NCERT Exemplar Class 11 Physics Solutions Chapter 11 Thermal Properties Of Matter

Vishal kumarUpdated on 10 Dec 2025, 06:03 PM IST

Chapter 11 on NCERT Exemplar Class 11 Physics is devoted to such concepts as heat transfer, measurement of temperature and thermal behaviour of various materials. Ever wondered why metal is colder than wood, yet they happen to be at the same temperature? In this chapter, there is a scientific explanation of such common observations. The students will get to know about heat conduction, convection, radiation, thermal expansion, and specific heat capacity, and the real-life applications, such as expansion joints in bridges, or how a thermos can keep liquids hot.

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NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ I
NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ II
NCERT Exemplar Class 11 Physics Solutions Chapter 11: Very Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 11: Short Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 11: Long Answer
NCERT Exemplar Class 11 Physics Solutions Chapter 11: Important Concepts and Formulas
Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 11 Thermal Properties of Matter
NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

The NCERT Exemplar Class 11 Physics Solutions Chapter 11 Thermal Properties of Matter gives step-by-step explanations that will enable the students to distinguish between heat and temperature. It also includes fundamentals like heat capacity, latent heat, law of cooling by Newton, and the Stefan-Boltzmann law. All these customised NCERT Exemplar solutions aim at improving the level of understanding, clarity of concepts and enabling students to perform better in board exams. Moreover, the NCERT Exemplar Class 11 Solutions Physics Chapter 11 Thermal Properties of Matter are helpful in learning because they provide the solutions to Numerical problems, multiple choice questions, and clearly organised concept explanations. This chapter is a major part of thermal physics preparation because these subjects are not only needed in examinations such as Class 11 but also in competitive exams such as JEE and NEET.

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NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ I

In NCERT Exemplar Class 11 Physics Chapter 11, the conceptual clarity of heat, temperature, thermal expansion and modes of heat transfer is tested in Multiple Choice Questions (MCQ I). These are some of the questions that assist the students in revising the major formulas in thermal physics to score better in board and competitive exams.

Question:11.1

A bimetallic strip is made of aluminium and steel $(\alpha_{ Al} > \alpha_{ steel})$. On heating, the strip will
a) remain straight
b) get twisted
c) will bend with aluminium on concave side
d) will bend with steel on concave side

Answer:

The answer is option (d) Will bend with steel on the concave side.
Explanation : Initially, Aluminium and Steel are fixed together in a bimetallic strip, and when you heat them, the expansion in steel will be smaller than the expansion in aluminium. Hence, the steel strip will be on the concave side, whereas the aluminium strip will be on the convex side.

Question:11.2

A uniform metallic rod rotates about its perpendicular bisector with constant angular speed. If it is heated uniformly to raise its temperature slightly
a) its speed of rotation increases
b) its speed of rotation decreases
c) its speed of rotation remains same
d) its speed increases because its moment of inertia increases

Answer:

When the rod is heated uniformly to raise its temperature slightly, the length of the rod will increase due to heating. This leads to an increase in its moment of inertia (I).
Since angular momentum, L=Iω
To maintain the angular momentum conserved, ω(angular speed ) decreases with an increase in moment of inertia.

Question:11.3

The graph between two temperature scales A and B is shown in the figure. Between upper fixed point and lower fixed point there are 150 equal division on scale A and 100 on scale B. The relationship for conversion between the two scales is given by
Graph between two temperatures
$a) \frac{t_{A}-180}{100}=\frac{t_{B}}{150}$
$b) \frac{t_{A}-30}{150}=\frac{t_{B}}{100}$
$c) \frac{t_{B}-180}{150}=\frac{t_{A}}{150}$
$d) \frac{t_{B}-40}{100}=\frac{t_{A}}{180}$

Answer:

The answer is the option $b) \frac{t_{A}-30}{150}=\frac{t_{B}}{100}$
Explanation: Now, from graph t_A, we know that,
Lower fixed point (LFP) = $30^{\circ}$
& upper fixed point (UFP) = $180^{\circ}$
Similarly, in the case of scale B,
UFP = $100^{\circ}$
& LFP = $0^{\circ}$
Thus, the formula,
$\frac{t_{A}-(LFP)}{(UFP)_{A}-(LFP)_{A}}= \frac{t_{B}-(LFP)_{B}}{(UFP)_{B}-(LFP)_{B}}$
$\frac{T_{A}-30}{180-30}= \frac{t_{B}-0}{100-0 } or \frac{ t_{A}-30}{150}= \frac{t_{B}}{100 }$

