Q.A glass flask has volume 1×10⁻⁴ m³. It is filled with a liquid at 30 ºC. If the temperature of the system is raised to 100 ºC, how much of the liquid will overflow? (Coefficient of volume expansion of glass is 1.2×10⁻⁵ (ºC)⁻¹ while that of the liquid is 75×10⁻⁵ (ºC)⁻¹.)

Showing the 12 most recent of 14 on this concept.

1. CBSE 2026Set ANNUAL1 markMCQ
   Q.When water is heated from 0 degrees C to 4 degrees C, the volume of water

   (a) remains the same

   (b) decreases

   (c) increases

   (d) none of these

   ›Reveal solutionSolution

   Water shows anomalous expansion: unlike most substances, its volume DECREASES (not increases) as it is warmed from 0 degrees C to 4 degrees C, because its density is maximum at 4 degrees C.

   Most substances expand uniformly on heating. Water is an exception in the narrow range 0 to 4 degrees C: as ice-cold water (just above 0 degrees C) is warmed, its molecules rearrange in a way (related to the breakdown of the open hydrogen-bonded ice-like structure) that causes the volume to CONTRACT, reaching a minimum volume -- and therefore MAXIMUM density -- exactly at 4 degrees C. Only above 4 degrees C does water begin to expand normally with further heating.

   This anomalous expansion of water is why ice floats and why lakes freeze from the top down, insulating aquatic life below (the densest water at 4 degrees C sinks to the bottom).

   ✓Final answer

   (b) decreases.

2. CBSE 2026Set ANNUAL1 markMCQ
   Q.When a uniform rod is heated, which of the following quantities of the rod will increase?

   (a) Mass

   (b) Moment of inertia

   (c) Weight

   (d) Centre of mass

   ›Reveal solutionSolution

   Heating a rod increases its length by thermal expansion; since I = integral of r^2 dm depends on how far the mass lies from the axis, a longer rod has a larger moment of inertia, even though its mass is unchanged.

   When a uniform rod is heated, it undergoes linear thermal expansion: its length increases according to L = L0(1 + alpha*deltaT), where alpha is the coefficient of linear expansion and deltaT is the rise in temperature.

   Let us check each option:

   1. Mass: No material is added to or removed from the rod on heating, so its mass stays exactly the same.
   2. Weight: Weight = mass times g. Since mass is unchanged and g (at a given location) is unchanged, weight stays the same.
   3. Centre of mass: For a uniform rod, the centre of mass lies at its geometric centre. Since the rod expands symmetrically about its centre, the centre of mass (relative to the rod itself) does not shift.
   4. Moment of inertia: For a uniform rod about a transverse axis, I = (1/12)ML^2 (about the centre) or (1/3)ML^2 (about one end). Since L increases while M stays constant, I increases as the square of the increased length.

   So heating increases the moment of inertia because the same mass is now distributed over a greater length, i.e., farther from the axis of rotation on average.

   ✓Final answer

   The correct option is (b) Moment of inertia — thermal expansion lengthens the rod, so its mass lies farther from the axis, increasing I = (1/12)ML^2 even though mass, weight, and centre of mass position are unchanged.

3. CBSE 2026Set ANNUAL1 mark
   Q.Write true or false: The density of water is maximum at 0 degrees C temperature.

   ›Reveal solutionSolution

   The statement is FALSE: water's density is maximum at 4°C (the anomalous expansion of water), not at 0°C.

   Most substances contract steadily as they are cooled, so their density steadily increases all the way down to their freezing point. Water is a notable exception, called the anomalous expansion of water: as water is cooled from higher temperatures toward 0°C, its density increases only until it reaches 4°C; below 4°C, further cooling causes water to expand slightly again, so its density decreases as it approaches 0°C (and ice, at 0°C, is even less dense, which is why ice floats). This is why deep lakes in cold climates have their densest, 4°C water sitting at the bottom in winter, insulating aquatic life from the colder water and ice above.

   ✓Final answer

   False — the density of water is maximum at 4 degrees C, not at 0 degrees C.

