NCERT Exemplar Solutions Class 11 Physics Thermal Properties of Matter — Multiple Choice Questions Q1–12

NCERT Exemplar Problems, Class 11 Physics — Thermal Properties of Matter, MCQ I and MCQ II. All 12 questions solved.

The book numbers these questions 11.1 to 11.12, because the exemplar counts an extra introductory unit; the matching textbook chapter is Chapter 10. Here they are Questions 1 to 12. Questions 1 to 8 are MCQ I (exactly one option is correct) and Questions 9 to 12 are MCQ II (one or more options may be correct). Four ideas carry the whole page:

• Thermal expansion. A length grows by $\Delta l = l\alpha\Delta T$, an area by $2\alpha$ and a volume by $3\alpha$ per kelvin, so anything that depends on a length squared (moment of inertia) or cubed (volume) changes by $2\alpha\Delta T$ or $3\alpha\Delta T$.
• Linear temperature scales. Any two scales are related by a straight line, so a reading is fixed by where it sits between the two fixed points.
• Heating and cooling rates. Heat enters or leaves through the surface while the heat needed depends on the mass, so the surface-to-mass ratio decides which body heats or cools first.
• Change of state. While a substance melts or boils its temperature stays fixed, so a heating curve has flat steps where two phases sit in equilibrium.

Key insight. Ask what scales with the surface and what scales with the volume or mass. Questions 2, 7, 8 and 10 are all settled by that one comparison.

Choose the correct answer from the given options. (MCQ I: one correct. MCQ II: one or more correct.)

Question 1

A bimetallic strip is made of aluminium and steel $(\alpha_{Al} > \alpha_{steel})$. On heating, the strip will

Solution. The two strips are riveted together along their length, so they cannot slide past each other. On heating, aluminium tries to grow more than steel. The only way a longer strip and a shorter strip can stay joined is for the pair to curve, with the longer strip on the outside of the arc. The outside of a curve is the convex side, so aluminium is convex and steel is on the concave side. Nothing in the problem produces a twist, which would need a turning effect about the length of the strip.

Question 2

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly

Solution. No external torque acts about the axis, so angular momentum $L = I\omega$ is conserved. Heating lengthens the rod, and since $I = \frac{1}{12}Ml^2$, the moment of inertia grows (by about $2\alpha\Delta T$ times itself). With $I\omega$ fixed, a larger $I$ means a smaller $\omega$, so the rod slows down. Option (d) gets the reasoning backwards: a larger moment of inertia lowers the speed, it does not raise it.

Question 3

The graph between two temperature scales A and B is shown in Fig. 11.1. Between upper fixed point and lower fixed point there are 150 equal divisions on scale A and 100 on scale B. The relationship for conversion between the two scales is given by

Solution. The upper fixed point reads $180$ on A and $100$ on B, and the gap between the fixed points is $150$ divisions on A. So the lower fixed point reads $180 – 150 = 30$ on A, and (from the graph, where the line meets the vertical axis) $0$ on B. Any reading is the same fraction of the way from the lower to the upper fixed point on both scales:

$$\frac{t_A – 30}{150} = \frac{t_B – 0}{100}.$$

Check: $t_B = 100$ gives $t_A = 180$, and $t_B = 0$ gives $t_A = 30$. Options (a), (c) and (d) all fail at least one of these two checks.

Question 4

An aluminium sphere is dipped into water. Which of the following is true?

Solution. The buoyant force is $V\rho_w g$, the weight of water displaced. Water behaves anomalously between $0^\circ$C and $4^\circ$C: it contracts on heating, so its density is greatest at $4^\circ$C. The sphere’s own volume barely changes over $4$ K (aluminium’s expansion is tiny next to the change in water’s density), so the buoyancy is larger at $4^\circ$C and smaller at $0^\circ$C. The radius only scales both cases by the same factor, so (d) is wrong.

Question 5

As the temperature is increased, the time period of a pendulum

Solution. The suspension lengthens on heating, so the distance from the point of support to the centre of the bob grows. The bob expands uniformly about its own centre, so its centre of mass stays at its centre. Since $T = 2\pi\sqrt{l/g}$, a longer effective length gives a longer period. Option (b) has the right cause but the wrong effect, and (c) and (d) invent a shift of the centre of mass that uniform expansion does not produce.

Question 6

Heat is associated with

Solution. Orderly motion, in which all the molecules move together, is just the motion of the body as a whole; a ball thrown across a room is not hotter for being thrown. What rises when a body is heated is the random, disordered motion of its molecules about their mean positions. So heat is associated with the kinetic energy of random motion only.

Question 7

The radius of a metal sphere at room temperature $T$ is $R$, and the coefficient of linear expansion of the metal is $\alpha$. The sphere is heated a little by a temperature $\Delta T$ so that its new temperature is $T + \Delta T$. The increase in the volume of the sphere is approximately

Solution. For an isotropic solid the coefficient of volume expansion is $\gamma = 3\alpha$, because each of the three dimensions grows by the fraction $\alpha\Delta T$. So

$$\Delta V = 3\alpha V\Delta T = 3\alpha\cdot\frac{4}{3}\pi R^3\cdot\Delta T = 4\pi R^3\alpha\Delta T.$$

The same follows from differentiating $V = \frac{4}{3}\pi R^3$: $\Delta V = 4\pi R^2\Delta R$ with $\Delta R = R\alpha\Delta T$. Option (c) forgets the factor $3$; (a) and (b) have the wrong dimensions for a volume.

