Q.A large steel wheel is to be fitted on to a shaft of the same material. At $27\ ^\circ\text{C}$, the outer diameter of the shaft is $8.70\ \text{cm}$ and the diameter of the central hole in the wheel is $8.69\ \text{cm}$. The shaft is cooled using 'dry ice'. At what temperature of the shaft does the wheel slip on the shaft? Assume coefficient of linear expansion of the steel to be constant over the required temperature range: $\alpha_{\text{steel}} = 1.20 \times 10^{-5}\ \text{K}^{-1}$.

Showing the 12 most recent of 30 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.How much should the temperature of a metal ring of diameter 100 cm be raised in order to fit it on a wooden disc of diameter 100.6 cm [coefficient of linear expansion of metal $=12\times10^{-6}\ ^{\circ}C^{-1}$] (A) 300°C (B) 400°C (C) 500°C (D) 600°C

   ›Reveal solutionSolution

   A ring's diameter (and circumference) both expand according to the same linear-expansion law as a straight rod — set the required diameter change equal to $\alpha L_0\Delta T$ and solve for $\Delta T$. Answer: (C) 500°C.

   Concept and Intuition

   For small temperature changes, any linear dimension of a solid — length, diameter, or circumference of a ring — expands according to $\Delta L=\alpha L_0\Delta T$, where $\alpha$ is the coefficient of linear expansion. A metal ring's diameter behaves exactly like a rod of the same original length for this purpose (all linear dimensions scale by the same factor $(1+\alpha\Delta T)$). To fit the ring over the larger wooden disc, its diameter must expand from 100 cm to at least 100.6 cm.

   Step-by-Step Solution

   1. Required change in diameter: $\Delta L=100.6-100=0.6\ cm$, with original diameter $L_0=100\ cm$.
   2. Linear expansion equation:

   $\Delta L=\alpha L_0\Delta T \quad\Rightarrow\quad \Delta T=\frac{\Delta L}{\alpha L_0}.$

   3. Substitute values ($\alpha=12\times10^{-6}\ ^\circ C^{-1}$): …

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2. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.Two rods made of metals A and B, each of length 20cm expand by 0.075cm and 0.045cm respectively, when heated from $0°C$ to $100°C$. A composite rod of same length is made with a portion of metal A and remaining with B expands by 0.060cm for same rise of temperature. Then the portion of composite rod made of A has initial length (A) 8 cm (B) 10 cm (C) 15 cm (D) 18 cm

   ›Reveal solutionSolution

   Each metal has its own linear expansion coefficient found from its solo expansion data; in a composite rod the two pieces expand independently and additively. Solving the linear equation for the split gives 10 cm of metal A.

   Concept and Intuition

   When a rod made of two different metals joined end-to-end (not bonded to constrain each other, just placed in series to make one longer rod) is heated, each segment expands according to its own coefficient of linear expansion $\alpha$, applied to its own original length. The total expansion of the composite rod is simply the sum of the two segments' expansions — there's no more physics than superposition here.

   Step-by-Step Solution

   1. From the solo-rod data (length $L_0=20$ cm each, $\Delta T = 100^\circ C$):

   $\alpha_A = \frac{\Delta L_A}{L_0 \Delta T} = \frac{0.075}{20\times100} = 3.75\times10^{-5}\ /^\circ C$

   $\alpha_B = \frac{\Delta L_B}{L_0\Delta T} = \frac{0.045}{20\times100} = 2.25\times10^{-5}\ /^\circ C$

   2. Let the composite rod (total length 20 cm) have length $x$ of metal A and $(20-x)$ of metal B. Each piece expands independently:

   $\Delta L_{composite} = x\,\alpha_A\,\Delta T + (20-x)\,\alpha_B\,\Delta T = 0.060\text{ cm}$