Question:11.4

An aluminium sphere is dipped into water. Which of the following is true?
a) buoyancy will be less in water at $0^{\circ}C$ than that in water at $4^{\circ}C$
b) buoyancy will be more in water at $0^{\circ}C$ than that in water at $4^{\circ}C$
c) buoyancy in water at $0^{\circ}C$ will be same as that in water at $4^{\circ}C$
d) buoyancy may be more or less in water at $4^{\circ}C$ depending on the radius of the sphere

Answer:

We know that,
Buoyant force (B.F.) =$V^{'}\rho lg$
Where, V’ = volume of displaced liquid by the dipped body
$\rho l$ = density if liquid
Now, at 4⁰C, the density of water is maximum; thus, the density of water and buoyant force will also be maximum at 4⁰C.
$B.F. \propto P_{l}$
Thus,
$\frac{F_{A}}{F_{0}}= \frac{\rho _{A}}{\rho _{0}}, where, \rho_{A}> \rho _{0}$
$\frac{F_{A}}{F_{0}}>1$or
$F_{A}>F_{0}$

Question:11.5

As the temperature is increased, the time period of a pendulum
a) increases as its effective length increases even though its centre of mass still remains at the centre of the bob
b) decreases as its effective length increases even though its centre of mass still remains at the centre of the bob
c) increases as its effective length increases due to shifting of centre of mass below the centre of the bob
d) decreases as its effective length remains same but the centre of mass shifts above the centre of the bob

Answer:

The answer is option (a) increases as its effective length increases, even though its centre of mass still remains at the centre of the bob.
Explanation: Due to linear expansion, the length will increase with temperature,
$T =\sqrt{\frac{ L}{g}} or T \propto L$
Hence, the effective length and T increase with increasing temperature.

Question:11.6

Heat is associated with
a) kinetic energy of random motion of molecules
b) kinetic energy of orderly motion of molecules
c) total kinetic energy of random and orderly motion of molecules
d) kinetic energy of random motion in some cases and kinetic energy of orderly motion in other

Answer:

The answer is option (a), kinetic energy of the random motion of molecules.
Explanation: Kinetic energy associated with the random motion of molecules increases as the vibration of molecules about their mean position increases with increasing temperature.

Question:11.7

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
$(a) 2 \pi R \alpha \Delta T$
$(b) \pi R^{2} \alpha \Delta T$
$(c) \frac{ \pi R^{3} \alpha \Delta T}{3}$
$(d) 4 \pi R^{3} \alpha \Delta T$

Answer:

The answer is the option
Explanation: Here, we know that,
$\alpha$ = coefficient of linear expansion
$3\alpha$= ϒ = coefficient of cubical expansion
Now, $V_{0} = \frac{4}{3}\pi R^{3}$
$\gamma = \frac{\Delta V}{V\Delta T}$
Thus, $\Delta V=\gamma V\Delta T$
i.e.,$\Delta V=3\alpha .\frac{4}{3}\pi R ^{3}\Delta T$
$=4\pi R ^{3}\Delta T$
Hence, opt (d).

Question:11.8

A sphere, a cube, and a thin circular plate, all of the same material and same mass are initially heated to same high temperature.
a) plate will cool fastest and cube the slowest
b) sphere will cool fastest and cube the slowest
c) plate will cool fastest and sphere the slowest
d) cube will cool fastest and plate the slowest

Answer:

The answer is option (c) Plate will cool fastest, and the sphere the slowest.
Explanation: On cooling, loss of heat is directly proportional to surface area exposed to the surroundings, the material of the object & temperature difference between the surroundings and the body.
Here, opt (c) is verified as the surface area of the sphere is minimum and thus it will cool slowest, while the surface area of the circular plate is maximum, and it will cool fastest.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ II

The conceptual and application-based multiple-choice question in the NCERT Exemplar Class 11 Physics Chapter 11 MCQ II is used to revise thermal physics on a very deep level in students. These questions are analytical with problem-solving abilities associated with heat transfer, thermal expansion and cooling laws. The NCERT Exemplar Class 11 Physics Solutions Chapter 11 provides an effective understanding of the concepts that can be applied to the CBSE exams as well as competitive tests.