4. CBSE 2026Set ANNUAL1 markMCQ
   Q.When water freezes, the distance between its molecules

   (a) Decreases

   (b) Increases

   (c) Becomes zero

   (d) Remains unchanged

   ›Reveal solutionSolution

   On freezing, water expands into an open structure, so molecular distance increases. Answer (B).

   Water shows anomalous behaviour: when it freezes into ice, the molecules arrange into an open, cage-like hexagonal crystal held together by hydrogen bonds. This structure occupies MORE volume than the same mass of liquid water.

   Because the volume increases while the number of molecules is unchanged, the average distance between molecules increases (which is why ice floats on water).

   ✓Final answer

   (B) Increases.

5. CBSE 2026Set ANNUAL1 markMCQ
   Q.A metal sheet with a circular hole is heated. The hole

   (a) gets smaller

   (b) gets larger

   (c) remains unchanged

   (d) gets deformed

   ›Reveal solutionSolution

   A hole in a heated plate expands like the surrounding material, so it gets larger. Answer (B).

   When a metal sheet is heated, every linear dimension increases by the same factor (1 + alpha_l dT). Imagine the disc of metal that would fill the hole; it too would expand. So the hole behaves as if it were filled with the same metal, and its diameter (and area) increases.

   ✓Final answer

   (B) gets larger.

6. CBSE 2025Set ANNUAL1 markMCQ
   Q.When water is heated from 0°C to 10°C, its volume (A) decreases (B) increases (C) first decreases then increases (D) none of these

   ›Reveal solutionSolution

   As water is heated from 0°C to 10°C, its volume first decreases (till 4°C) and then increases.

   Most substances expand continuously on heating, but water shows an anomalous behaviour between 0°C and 4°C: as it is heated from 0°C, its volume actually decreases until 4°C (where density is maximum), because of the breaking down of the ice-like hydrogen-bonded structure present in cold water. Above 4°C, water behaves normally and its volume increases with further heating.

   So heating from 0°C to 10°C, the volume first decreases (0°C → 4°C) then increases (4°C → 10°C).

   ✓Final answer

   (C) first decreases then increases.

7. CBSE 2025Set ANNUAL1 markMCQ
   Q.The top of lake is frozen at the atmospheric temperature -10°C. The temperature at the bottom of the lake at that time is most likely to be (A) 0°C (B) -4°C (C) 4°C (D) -10°C

   ›Reveal solutionSolution

   Even though the surface is frozen at -10°C, the water at the bottom of the lake stays at about 4°C.

   Water is densest at 4°C (anomalous expansion). As a lake cools in winter, surface water cools, becomes denser, and sinks — until the whole lake reaches 4°C. Further surface cooling below 4°C makes that top layer less dense (due to anomalous expansion), so it stays on top and eventually freezes into ice at 0°C, while the layer beneath the ice stays close to 4°C — the densest, and therefore the bottom-most, layer. The ice layer on top also insulates the water below from the very cold (-10°C) air, so the bottom stays near 4°C rather than freezing.

   ✓Final answer

   (C) 4°C.

8. CBSE 2025Set ANNUAL1 mark
   Q.Write relation between coefficient of linear expansion (α), coefficient of area expansion (β) and coefficient of volume expansion (γ).

   ›Reveal solutionSolution

   Because area scales with the square of length and volume with the cube, the area-expansion coefficient is twice, and the volume-expansion coefficient is three times, the linear-expansion coefficient of the same material.

   Let a solid have linear dimension $L$, area $A \propto L^2$, and volume $V \propto L^3$.

   For small temperature changes, differentiating each relation with respect to temperature and dividing through gives the standard result (derivable from $L' = L(1+\alpha\Delta T)$, $A' = A(1+\beta\Delta T)$, $V' = V(1+\gamma\Delta T)$, and expanding $A' = L'^2$, $V' = L'^3$ while dropping higher-order terms in $\Delta T$):

   $\beta = 2\alpha$

   $\gamma = 3\alpha$

   So: $\alpha : \beta : \gamma = 1 : 2 : 3$, or equivalently $\alpha = \beta/2 = \gamma/3$.