   3. Substitute $\Delta T = 100$: …

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3. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.Which of the following graphs shows the variation of volume expansion coefficient of copper with temperature? (A) [FIGURE: graph of $\alpha_v$ vs $T$ — a curve starting high and decreasing, concave up, approaching a low value] (B) [FIGURE: graph of $\alpha_v$ vs $T$ — a curve starting near zero and increasing, concave up (rising steeply)] (C) [FIGURE: graph of $\alpha_v$ vs $T$ — a straight line through the origin, increasing linearly] (D) [FIGURE: graph of $\alpha_v$ vs $T$ — an S-shaped (sigmoid) curve rising from the origin and flattening out]

   ›Reveal solutionSolution

   This tests knowledge of how the thermal (volume) expansion coefficient of a solid metal actually varies with absolute temperature — it is not linear, not a simple decaying curve, and not an unbounded exponential rise; it saturates, giving an S-shaped curve.

   Concept and Intuition

   Thermal expansion in a solid arises from the anharmonicity of interatomic bonds — as atoms vibrate more energetically, the average interatomic spacing increases. The volumetric expansion coefficient $\alpha_v = \frac{1}{V}\left(\frac{\partial V}{\partial T}\right)_P$ is, via the Grüneisen relation, proportional to the lattice specific heat capacity $C_V(T)$:

   $\alpha_v(T) = \frac{\gamma_G\, C_V(T)}{K V},$

   where $\gamma_G$ (Grüneisen parameter), bulk modulus $K$ and volume $V$ vary slowly with temperature compared to $C_V(T)$.

   The specific heat of a solid follows the well-known Debye behaviour: it is essentially zero at $T \to 0\,$K (vibrational modes are frozen out quantum-mechanically), rises rapidly through an intermediate temperature range, and then saturates to the classical Dulong–Petit value at high temperature (all vibrational modes fully excited classically). Since $\alpha_v$ tracks $C_V(T)$, it shows exactly the same qualitative shape: starts near zero, rises, and flattens — an S-shaped (sigmoid) curve.

   Step-by-Step Solution

   1. Recognise that $\alpha_v(T)$ for a solid metal is not an independent, freely-chosen function — it is tied to the lattice heat capacity through the Grüneisen relation.
   2. Recall the Debye/Einstein model result for $C_V(T)$: near $0$ K it goes as $T^3$ (Debye's law), so it starts at (near) zero and curves upward slowly at first.
   3. At intermediate temperatures $C_V(T)$ rises quickly. …

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4. AP EAPCET 2026Set eng-2026-05-15-FN1 markMCQ
   Q.If the temperature of a steel solid sphere of mass 4 kg and radius 5 cm is increased by $10^{0}C$, then the increase in the moment of inertia of the sphere about its diameter is (Coefficient of linear expansion of steel = $1.2\times10^{-5}\ K^{-1}$) (A) $3.6\ g\ cm^2$ (B) $4.8\ g\ cm^2$ (C) $2.4\ g\ cm^2$ (D) $9.6\ g\ cm^2$

   ›Reveal solutionSolution

   Heating a solid sphere expands its radius; since $I\propto R^2$, a fractional change $dR/R$ in radius produces twice that fractional change in $I$. Plugging in numbers gives $\Delta I = 9.6\ g\,cm^2$.

   Concept and Intuition

   The moment of inertia of a solid sphere about its diameter is $I=\dfrac{2}{5}MR^2$. Thermal expansion changes only the geometric size $R$ (mass is conserved), so differentiating: $dI = \dfrac{4}{5}MR\,dR$, which can be written as $\dfrac{dI}{I}=\dfrac{2\,dR}{R}$. Since linear expansion gives $\dfrac{dR}{R}=\alpha\Delta T$, we get the handy result $\dfrac{dI}{I}=2\alpha\Delta T$ — the same trick used for area/volume expansion coefficients ($2\alpha$ for area, $3\alpha$ for volume; here it's the "area-like" $R^2$ dependence).

