101. A rod of length \(2.5\,\text{m}\) has \(\alpha=1.6\times10^{-5}\,\text{K}^{-1}\). It is heated from \(30^\circ\text{C}\) to \(130^\circ\text{C}\). What is its final length?
ⓐ. \(2.5004\,\text{m}\)
ⓑ. \(2.5400\,\text{m}\)
ⓒ. \(2.5040\,\text{m}\)
ⓓ. \(2.4960\,\text{m}\)
Correct Answer: \(2.5040\,\text{m}\)
Explanation: \( \textbf{Original length:} \) \(L=2.5\,\text{m}\).
\( \textbf{Coefficient:} \) \(\alpha=1.6\times10^{-5}\,\text{K}^{-1}\).
The temperature change is
\[
\Delta T=130^\circ\text{C}-30^\circ\text{C}=100\,\text{K}
\]
First find the increase in length:
\[
\Delta L=\alpha L\Delta T
\]
\[
\Delta L=(1.6\times10^{-5})(2.5)(100)
\]
\[
\Delta L=4.0\times10^{-3}\,\text{m}
\]
The final length is
\[
L'=L+\Delta L
\]
\[
L'=2.5+0.0040=2.5040\,\text{m}
\]
The value \(0.0040\,\text{m}\) is the increase, not the final length.
\( \textbf{Final answer:} \) The final length is \(2.5040\,\text{m}\).
102. Linear expansion formulas are usually written for small temperature changes. The main reason is that:
ⓐ. \(\alpha\) is nearly constant over the range
ⓑ. heat has no unit for large temperature changes
ⓒ. all solids stop expanding at low temperature
ⓓ. length cannot be measured at high temperature
Correct Answer: \(\alpha\) is nearly constant over the range
Explanation: The relation \(\Delta L=\alpha L\Delta T\) is a simple linear approximation. It assumes that the coefficient of linear expansion \(\alpha\) remains nearly constant over the temperature interval being considered. For a very large temperature range, \(\alpha\) may change with temperature, and the expansion may not remain exactly linear. This does not mean solids cannot expand at high temperature. It also does not mean heat loses its unit. The formula is most reliable when the material remains in the same state and the temperature range is small enough for \(\alpha\) to be treated as constant.
103. An arrangement uses two rods of equal original length joined end to end. Rod \(P\) has coefficient \(\alpha\), and rod \(Q\) has coefficient \(2\alpha\). Both experience the same temperature rise \(\Delta T\). If rod \(P\) expands by \(x\), the total increase in length of the joined arrangement is:
ⓐ. \(4x\)
ⓑ. \(2x\)
ⓒ. \(x\)
ⓓ. \(3x\)
Correct Answer: \(3x\)
Explanation: \( \textbf{For rod \(P\):} \) The increase in length is given as \(x\).
For equal original length \(L\) and the same temperature rise \(\Delta T\),
\[
\Delta L=\alpha L\Delta T
\]
Rod \(Q\) has coefficient \(2\alpha\), while \(L\) and \(\Delta T\) are the same as for rod \(P\).
Therefore, rod \(Q\)'s expansion is
\[
\Delta L_Q=2\alpha L\Delta T=2x
\]
The total increase of the joined arrangement is the sum of the increases:
\[
\Delta L_{\text{total}}=x+2x=3x
\]
The rods are joined end to end, so their length changes add directly.
\( \textbf{Final answer:} \) The total increase in length is \(3x\).
104. Area expansion of a thin isotropic metal sheet is described by:
ⓐ. \(\Delta A=\frac{\beta\Delta T}{A}\)
ⓑ. \(\Delta A=A+\beta+\Delta T\)
ⓒ. \(\Delta A=\beta A\Delta T\)
ⓓ. \(\Delta A=\alpha L\Delta T\)
Correct Answer: \(\Delta A=\beta A\Delta T\)
Explanation: Area expansion deals with the change in surface area of a solid sheet when its temperature changes. The coefficient of area expansion is denoted by \(\beta\). For small temperature changes, the change in area is proportional to the original area \(A\), the temperature change \(\Delta T\), and the coefficient \(\beta\). This gives \(\Delta A=\beta A\Delta T\). The linear expansion formula uses \(\alpha\) and length \(L\), so it is not the direct relation for area change. For isotropic solids, \(\beta\) is approximately related to \(\alpha\), but the area formula must still use the original area.