Question:11.9

Mark the correct options:
a) A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z may be in thermal equilibrium with each other
b) A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other
c) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z must be in thermal equilibrium with each other
d) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z may be in thermal equilibrium with each other

Answer:

The answer is option (b). A system X is in thermal equilibrium with Y but not with Z. Systems Y and Z are not in thermal equilibrium with each other, and option (d) A system X is neither in thermal equilibrium with Y nor with Z. The systems Y and Z may be in thermal equilibrium with each other.
Explanation: (a) Here, Y and Z are not in thermal equilibrium since,
$T_{x}= T_{y}$ & $T_{x}$ is not equal to $T_{z}$.
Thus, $T_{x} = T_{y} \neq T_{z}.$ Hence, it is incorrect.
(b) Systems Y and Z are not in thermal equilibrium since,
$T_{x} = T_{y}$ and $T_{x} \neq T_{z}$
Thus, $T_{y} \neq T_{z}$. Thus, (b) is correct.
(c) Y and Z are not in thermal equilibrium because,
$T_{x} \neq T_{y}$ and $T_{x} \neq T_{z}$
Thus, $T_{y} \neq T_{z}$. Hence, opt (c) is incorrect.
(d) Here,
$T_{x} \neq T_{y}$ & $T_{x} \neq T_{z}$.
Here, $T_{z}$ may be equal to $T_{y}$. Hence, opt (d) is correct.

Question:11.10

Gulab Jamuns’ (assumed to be spherical) are to be heated in an oven. They are available in two sizes, one twice bigger 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:
a) both size gulab jamuns will get heated in the same time
b) smaller gulab jamuns are heated before bigger ones
c) smaller pizzas are heated before bigger ones
d) bigger pizzas are heated before smaller ones

Answer:

The answer is option (b) Smaller gulab jamuns are heated before bigger ones, and (c) Smaller pizzas are heated before bigger ones.
Explanation: Here, the smaller ones have a small thickness, through which heat to transfer up to depth gets heated before the bigger ones that have a larger depth.
Hence, opt (b) & (c).

Question:11.11

Refer to the plot of temperature versus time showing the changes in the state of ice on heating. Which of the following is correct?
Temperature - Time Graph
a) the region AB represents ice and water in thermal equilibrium
b) at B water starts boiling
c) at C all the water gets converted into steam
d) C to D represents water and steam in equilibrium at boiling point

Answer:

The answer is option (a) The region AB represents ice and water in thermal equilibrium, and (d) C to D represents water and steam in equilibrium at the boiling point.
Explanation: If the temperature of a substance does not change on applying heat continuously, then its state changes.
Here, AB = $0^{\circ}C$ & CD = $100^{\circ}C$ , thus,
(i) AB represents ice and water up to B &
(ii) CD represents water & steam.

Question:11.12

A glass full of hot milk is poured on the table. It begins to cool gradually. Which of the following is correct?
a) The rate of cooling is constant till milk attains the temperature of the surroundings
b) the temperature of milk falls off exponentially with time
c) While cooling, there is a flow of heat from milk to the surroundings as well as from the surroundings to the milk, but the net flow of heat is from milk to the surroundings, and that is why it cools
d) all three phenomena, conduction, convection, and radiation, are responsible for the loss of heat from milk to the surroundings

Answer:

The answer is the option (b) The temperature of milk falls off exponentially with time, (c) While cooling, there is a flow of heat from milk to the surrounding as well as from surrounding to the milk but the net flow of heat is from milk to the surrounding and that is why it cools and (d) All three phenomenon, conduction, convection and radiation are responsible for the loss of heat from milk to the surroundings.
Explanation:

Loss of heat is proportional to the temperature difference between the surroundings and the body. As milk cools, the rate of cooling decreases with time. Thus, opt (a) is incorrect.
The heat of milk falls exponentially by Newton’s law of cooling.
As compared to the heat lost by milk to the surroundings while cooling, a very small amount of heat also flows from the surroundings to milk. Hence, it is correct.
By conduction, convection, and radiation, when hot milk is spread on a table, it transfers heat to the surroundings. Hence, it is correct.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Very Short Answer

The Very Short Answer questions of NCERT Exemplar Class 11 Physics Chapter 11 focus on quick conceptual understanding of thermal properties of matter. These questions help students recall the basics of heat transfer, thermal expansion, and temperature in just a few lines. Practising them strengthens clarity and boosts confidence during exams.