   Step-by-Step Solution

   1. Convert to cgs for convenience: $M=4000$ g, $R=5$ cm.
   2. $I = \dfrac{2}{5}MR^2 = \dfrac{2}{5}\times4000\times25 = 40000$ g·cm². …

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5. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.Which of the following graphs is correctly drawn between temperature (t) and the density

   (d) of water? (A) [FIGURE] (a graph of density d on the y-axis vs temperature t°C on the x-axis: the curve starts at d = 1.013 at t = 0, decreases to a minimum of d = 1.000 at t = 4, then increases again as t rises toward 8) (B) [FIGURE] (a graph of density d on the y-axis vs temperature t°C on the x-axis: the curve starts at d = 1.000 at t = 0, rises to a maximum of d = 1.013 at t = 4, then decreases as t rises toward 8) (C) [FIGURE] (a straight-line graph of density d on the y-axis vs temperature t°C on the x-axis: the line rises linearly, crossing zero density near t = -4, passing through about d = 0.5 near t = -2 and d = 1.0 near t = 4) (D) [FIGURE] (a graph of density d on the y-axis vs temperature t°C on the x-axis: the curve starts near d = 1.00 at low t and decreases steeply, asymptotically approaching about d = 0.2 as t increases toward 8)

   ›Reveal solutionSolution

   This tests the well-known anomalous behaviour of water's density with temperature, which peaks at 4°C. Answer: option (B), the hump-shaped curve with maximum at $t=4°C$.

   Concept and Intuition

   Most liquids simply expand (density decreases) as temperature rises. Water is anomalous between 0°C and 4°C: as it warms from 0°C, hydrogen-bonded ice-like clusters break down and the water molecules pack more efficiently, so density increases up to 4°C. Above 4°C, normal thermal expansion takes over and density decreases as usual. This means density is at its maximum at 4°C, lower on both sides.

   Step-by-Step Solution

   1. Identify the physical fact: water's density-temperature graph has a maximum at $t=4°C$ (this is why lakes freeze from the top — 4°C water sinks to the bottom).
   2. This means the graph must rise from $t=0$ to $t=4$, then fall from $t=4$ onward — a hump (∩)-shaped curve, not a U-shape, not monotonic, and not a straight line.
   3. Option (A) shows a U-shape (minimum at 4°C) — this is the opposite of the real physical behaviour, so it's wrong.
   4. Option (B) shows exactly the hump shape (rises to a peak at $t=4$, then falls) — matches the real anomalous expansion of water. …

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6. AP EAPCET 2023Set eng-2023-05-17-AN1 markMCQ
   Q.A steel tape of 300 cm length is graduated at 27 °C. The length of a steel rod measured using the tape is found to be 110 cm at 50 °C. The actual length of steel rod at 50 °C is ($\alpha_{steel} = 1.2\times10^{-5}\ \mathrm{K^{-1}}$) (A) 110.03 cm (B) 110.10 cm (C) 110.07 cm (D) 110.62 cm

   ›Reveal solutionSolution

   A tape calibrated at $27°C$ but used at a hotter $50°C$ has physically expanded, so its markings under-read the true length; correcting for that expansion gives 110.03 cm.

   Concept and Intuition

   A measuring tape is manufactured with its scale markings correct at a specific reference temperature (here $27°C$). If used at a higher temperature, the metal expands, so the distance between successive markings (say, between the '0' and '1 cm' marks) becomes physically longer than 1 true cm. This means when the tape reads a certain length, the object it's measuring is actually slightly longer than that reading, because each 'cm' on the expanded tape now corresponds to more than a true cm.

   Step-by-Step Solution

   1. Temperature rise from calibration: $\Delta T = 50-27 = 23°C$.
   2. Expansion factor of the tape's own markings: $(1+\alpha\Delta T) = 1 + (1.2\times10^{-5})(23) = 1+2.76\times10^{-4} = 1.000276$.
   3. Since the tape's markings have stretched, the object's true length is the tape's reading multiplied by this same expansion factor: $L_{true} = L_{measured}\times(1+\alpha\Delta T) = 110\times1.000276$. …

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7. AP EAPCET 2022Set ap-2022-07-11-FN1 markMCQ
   Q.If we consider a rectangular sheet of the solid, the coefficient of areal expansion is (A) Half of its coefficient of linear expansion (B) Thrice of its coefficient of linear expansion (C) Twice of its coefficient of linear expansion (D) Square root of its coefficient of linear expansion

   ›Reveal solutionSolution

   For an isotropic solid, if the linear expansion coefficient is $\alpha$, the areal expansion coefficient $\beta = 2\alpha$ and the volumetric expansion coefficient $\gamma = 3\alpha$.