105. For an isotropic solid, the coefficient of area expansion \(\beta\) is approximately:
ⓐ. \(2\alpha\)
ⓑ. \(3\alpha\)
ⓒ. \(\alpha\)
ⓓ. \(\frac{\alpha}{2}\)
Correct Answer: \(2\alpha\)
Explanation: In an isotropic solid, expansion is the same in all directions. Area involves two perpendicular length directions. If each linear dimension increases fractionally by about \(\alpha\Delta T\), the fractional increase in area is approximately twice that value for small temperature changes. Therefore, \(\beta\approx2\alpha\). This approximation ignores the very small second-order term \((\alpha\Delta T)^2\). The result applies to isotropic solids and small temperature changes, not to every material under all conditions.
106. A square metal plate is heated uniformly. What happens to a circular hole at the centre of the plate?
ⓐ. The hole expands as if it were made of the same metal
ⓑ. The hole disappears because metal fills it during heating
ⓒ. The hole contracts because surrounding metal expands inward
ⓓ. The hole remains exactly unchanged while the outer boundary expands
Correct Answer: The hole expands as if it were made of the same metal
Explanation: When a plate is heated uniformly, all its linear dimensions increase. A hole in the plate behaves as though the material occupying the hole also expanded. Therefore, the diameter and area of the hole increase. The surrounding metal does not simply expand inward to close the hole. This result is consistent with the idea that the whole geometrical pattern of the plate expands. The same reasoning is used for rings, cavities, and holes in heated solids.
107. The coefficient of area expansion \(\beta\) is best interpreted as:
ⓐ. total final area after heating
ⓑ. decrease in mass per unit rise in temperature
ⓒ. fractional area change per unit temperature rise
ⓓ. increase in area per unit heat supplied to the sheet
Correct Answer: fractional area change per unit temperature rise
Explanation: Area expansion describes how the surface area of a solid changes when its temperature changes. The coefficient \(\beta\) is defined through \(\Delta A=\beta A\Delta T\). Rearranging gives \(\beta=\frac{\Delta A}{A\Delta T}\), so it measures fractional change in area per unit temperature rise. It is not the final area itself. It also does not describe heat supplied directly, because expansion depends on temperature change and material response. The word fractional is important because \(\Delta A\) must be compared with the original area \(A\).
108. A metal sheet has original area \(A\) and coefficient of area expansion \(\beta\). After a small temperature rise \(\Delta T\), its final area is:
ⓐ. \(A'=A(1+\beta\Delta T)\)
ⓑ. \(A'=\frac{A}{1+\beta\Delta T}\)
ⓒ. \(A'=\beta A\Delta T\)
ⓓ. \(A'=A(1-\beta\Delta T)\)
Correct Answer: \(A'=A(1+\beta\Delta T)\)
Explanation: The area expansion formula gives the change in area:
\[
\Delta A=\beta A\Delta T
\]
The final area is the original area plus the increase in area:
\[
A'=A+\Delta A
\]
Substituting the expression for \(\Delta A\),
\[
A'=A+\beta A\Delta T
\]
Taking \(A\) common,
\[
A'=A(1+\beta\Delta T)
\]
The expression \(\beta A\Delta T\) alone gives only the extra area, not the total area after heating. For normal heating of most solids, \(\Delta T\) is positive and the final area is larger than the original area.
109. A thin metal plate of area \(0.50\,\text{m}^2\) is heated through \(50\,\text{K}\). If \(\beta=4.0\times10^{-5}\,\text{K}^{-1}\), what is the increase in area?
ⓐ. \(2.0\times10^{-3}\,\text{m}^2\)
ⓑ. \(1.0\times10^{-2}\,\text{m}^2\)
ⓒ. \(4.0\times10^{-3}\,\text{m}^2\)
ⓓ. \(1.0\times10^{-3}\,\text{m}^2\)
Correct Answer: \(1.0\times10^{-3}\,\text{m}^2\)
Explanation: \( \textbf{Given data:} \) \(A=0.50\,\text{m}^2\), \(\Delta T=50\,\text{K}\), and \(\beta=4.0\times10^{-5}\,\text{K}^{-1}\).