Question 8

A sphere, a cube and a thin circular plate, all of same material and same mass are initially heated to same high temperature.

Solution. Same material and same mass means the same volume and the same heat content to lose for each degree of cooling. The rate of heat loss is proportional to the surface area. For a fixed volume the sphere has the least surface area of any shape; a cube of the same volume has about $24\%$ more; and a thin plate, spread out, has far more than either. So the plate loses heat fastest and the sphere slowest.

Question 9

Mark the correct options:

Solution. By the zeroth law, being in thermal equilibrium means having the same temperature, and that relation passes along a chain.

• (a), (b): $X$ has the temperature of $Y$ but not of $Z$. If $Y$ and $Z$ were in equilibrium, $Z$ would share $Y$’s temperature and so $X$’s, which contradicts the data. So $Y$ and $Z$ cannot be in equilibrium: (b) is true and (a) is false.
• (c), (d): $X$ differs from both $Y$ and $Z$, which says nothing about how $Y$ and $Z$ compare. They may or may not share a temperature, so “may” in (d) is right and “must” in (c) is wrong.

Question 10

‘Gulab Jamuns’ (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger (in radius) than the other. Pizzas (assumed to be discs) are also to be heated in oven. They are also in two sizes, one twice big (in radius) than the other. All four are put together to be heated to oven temperature. Choose the correct option from the following:

Solution. The heating time is roughly the heat needed divided by the rate at which heat enters, that is (mass) over (surface area).

• Gulab jamuns. Mass grows as $r^3$, surface area as $r^2$, so the time grows as $r$. Doubling the radius roughly doubles the time: the smaller ones are ready first. (b) is true and (a) is false.
• Pizzas. A pizza is a thin disc of the same thickness $t$ in both sizes, and it takes in heat through its two flat faces. Mass grows as $\pi r^2 t$ and face area as $2\pi r^2$, so the ratio does not depend on $r$. Both sizes are heated in about the same time, so neither (c) nor (d) holds.

The rim does add a little area ($2\pi r t$), which favours the small pizza slightly, but for a thin disc it is negligible next to the faces. The key’s answer is (b) alone, which agrees with this.

Question 11

Refer to the plot of temperature versus time (Fig. 11.2) showing the changes in the state of ice on heating (not to scale). Which of the following is correct?

Solution. The flat parts of a heating curve are changes of state, where the heat supplied goes into latent heat and the temperature stays put.

• AB is flat at $0^\circ$C: ice is melting, with ice and water together in equilibrium. (a) is true.
• At B the last of the ice has melted; BC is water warming from $0^\circ$C to $100^\circ$C. Boiling cannot start at $0^\circ$C, so (b) is false.
• C is where boiling begins. Water turns to steam all along CD and is all steam only at D, after which the steam heats up (DE). So (c) is false and (d) is true.

Question 12

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?

Solution. By Newton’s law of cooling, the rate of cooling is proportional to the temperature difference with the surroundings, $\dfrac{dT}{dt} = -k(T – T_s)$.

• (a) The rate falls as the difference shrinks, so it is not constant. False.
• (b) Integrating gives $T – T_s = (T_0 – T_s)e^{-kt}$: the temperature falls exponentially towards the surroundings’ temperature. Strictly it is the excess temperature that decays exponentially, which is how this option is meant. True.
• (c) Every body radiates and absorbs at once; milk hotter than its surroundings simply emits more than it receives, so the net flow is outwards. True.
• (d) Milk spilt on a table loses heat by conduction into the table, by convection to the air above and by radiation. True.

Common mistakes

• Putting the more expansive metal on the concave side in Question 1. The strip that grows more must go round the outside of the curve.
• Thinking heating speeds up the rod in Question 2. Angular momentum is fixed, so a larger moment of inertia means a smaller angular speed.
• Taking the lower fixed point of scale A as $0$ in Question 3. It is $180 – 150 = 30$.
• Forgetting that water is densest at $4^\circ$C in Question 4. Between $0^\circ$C and $4^\circ$C water contracts on heating.
• Using $\alpha$ instead of $3\alpha$ for a volume in Question 7. The factor $3$ comes from three dimensions each expanding.
• Thinking the sphere cools fastest in Question 8. It has the least surface for its volume, so it cools slowest.
• Ticking (c) in Question 10. A thin pizza takes in heat through its faces in proportion to its mass, so both sizes take about the same time.
• Reading C as the end of boiling in Question 11. Boiling starts at C and ends at D.

Practise next

• Problem. A steel rod ($\alpha = 1.2\times10^{-5}\ \text{K}^{-1}$) is $1$ m long at $20^\circ$C. By how much does it lengthen when heated to $70^\circ$C?

• Problem. At what temperature do the Kelvin and Fahrenheit scales show the same number? Use $T_F = \frac{9}{5}(T_K – 273.15) + 32$.

• Problem. A metal sphere of radius $10$ cm with $\alpha = 2\times10^{-5}\ \text{K}^{-1}$ is heated through $100$ K. Find the increase in its volume.

• Exemplar Thermal Properties of Matter: Very Short, Short and Long Answer Questions — where the same ideas are applied to calorimetry, constant-length scales, thermal stress and radiation.
• NCERT textbook Chapter 10 — Thermal Properties of Matter for the theory of expansion, calorimetry, change of state and heat transfer.
• Exemplar Mechanical Properties of Fluids — the previous unit, whose buoyancy ideas come back in Question 4.
• Exemplar Thermodynamics — the next unit, which builds on the zeroth law of Question 9.