Question:11.13

Is the bulb of a thermometer made of diathermic or adiabatic wall?
Diathermic walls are used in the bulb of thermometer as they allow to conduct heat through it into mercury of bulb, whereas adiabatic walls does not allow to pass heat through it.
A student records the initial length l, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

S.No	l (m)	$\Delta T (^{\circ}C)$	$\Delta l (m)$
1	2	10	$4 \times 10^{-4}$
2	1	10	$4 \times 10^{-4}$
3	2	20	$2 \times 10^{-4}$
4	3	10	$6 \times 10^{-4}$

If the first observation is correct, what can you say about observations 2, 3, and 4.

Answer:

Here, it is given that there is a rod and a material that has linear expansion ‘$\alpha$’, remains the same.
$\Delta t$ is the same here.
Now, from observation no. 1
$\alpha =\frac{\Delta L}{L \Delta t}$
$\alpha=\frac{4 \times 10^{-4}}{2 \times 10}= 2 \times 10^{-5} /C^0$
Now, for observation no. 2,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 1 \times 10$
$= 2 \times 10^{-4}m$ viz. not equal to $4 \times 10^{-4}m$
Now, for observation no. 3,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 2 \times 20$
$= 8 \times 10^{-4}$ viz. not equal to $= 2 \times 10^{-4}m$
Now, for the last observation, i.e., no. 4,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 3 \times 10$
$= 6 \times 10^{-4 }= 6 \times 10^{-4}m.$

Question:11.14

A student records the initial length l, change in temperature $\Delta T$ and change in length $\Delta l$ of a rod as follows:

S.No	l (m)	$\Delta T (^{\circ}C)$	$\Delta l (m)$
1	2	10	$4 \times 10^{-4}$
2	1	10	$4 \times 10^{-4}$
3	2	20	$2 \times 10^{-4}$
4	3	10	$6 \times 10^{-4}$

If the first observation is correct, what can you say about observations 2, 3, and 4.

Answer:

$\Delta t$ is the same here.
Now, from observation no. 1
$\alpha =\frac{\Delta L}{L \Delta t}$
$\frac{4 \times 10^{-4}}{2 \times 10}= 2 \times 10^{-5} C^{\circ}-1$
Now, for observation no. 2,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 1 \times 10$
$= 2 \times 10^{-4}m$ viz. not equal to $4 \times 10^{-4}m$
Now, for observation no. 3,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 2 \times 20$
$= 8 \times 10^{-4}$ viz. not equal to $= 2 \times 10^{-4}m$
Now, for the last observation, i.e., no. 4,
$\Delta L=\alpha L \Delta t$
$= 2 \times 10^{-5} \times 3 \times 10$
$= 6 \times 10^{-4 }= 6 \times 10^{-4}m.$

Question:11.15

Why does a metal bar appear hotter than a wooden bar at the same temperature? Equivalently it also appears cooler than wooden bar if they are both colder than room temperature.

Answer:

The rate of transferring heat in metal is larger than that in wood, as the conductivity of the metal bar is extremely high compared to that of wood.
Also, metal requires exceedingly small quantities of heat than wood to change temperature, as the specific heat of metal is exceptionally low compared to that of wood.
Thus, due to larger conductivity and smaller specific heat, metal becomes colder and hotter than wood when kept in cold and hot regions, respectively.

Question:11.16

Calculate the temperature which has same numerical value on Celsius and Fahrenheit scale.

Answer:

Let us consider that the required temperature is,
$x^{\circ}C = x^{\circ}F$
Now, we know that,
$\frac{C}{100}=\frac{F-32}{180}$
Thus,
$\frac{x}{5}=\frac{x-32}{9}$
Thus, $5x -160 = 9x$
-9x +5x = 160
-4x = 160x
Thus,$x = -40^{\circ}$
Thus, $-40^{\circ}F = -40^{\circ}C$

Question:11.17

These days people use steel utensils with copper bottom. This is supposed to be good for uniform heating of food. Explain this effect using the fact that copper is the better conductor.