\( \textbf{Required quantity:} \) Increase in area \(\Delta A\).
For area expansion,
\[
\Delta A=\beta A\Delta T
\]
This relation applies because the question asks for the change in surface area of a heated plate.
Substitute the values:
\[
\Delta A=(4.0\times10^{-5})(0.50)(50)
\]
First multiply \(0.50\times50=25\).
So,
\[
\Delta A=(4.0\times10^{-5})(25)
\]
\[
\Delta A=100\times10^{-5}\,\text{m}^2
\]
\[
\Delta A=1.0\times10^{-3}\,\text{m}^2
\]
The unit \(\text{K}^{-1}\) cancels with \(\text{K}\), leaving \(\text{m}^2\).
\( \textbf{Final answer:} \) The increase in area is \(1.0\times10^{-3}\,\text{m}^2\).
110. A square metal plate is heated uniformly. Its side length increases by a small fraction \(x\). The approximate fractional increase in its area is:
ⓐ. \(3x\)
ⓑ. \(\frac{x}{2}\)
ⓒ. \(2x\)
ⓓ. \(x\)
Correct Answer: \(2x\)
Explanation: A square plate has area \(A=l^2\). If each side increases by a small fraction \(x\), the new side is approximately \(l(1+x)\). The new area becomes \(l^2(1+x)^2\). For small \(x\), \((1+x)^2\approx1+2x\), because the term \(x^2\) is very small. Therefore, the fractional increase in area is approximately \(2x\). This is the reason why, for isotropic solids, \(\beta\approx2\alpha\).
111. A graph is drawn with increase in area \(\Delta A\) on the vertical axis and temperature rise \(\Delta T\) on the horizontal axis for a fixed metal sheet. The graph is a straight line through the origin. Its slope represents:
ⓐ. \(\beta A\)
ⓑ. \(\frac{\beta}{A}\)
ⓒ. \(\frac{A}{\beta}\)
ⓓ. \(A+\beta\)
Correct Answer: \(\beta A\)
Explanation: \( \textbf{Area expansion relation:} \)
\[
\Delta A=\beta A\Delta T
\]
For a graph of \(\Delta A\) against \(\Delta T\), the equation has the form \(y=mx\).
Here, \(\Delta A\) is the vertical quantity and \(\Delta T\) is the horizontal quantity.
The coefficient multiplying \(\Delta T\) is \(\beta A\).
Therefore, the graph slope is
\[
\frac{\Delta A}{\Delta T}=\beta A
\]
For the same material, a sheet of larger original area gives a steeper graph.
\( \textbf{Final answer:} \) The slope represents \(\beta A\).
112. A circular hole is made in a thin metal sheet. The sheet is heated uniformly through a small temperature rise. If the original area of the hole is \(A_h\), the increase in the hole’s area is best represented as:
ⓐ. \(\Delta A_h=\beta\Delta T\) only
ⓑ. \(\Delta A_h=\frac{A_h}{\beta\Delta T}\)
ⓒ. \(\Delta A_h=\beta A_h\Delta T\)
ⓓ. \(\Delta A_h=-\beta A_h\Delta T\)
Correct Answer: \(\Delta A_h=\beta A_h\Delta T\)
Explanation: A hole in a uniformly heated sheet expands as if the hole were filled with the same material. This means the dimensions of the hole follow the same expansion pattern as the surrounding sheet. Therefore, the area of the hole increases on heating. The area increase is proportional to the original hole area \(A_h\), the coefficient of area expansion \(\beta\), and the temperature rise \(\Delta T\). The negative sign would represent contraction, not normal heating expansion. The original area cannot be omitted because a larger hole undergoes a larger actual increase in area.