   Concept and Intuition

   A rectangular sheet expands in both its length and breadth on heating; since area is a product of two linear dimensions, each of which independently grows by a factor $(1+\alpha\Delta T)$, the area grows (to first order) by twice that fractional change.

   Step-by-Step Solution

   1. Let sides be $l_1, l_2$ initially, so $A = l_1 l_2$.
   2. Each dimension expands: $l_1' = l_1(1+\alpha\Delta T)$, $l_2' = l_2(1+\alpha\Delta T)$.
   3. New area: $A' = l_1 l_2 (1+\alpha\Delta T)^2 \approx l_1 l_2(1+2\alpha\Delta T)$ (neglecting the tiny $\alpha^2$ term). …

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8. AP EAPCET 2022Set ap-2022-07-12-AN1 markMCQ
   Q.A steel rod of length 5 m and radius 2 cm kept at room temperature is heated to 10 °C above the room temperature. The % change in the volume is (Coefficient of linear expansion of steel $= 10\times10^{-6}$ °C$^{-1}$) (A) 0.03 (B) 0.01 (C) 0.9 (D) 1.2

   ›Reveal solutionSolution

   For small temperature changes, the volumetric expansion coefficient is three times the linear expansion coefficient; multiplying by $\Delta T$ gives the fractional (and hence percentage) volume change.

   Concept and Intuition

   For an isotropic solid, if the coefficient of linear expansion is $\alpha$, the coefficient of volume expansion is $\gamma = 3\alpha$ (since $V \propto L^3$ and differentiating gives a factor of 3). Note that the length (5 m) and radius (2 cm) of the rod are irrelevant here — only $\alpha$ and $\Delta T$ matter for percentage volume change.

   Step-by-Step Solution

   1. $\gamma = 3\alpha = 3\times10\times10^{-6} = 30\times10^{-6}\ ^\circ\text{C}^{-1}$. …

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9. AP EAPCET 2022Set ap-2022-07-12-FN1 markMCQ
   Q.A steel rod of length 5 m and radius 2 cm kept at room temperature is heated to $10\,°C$ above the room temperature. Then the change in the cross-sectional area of the rod. (coefficient of linear expansion of steel = $10\times10^{-6}\ °C^{-1}$) (A) 0.01 % (B) 0.02 % (C) 0.03 % (D) 0.09 %

   ›Reveal solutionSolution

   Since area scales as the square of a linear dimension, its thermal-expansion coefficient is exactly twice the linear coefficient. Answer: $0.02\%$.

   Concept and Intuition

   When every linear dimension of an object expands by a fractional amount $\alpha\Delta T$, an area (which scales as length²) expands by approximately twice that fraction, since $A=\pi r^2$ and a small relative change in $r$ doubles when squared: $(1+\alpha\Delta T)^2\approx 1+2\alpha\Delta T$.

   Step-by-Step Solution

   1. The rod's cross-section is a circle of radius $r$, so $A=\pi r^2$.
   2. With temperature rise $\Delta T$, radius becomes $r' = r(1+\alpha\Delta T)$.
   3. New area: $A' = \pi r'^2 = \pi r^2(1+\alpha\Delta T)^2 \approx A(1+2\alpha\Delta T)$ for small $\alpha\Delta T$.
   4. So $\dfrac{\Delta A}{A} = 2\alpha\Delta T = 2(10\times10^{-6})(10) = 2\times10^{-4}$. …

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10. AP EAPCET 2022Set eng-2022-07-04-AN1 markMCQ
    Q.A metal tape is calibrated at $25\,^\circ\mathrm{C}$. On a cold day when the temperature is $-15\,^\circ\mathrm{C}$, the percentage error in the measurement of length is (Coefficient of linear expansion of metal $= 1\times10^{-5}\,^\circ\mathrm{C^{-1}}$) (A) 0.04 % (B) 0.05 % (C) 0.1 % (D) 0.08 %

    ›Reveal solutionSolution

    The tape's length changes by $\alpha\Delta T=1\times10^{-5}\times40=4\times10^{-4}$, i.e. a 0.04% error, because it is used at a temperature different from its calibration temperature.