Answer:

Copper base transfers heat quickly to food from the bottom and also needs a low quantity of heat as compared to steel because, as compared to steel, copper has higher conductivity and low specific heat. Thus, heat is supplied from the burner to the food in the utensil quickly and in a large amount when the base is copper.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Short Answer

The well-structured short-answer solutions facilitate the understanding of heat transfer, temperature variations and thermal properties of matter. These NCERT Exemplar Class 11 Physics Chapter 11 Short Answer questions will provide the clarity of concepts and will enable the students to cope with both numerical and theoretical problems effectively. These are good exercises that build the base for board exams and competitive entrance tests.

Question:11.18

Find out the increase in moment of inertia I of a uniform rod about its perpendicular bisector when its temperature is slightly increased by $\Delta T$.

Answer:

Now, we know that the moment of inertia (I) of a rod when its axis is along the perpendicular bisector is =
$\frac{1}{12}ML^{2}$
Now,$\Delta L = \alpha .L\Delta T$
Thus, $I^{'} = \frac{1}{12} M (L + \Delta L)^{2}$
$= \frac{1}{12} M (L ^{2}+\Delta L^{2}+2L \Delta L)^{2}$
Since $\Delta L^{2}$ is a very small term, we will neglect it,
$= \frac{M}{12} (L ^{2}+2L \Delta L)$
$= \frac{ML^{2}}{12}+\frac{ML\Delta L}{6}\times \frac{2L}{2L}$
$= \frac{ML^{2}}{12}+\frac{ML^{2}}{12}. \frac{2\Delta L}{L}$
$I^{'} = I (1 + 2\alpha \Delta T)$
Thus, the new moment of inertia will increase by $2I\alpha \Delta T$.

Question:11.19

During summers in India, one of the common practice to keep cool is to make ice balls of crushed ice, dip it in flavoured sugar syrup and sip it. For this a stick is inserted into crushed ice and is squeezed in the palm to make it into the ball. Equivalently in winter, in those areas where it snows, people make snowballs and throw around. Explain the formation of ball out of crushed ice or snow in the light P-T diagram of water.

Answer:

From the P-T graph, we know that at decreasing pressure in the liquid state at $0^{\circ}C$ and 1 atm, water takes on the ice state & increasing pressure from $0^{\circ}C$ to 1 atm takes water to the ice state.
If we squeeze crushed ice, some parts of it melt into water at $0^{\circ}C$ and fill the gap between ice flakes. Also, when we squeeze crushed ice, its melting point increases, and the water present in the ice flakes freezes, binding them together and making the ball more stable.

Question:11.20

100 g of water is supercooled to $-10^{\circ}C$. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?

Answer:

Given: Mass of water = 100g
Ice mixes with water at $-10^{\circ}C$
Now, the heat required by -$-10^{\circ}C$ ice to $0^{\circ} C$ ice $= ms\Delta t$
$= 100 \times 1 \times [0 - (-10)]$
Thus, Q = 1000cal
Thus,
$m = \frac{Q}{L}$
$=\frac{1000}{80}$
= 12.5 gm
Thus, there is 12.5 gm of water and ice in the mixture, and hence its temperature remains $0^{\circ} C$.

Question:11.21

One day in the morning, Ramesh filled up 1/3 bucket of hot water from geyser, to take bath. Remaining 2/3 was to be filled by cold water to bring mixture to a comfortable temperature. Suddenly Ramesh had to attend something which would take sometimes, say 5-10 minutes before he could take bath. Now he has two options:
a) fill the remaining bucket completely by cold water and then attend the work
b) first attend to the work and fill the remaining bucket just before taking bath.
Which option do you think would have kept water warmer? Explain.

Answer:

According to Newton’s law of cooling,
The rate of cooling(loss of heat) is proportional to the Temperature difference between the body & surroundings.
Thus, for the option

Since the temperature of the surroundings and water is small, water should be kept warmer; thus, the amount of heat energy lost here is less.
Here, a large amount of heat energy is lost since the temperature difference between the water and the surroundings is large.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Long Answer

The Thermal Properties of Matter Class 11 NCERT Exemplar goes deeper into such topics as heat transfer, thermal expansion, and calorimetry with long answer questions. These NCERT Exemplar Class 11 Physics Chapter 11 Solutions provide step-by-step explanations, which not only help the students grasp the logic but also help them find solutions to numericals and conceptual questions. Long-answer questions are easier to master using such well-organised explanations in order to pass the board exams and compete.