113. Study the table and identify the row that correctly connects linear and area expansion for an isotropic solid.
| Row | Condition | Approximate relation |
|---|
| P | Small temperature change, isotropic solid | \(\beta\approx2\alpha\) |
| Q | Small temperature change, isotropic solid | \(\beta\approx\frac{\alpha}{2}\) |
| R | Any material under all conditions | \(\beta=0\) |
| S | Area expansion of a sheet | \(\beta\) has unit \(\text{m}^2\) |
ⓐ. Row S only
ⓑ. Row R only
ⓒ. Row Q only
ⓓ. Row P only
Correct Answer: Row P only
Explanation: For an isotropic solid, expansion is the same in every direction. Since area involves two independent length directions, the fractional change in area is approximately twice the fractional change in length for small temperature changes. Therefore, \(\beta\approx2\alpha\). The relation is approximate because very small second-order terms are neglected. Row R is wrong because area expansion is not zero for ordinary expanding solids. Row S is wrong because \(\beta\) has unit \(\text{K}^{-1}\), not \(\text{m}^2\).
114. A rectangular metal sheet has dimensions \(20\,\text{cm}\times30\,\text{cm}\). It is heated so that each linear dimension increases by \(0.1\%\). The approximate percentage increase in area is:
ⓐ. \(0.1\%\)
ⓑ. \(0.2\%\)
ⓒ. \(1.0\%\)
ⓓ. \(0.05\%\)
Correct Answer: \(0.2\%\)
Explanation: \( \textbf{Given fractional linear increase:} \) Each linear dimension increases by \(0.1\%\).
For a rectangular sheet, area depends on two perpendicular lengths.
If each length increases by a small fraction \(x\), the fractional area increase is approximately \(2x\).
Here,
\[
x=0.1\%
\]
So the approximate percentage increase in area is
\[
2x=2(0.1\%)=0.2\%
\]
The actual dimensions \(20\,\text{cm}\) and \(30\,\text{cm}\) are not needed for the percentage change because both sides have the same fractional increase.
The small product term \(x^2\) is neglected in the usual first-order expansion approximation.
\( \textbf{Final answer:} \) The approximate percentage increase in area is \(0.2\%\).
115. Volume expansion of an isotropic solid is described by:
ⓐ. \(\Delta V=\alpha L\Delta T\)
ⓑ. \(\Delta V=\frac{\gamma\Delta T}{V}\)
ⓒ. \(\Delta V=\beta A\Delta T\)
ⓓ. \(\Delta V=\gamma V\Delta T\)
Correct Answer: \(\Delta V=\gamma V\Delta T\)
Explanation: Volume expansion deals with the change in volume of a body when its temperature changes. The coefficient of volume expansion is denoted by \(\gamma\). For small temperature changes, the change in volume is proportional to original volume \(V\), temperature change \(\Delta T\), and the coefficient \(\gamma\). Thus the relation is \(\Delta V=\gamma V\Delta T\). The formulas using \(\alpha\) and \(\beta\) apply to length and area changes respectively. Selecting the correct coefficient depends on whether the question asks about length, surface area, or volume.
116. For a solid of original volume \(V\), coefficient of volume expansion \(\gamma\), and temperature rise \(\Delta T\), the final volume is:
ⓐ. \(V'=V(1+\gamma\Delta T)\)
ⓑ. \(V'=V(1-\gamma\Delta T)\)
ⓒ. \(V'=\gamma V\Delta T\)
ⓓ. \(V'=\frac{V}{\gamma\Delta T}\)
Correct Answer: \(V'=V(1+\gamma\Delta T)\)
Explanation: The relation \(\Delta V=\gamma V\Delta T\) gives only the change in volume. The final volume is obtained by adding the original volume and the volume increase. Therefore,
\[
V'=V+\Delta V
\]
Substitute \(\Delta V=\gamma V\Delta T\):
\[
V'=V+\gamma V\Delta T
\]
Taking \(V\) common,
\[
V'=V(1+\gamma\Delta T)
\]
For a normal solid heated through a small positive \(\Delta T\), the final volume is larger than the original volume. Confusing \(V'\) with \(\Delta V\) is a common formula-selection error.
