed mass, thermal expansion generally lowers its density.
Statement V: Water has maximum density near \(4^\circ\text{C}\).
Statement VI: For a small temperature change in an isotropic solid, the volume-expansion coefficient is approximately \(2\alpha\), where \(\alpha\) is the linear-expansion coefficient.
Which combination is correct?
ⓐ. I, II, III, IV and V only
ⓑ. I, II, IV and VI only
ⓒ. II, III, V and VI only
ⓓ. I, II, III, IV, V and VI
Correct Answer: I, II, III, IV and V only
Explanation: Statement I is correct because every linear dimension of a uniformly heated free isotropic plate scales upward, including the dimensions of a hole.
Statement II is correct because thermal stress requires mechanical restraint; free expansion itself does not create elastic stress.
Statement III is also correct in the fully constrained one-dimensional idealisation:
\[
\sigma=Y\alpha\Delta T.
\]
The rod length does not appear in this expression.
Statement IV is correct because ordinary thermal expansion increases volume while mass remains unchanged, so
\[
\rho=\frac{m}{V}
\]
generally decreases.
Statement V is correct and expresses the well-known anomalous behaviour of water near \(4^\circ\text{C}\).
Statement VI is incorrect.
For small isotropic expansion,
\[
\alpha_A\approx2\alpha
\]
for area expansion, whereas
\[
\alpha_V\approx3\alpha
\]
for volume expansion.
Hence Statements I, II, III, IV and V only are correct.