Question:11.22

We would like to prepare a scale whose length does not change with temperature. It is proposed to prepare a unit scale of this type whose length remains, say 10 cm. We can use a bimetallic strip made of brass and iron each of different length whose length would change in such a way that difference between their lengths remain constant. If $\alpha _{iron }= \frac{1.2 \times 10^{-5}}{K}$ and $\alpha _{brass}= \frac{1.8 \times 10^{-5}}{K}$, what should we take as the length of each strip?

Answer:

We can use iron and brass to make the required scale. Here, one end will be connected with brass and the other end will be only of iron at a distance of 10 cm at any temperature. Let us consider the initial length of iron and brass to be
$\alpha _{iron }= \frac{1.2 \times 10^{-5}}{K}$ &
$\alpha _{brass}= \frac{1.8 \times 10^{-5}}{K}$
Now, $L_{11 }- L_{1B }= 10cm$ …… (i)
Thus, $\alpha =\frac{ \Delta L}{L_{0}\Delta T }or\frac{ L_{2}-L_{1}}{L_{1}\Delta T}$
Now, $L_{2} = L_{1} + L_{1}\alpha \Delta T$
$= L_{1} (1+\alpha _{B}\Delta T)$
If the rod is heated, then the length will become,
$L_{21} and L_{2B}$
Now, $L_{21}- L_{2B}=10cm$
$L_{11}(1 + \alpha _{1}\Delta T) - L_{1B} (1+\alpha _{B}\Delta T) = 10$
$L_{11} + \alpha _{1}L_{11}\Delta T - L_{1B} - L_{1B} \alpha _{B}\Delta T = 10$
Thus, from (i),
$10 + (\alpha _{1}L_{11} -\alpha _{B}L_{1B}) \Delta T = 10 or \alpha _{1}L_{11} - \alpha _{B}L_{1B} = 0$
$\alpha _{1}L_{11} = \alpha _{B}L_{1B}$
$\frac{L_{11}}{L_{1B}}= \frac{\alpha _{B}}{\alpha _{1}}$
$= \frac{1.8 \times 10^{-5}}{1.2 \times 10^{-5}}$
$= \frac{3}{2}$
Now, let$L_{11} = 3x$ and $L_{1B} = 2x$
$L_{11} - L_{1} = 10$
3x – 2x = 10
Thus, x = 10.
Therefore, length of iron rod, (3x) = 30 & length of brass rod, (2x) = 20
Thus, the difference between the second ends will be 10 cm.

Question:11.23

We would like to make a vessel whose volume does not change with temperature. We can use brass and iron $\left (\beta _{brass} = \frac{6 \times 10^{-6}}{K} and \beta _{iron} = \frac{3.55 \times 10^{-5}}{K} \right )$to create a volume of 100 cc. How do you think you can achieve this?

Answer:

For the above situation, we need to make a double container whose volume difference will be 100cc.
Let us consider $V_{1i}$& $V_{1b}$ as the initial volumes of the iron and brass container, i.e., $V_{1i}- V_{1b} =100cc$.
After heating by $\Delta T K$, the difference will be the same, but the new volumes will be V_2i &V_2b.
Now, $V_{2i}- V_{2b} =100cc$
$\gamma = \frac{\Delta V}{V\Delta T}$
Therefore, $V_{2} - V_{1} = \gamma V_{1}\Delta T$
$V_{2} =V_{1}+ \gamma V_{1}\Delta T$
$= V_{1} (1 + \gamma \Delta T)$
$V_{2i}= V_{1i} (1 + \gamma_{i} \Delta T)$
$V_{2b} = V_{1b} (1 + \gamma _{b}\Delta T)$
$V_{1i} + V_{1i}\gamma _{i}\Delta T - (V_{ib} + V_{ib}\gamma _{b}\Delta T) = 100 cc.$
$100 + (V_{1i}\gamma _{i} - V_{1b}\gamma _{b}) \Delta T = 100$
$V_{1i}\gamma _{i} - V_{1b}\gamma _{b} = 0$
$\frac{V_{1i}}{V_{1b} }= \frac{\gamma _{b}}{\gamma _{i}}$
$=\frac{ 6 \times 10^{-5}}{3.55 \times 10^{-5}}$
$=\frac{ 6 }{3.55 }$
$=\frac{ 120}{71 }$
Now, let $V_{1i} = 120x , V_{1b} = 71x$
& $V_{1i} - V_{1b} = 100$
Thus, 120x – 71x = 100
49x = 100
$x =\frac{100}{49}$
= 2.04
Thus, $V_{1i} = 120 \times 2.04 = 245 cc$
$V_{1b} = 71 \times 2.04 = 145 cc$