    Concept and Intuition

    A metal tape marked at one temperature contracts or expands when used at another temperature, per $\Delta L=L_0\alpha\Delta T$. The tape's markings no longer correspond to true lengths, giving a systematic percentage error equal to $\alpha\Delta T$ in magnitude.

    Step-by-Step Solution

    1. $\Delta T = (-15) - 25 = -40^\circ\mathrm C$.
    2. Fractional error $=\left|\alpha\Delta T\right| = 1\times10^{-5}\times40=4\times10^{-4}$. …

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11. AP EAPCET 2022Set eng-2022-07-05-FN1 markMCQ
    Q.An iron sphere having diameter D and mass M is immersed in hot water so that the temperature of the sphere increases by $\delta T$. If $\alpha$ is the coefficient of linear expansion of the iron then the change in the surface area of the sphere is (A) $\pi D^2 . \alpha . \delta T\,(\alpha.\delta T - 4)$ (B) $\pi D^2 . \alpha . \delta T\,(\alpha.\delta T + 4)$ (C) $\pi D^2 . \alpha . \delta T\,(\alpha.\delta T - 2)$ (D) $\pi D^2 . \alpha . \delta T\,(\alpha.\delta T + 2)$

    ›Reveal solutionSolution

    This tests superficial (areal) thermal expansion, derived exactly (not using the small-expansion approximation) from linear expansion of the diameter.

    Concept and Intuition

    When a sphere's diameter expands linearly by a factor $(1+\alpha\delta T)$, its surface area (which scales as the square of a linear dimension) expands by the square of that factor. Expanding this square exactly (not dropping the $\alpha^2\delta T^2$ term) gives the precise change in area.

    Step-by-Step Solution

    1. Original surface area: $A=\pi D^2$.
    2. New diameter after heating: $D'=D(1+\alpha\delta T)$.
    3. New surface area: $A'=\pi D'^2=\pi D^2(1+\alpha\delta T)^2=\pi D^2\left(1+2\alpha\delta T+\alpha^2\delta T^2\right)$.
    4. Change in area: $\Delta A=A'-A=\pi D^2\left(2\alpha\delta T+\alpha^2\delta T^2\right)$.
    5. Factor out $\alpha\delta T$: $\Delta A=\pi D^2\cdot\alpha\delta T\,(\alpha\delta T+2)$.

    Common Mistakes …

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12. AP EAPCET 2022Set eng-2022-07-07-FN1 markMCQ
    Q.The coefficient of volume expansion of a material is $5\times10^{-4}\ {}^\circ\mathrm{C}^{-1}$. The fractional change in its density for a 40 $^\circ$C rise in temperature is nearly (A) 0.01 (B) 0.02 (C) 0.03 (D) 0.04

    ›Reveal solutionSolution

    This tests the relation between volume expansion and density change; the density decreases fractionally by 0.02 (2%) for this temperature rise.

    Concept and Intuition

    Mass is conserved as a substance is heated, but its volume expands, so density must fall. Since $\rho = m/V$, and $V$ increases by a fractional amount $\Delta V/V = \gamma\Delta T$ (definition of the volume expansion coefficient), a first-order (small-change) argument gives

    $\rho' = \frac{m}{V(1+\gamma\Delta T)} \approx \rho(1-\gamma\Delta T)$

    so the fractional decrease in density is approximately $\gamma\Delta T$, valid when $\gamma\Delta T \ll 1$.

    Step-by-Step Solution

    1. Given $\gamma = 5\times10^{-4}\ {}^\circ\text{C}^{-1}$ and $\Delta T = 40\ {}^\circ\text{C}$.
    2. Fractional volume change: $\dfrac{\Delta V}{V} = \gamma\Delta T = 5\times10^{-4}\times40 = 0.02$.
    3. To first order, fractional density change has the same magnitude (opposite sign): …

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