117. The coefficient of volume expansion \(\gamma\) has the same unit as \(\alpha\) and \(\beta\) because:
ⓐ. it measures temperature without using a scale
ⓑ. fractional volume change per unit temperature rise
ⓒ. heat supplied per unit mass and unit temperature rise
ⓓ. it measures volume directly
Correct Answer: fractional volume change per unit temperature rise
Explanation: From \(\Delta V=\gamma V\Delta T\), the ratio \(\frac{\Delta V}{V}\) is a fractional change and has no unit. Therefore, \(\gamma\Delta T\) must also be dimensionless. Since \(\Delta T\) has unit \(\text{K}\), \(\gamma\) must have unit \(\text{K}^{-1}\). The same reasoning applies to \(\alpha\) and \(\beta\), because they also represent fractional expansion per unit temperature rise. The coefficient is not a volume, so its unit is not \(\text{m}^3\). The unit \(\text{K}^{-1}\) shows how expansion is linked to each unit rise in temperature.
118. For an isotropic solid, the coefficient of volume expansion \(\gamma\) is approximately:
ⓐ. \(3\alpha\)
ⓑ. \(\alpha\)
ⓒ. \(\frac{\alpha}{3}\)
ⓓ. \(2\alpha\)
Correct Answer: \(3\alpha\)
Explanation: In an isotropic solid, expansion is equal in all directions. Volume depends on three mutually perpendicular linear dimensions. If each linear dimension has a small fractional increase \(\alpha\Delta T\), then the fractional increase in volume is approximately \(3\alpha\Delta T\). Therefore, \(\gamma\approx3\alpha\). This is a first-order approximation valid for small temperature changes. The result is not obtained by using area expansion alone; it comes from the three-dimensional nature of volume.
119. A metal cube of volume \(2.0\times10^{-3}\,\text{m}^3\) is heated through \(100\,\text{K}\). If \(\gamma=3.0\times10^{-5}\,\text{K}^{-1}\), what is the increase in volume?
ⓐ. \(6.0\times10^{-5}\,\text{m}^3\)
ⓑ. \(3.0\times10^{-5}\,\text{m}^3\)
ⓒ. \(6.0\times10^{-6}\,\text{m}^3\)
ⓓ. \(3.0\times10^{-6}\,\text{m}^3\)
Correct Answer: \(6.0\times10^{-6}\,\text{m}^3\)
Explanation: \( \textbf{Given data:} \) \(V=2.0\times10^{-3}\,\text{m}^3\), \(\Delta T=100\,\text{K}\), and \(\gamma=3.0\times10^{-5}\,\text{K}^{-1}\).
\( \textbf{Required quantity:} \) Increase in volume \(\Delta V\).
For cubical or volume expansion,
\[
\Delta V=\gamma V\Delta T
\]
This is suitable because the body’s volume change is being asked.
Substitute the values:
\[
\Delta V=(3.0\times10^{-5})(2.0\times10^{-3})(100)
\]
First combine \(3.0\times2.0=6.0\).
Combine the powers and the temperature factor:
\[
10^{-5}\times10^{-3}\times100=10^{-8}\times10^2=10^{-6}
\]
So,
\[
\Delta V=6.0\times10^{-6}\,\text{m}^3
\]
The result is positive because the cube is heated and expands normally.
\( \textbf{Final answer:} \) The increase in volume is \(6.0\times10^{-6}\,\text{m}^3\).
120. A solid sphere is heated uniformly. Its radius increases slightly. The volume increases because:
ⓐ. only the surface colour changes
ⓑ. the mass must increase in the same ratio
ⓒ. the temperature unit changes from \(\text{K}\) to \(\text{m}^3\)
ⓓ. expansion occurs in all three dimensions
Correct Answer: expansion occurs in all three dimensions
Explanation: A sphere expands uniformly when heated if the material is isotropic and the heating is uniform. The radius increases, and the change affects the whole three-dimensional size of the sphere. Volume expansion is therefore connected with expansion in all three spatial dimensions. The mass of the sphere does not need to increase during ordinary thermal expansion. A change in colour, if any, is not the reason for volume change. This is why volume expansion uses \(\gamma\), not only the linear coefficient \(\alpha\).