Question:11.24

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at a temperature of $57^{\circ}C$ is drunk. You can take body temperature to be $37^{\circ}C$ and 1.7 x 10^-5 ⁰C, bulk modulus for copper =$140 \times 10^9 N/m^{2}.$

Answer:

Let $\Delta T$ be the change in temperature, $\Delta T = 57 - 37 = 20^{\circ}C$
Let $\alpha$ be linear expansion of body, $\alpha = \frac{1.7 \times 10^{-5}}{K}$
& ϒ be the cubical expansion $=3\alpha =3\times 1.7 \times 10^{-5}$
$= \frac{5.1 \times 10^{-5}}{K}$
Let V be the volume of the cavity and $\Delta V$ be the increase in its volume which is a result of an increase in temperature by $\Delta T$.
$\Delta V = \gamma V. \Delta T$
$\frac{\Delta V}{V} = \gamma \Delta T$
Now, we know that,
Thermal stress production = B x Volumetric strain
$=B.\frac{\Delta V}{V}$
$=B.\gamma \Delta T$
$= 140 \times 10^9 \times 5.1 \times 10^{-5} \times 20$
$= 1.428 \times 10^{8} Nm^{-2}$
Thus, the stress is $1.01 \times 10^{5} Nm^{-2}$, viz., $10^{3}$ times of atmospheric pressure.

Question:11.25

A rail track made of steel having length 10 m is clamped on a railway line at its two ends. On a summer day due to rise in temperature by $20^{\circ}C$, it is deformed as shown in the figure. Find x if $\alpha {steel} = \frac{1.2 \times 10^{-5}}{^{\circ}C}$.
Numerical on strain

Answer:

Given data:
$\alpha {steel} = \frac{1.2 \times 10^{-5}}{^{\circ}C}$
L₀ = 10m
& $\Delta T = 20^{\circ}C$
Now, by using Pythagoras' theorem,
$x^{2} = \left [(L + \Delta L){\frac{1}{2}} \right ]^{2} - \left (\frac{L}{2} \right )^{2}$
$= \frac{1}{4} [L^{2} + \Delta L^{2} + 2L\Delta L] -\frac{ L^{2}}{4}$
Thus, $x = \frac{L^{2}}{4} + \frac{\Delta L^{2}}{4} + \frac{2L\Delta L}{4} -\frac{ L^{2}}{4}$ ……… (neglecting $\Delta L^{2} since \Delta L^{2}<<L$)
$x^{2 }= \frac{2L\Delta L}{4}$
Thus, $x= \frac{1}{2} \sqrt{2L\Delta L}$
$\Delta L = L_{c} \alpha .\Delta T$
$= 10 \times 1.2 \times 10^{-5} \times 20$
$= 240 \times 10^{-5}$
Thus, $x= \frac{1}{2} \sqrt{2\times 10\times 240 \times 10^{-5}}$
$= \frac{40}{2}\times \sqrt{0.3} \times 10^{-2}$
$= 10.8 \times 10^{-2}m$
Thus, $x = 10.8 cm$

Question:11.26

A thin rod having length Lo at $0^{\circ}C$ and coefficient of linear expansion α has its two ends maintained at temperatures $\theta _{1}$and $\theta _{2}$, respectively. Find its new length.

Answer:

From one end to another end, the temperature of the rod varies from $\theta _{1}$ to $\theta _{2}$
Mean temperature of rod= $\frac{\theta _{1} + \theta _{2}}{2 ^{\circ}C}$

A thin rod

Here, the rate of flow of heat from A to C to B is equal
Thus, $\theta _{1} > \theta >\theta _{2}$
Thus, $\frac{d\theta}{dt} = \frac{KA (\theta _{1} - \theta )}{\frac{L_{0}}{2}}$
$= \frac{KA (\theta - \theta _{2})}{ {\frac{L_{0}}{2}}}$
Here, K is the coefficient of thermal conductivity.
Therefore, $\theta _{1} - \theta = \theta - \theta _{2}$
$\theta = \theta _{1} + \frac{\theta _{2}}{2}$
Thus, $L = L_{0} (1+\alpha \theta )$
$= L_{0} \left [1 + \alpha \left (\frac{ \theta _{1} +\theta _{2}}{2} \right ) \right ]$