   ✓Final answer

   $\beta = 2\alpha$ and $\gamma = 3\alpha$ (i.e. $\alpha : \beta : \gamma = 1 : 2 : 3$).

9. CBSE 2024Set ANNUAL1 markMCQ
   Q.There is a concentric spherical cavity in a solid ball of metal. If the ball is heated, then volume of cavity (A) will decrease (B) will increase (C) will remain unchanged (D) none of these

   ›Reveal solutionSolution

   The cavity inside a heated solid expands, just like a solid piece of the same size would.

   A useful way to think about a cavity in a solid is to imagine it filled with the same material. When the whole solid (including that imagined filling) is heated, it expands uniformly according to its coefficient of volume expansion. Removing the imagined filling doesn't change how the surrounding solid expands — so the cavity itself grows in the same way a solid piece of that size and shape would, i.e. the cavity's volume increases on heating.

   ✓Final answer

   (B) will increase.

10. CBSE 2024Set SET-AP55001 mark
    Q.Write the relation between the coefficient of linear expansion (alpha), the coefficient of areal (superficial) expansion (beta), and the coefficient of volume (cubical) expansion (gamma).

    ›Reveal solutionSolution

    For an isotropic solid, the areal and cubical expansion coefficients are simple multiples of the linear expansion coefficient: β = 2α and γ = 3α.

    Consider a cube of side L that expands uniformly on heating by ΔT. Each side changes as ΔL = αLΔT, so the linear expansion coefficient is α = ΔL/(LΔT).

    For area, A = L^2, so dA = 2L dL, giving ΔA/A = 2(ΔL/L) = 2αΔT — so the areal expansion coefficient β = ΔA/(AΔT) = 2α.

    For volume, V = L^3, so dV = 3L^2 dL, giving ΔV/V = 3(ΔL/L) = 3αΔT — so the cubical expansion coefficient γ = ΔV/(VΔT) = 3α.

    So the relation is: α : β : γ = 1 : 2 : 3, i.e., β = 2α and γ = 3α.

    ✓Final answer

    β = 2α and γ = 3α (α : β : γ = 1 : 2 : 3).

11. CBSE 2024Set SET-NDP60001 markMCQ
    Q.At which temperature is the density of water maximum:

    (a) 4°C

    (b) 0°C

    (c) -25°C

    (d) -27°C

    ›Reveal solutionSolution

    Because of anomalous expansion, water's density is maximum at 4°C, not at 0°C.

    Most substances contract steadily as they cool and expand steadily as they heat. Water is unusual: as it is cooled from higher temperatures down to 4°C, it contracts (density rises) like a normal liquid, but on cooling further from 4°C to 0°C, it actually expands again (density falls) — this is called the anomalous expansion of water. As a result, water's density is at its highest at 4°C, and is lower both above and below this temperature (including as ice at 0°C, which is why ice floats on water). This behaviour is important ecologically: it is why the bottom of a frozen lake stays at 4°C, allowing aquatic life to survive under the ice.

    ✓Final answer

    The correct option is (a) 4°C — water has its maximum density at 4°C.

12. CBSE 2024Set ANNUAL1 mark
    Q.State whether true or false: At 4 degrees Celsius, the volume of water is minimum.

    ›Reveal solutionSolution

    True. Water shows anomalous expansion: its density is maximum (so volume is minimum, for a fixed mass) at 4 degrees C.

    Most substances contract steadily on cooling. Water is anomalous: as it is cooled from higher temperatures down to 4 degrees C, its volume decreases (density rises) like a normal liquid. But on cooling further from 4 degrees C to 0 degrees C, water actually starts to expand again (density falls) as hydrogen-bonded ice-like structures begin forming.

    As a result, water has its maximum density (and hence minimum volume, for a given mass) at exactly 4 degrees C.

    ✓Final answer

    True — the volume of water (for a fixed mass) is minimum, and its density is maximum, at 4 degrees C.