Question:11.27

According to Stefan’s law of radiation, a black body radiates energy $\sigma T^{4}$ from its unit surface area every second where T is the surface temperature of the black body and $\sigma = 5.67 \times 10^{-8} W/m^{2}K^{4}$ is known as Stefan’s constant. A nuclear weapon may be thought of as a ball of radius 0.5 m. When denoted, it reaches temperature of 106K and can be treated as a black body.
a) estimate the power it radiates
b) if surrounding has water at $30^{\circ}C$, how much water can 10% of the energy produced evaporate in 1 sec?
c) if all this energy U is in the form of radiation, corresponding momentum is $p=\frac{U}{c}$. How much momentum per unit time does it impart on unit area at a distance of 1 km?

Answer:

Given: E = $\sigma T^{4}$ per sec per sq. m
Thus, total E is equal to radiated from all surface area A per second will be the power radiated by a nuclear weapon
Thus, P = $\sigma A T^{4}$
Now, $\sigma = 5.67 \times 10^{-8} W/m^{2}K^{4}$ &
$P = 5.67 \times 10^{-8} \times (4 \times \pi R^{2}) (10^{6})^{4}$
$= 5.67 \times 4 \times 3.14 \times \times 0.5 \times 0.5 \times 10^{-8} \times 10^{24}$ …. (R = 0.5m, T = 106K)
$= 5.67 \times 3.14 \times 10^{24-8} \times 1$
Thus, $P \cong 18 \times 10^{16} Watt$
$= 1.8 \times 10^{17}J/s$ …… (i)
(b) $P = 18 \times 10^{16} Watt$ from (i)
Now, for the evaporation of water, 10% of this power is required.
Thus, $E = \frac{10}{100}\times 18 \times 10^{16} Watt$
$= 1.8 \times 10^{17}J/s$
Now, energy required by m kg of water at $30^{\circ}C$ to evaporate at $100^{\circ}C$ = E required to heat up water from $30 ^{\circ} C +100^{\circ}C$+ E required to evaporate water into vapour
$= mS_{w} (T_{2} - T_{1}) + mL$
$= m[S_{w}(T_{2} - T_{1}) + L]$
Thus,
$18 \times 10^{16 }= m [ 4180 (100-30) + 22.6 \times 10^{5}]$
$m = \frac{18 \times 10^{16}}{25.5 \times 10^{5}}$
$\cong 7 \times 10^{9}kg$
(c) Now, we know that,
Momentum per unit time $p^{'}=\frac{U}{c}$
$= \frac{18 \times 10^{16}}{3 \times 10^{8}}$

Thus,
$p^{'}= 6 \times 10^{8} kg ms^{-1 }$
Required momentum per unit time
$=\frac{6 \times 10^{8}}{4 \times 3.14 \times (10^{3})^{2}}$
$\frac{6 \times 10^{8}}{4 \times 3.14 \times 10^{6} } = 47.77 \frac{kgms^{-1}}{m^{2} }$

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Important Concepts and Formulas

Learning Thermal Properties of Matter is necessary to know how heat affects various materials and why objects cool, conduct or expand differently. The NCERT Exemplar Class 11 Physics Solutions are clear on main concepts, derivations and numerical formulas, enabling the student to form a solid foundation in the basis to board examinations, JEE and NEET.

1. Temperature and Heat:

Heat is the energy transferred due to a temperature difference, while temperature is the measure of the average kinetic energy of particles.
Heat is measured in Joules (J), and temperature in Kelvin (K).
Formula:

$
Q=m \cdot s \cdot \Delta T
$

where $Q=$ heat absorbed, $m=$ mass, $s=$ specific heat capacity.

2. Specific Heat Capacity (s):

It is the amount of heat required to raise the temperature of 1 kg of a substance by 1 K.
Formula:

$
s=\frac{Q}{m \Delta T}
$

Water has a high specific heat capacity, which helps regulate the climate.

3. Calorimetry:

Used to determine the unknown specific heat capacity by heat exchange.
Principle of Calorimetry:

Heat lost = Heat gained

$
m_1 s_1 \Delta T_1=m_2 s_2 \Delta T_2
$

4. Change of State (Latent Heat):

Heat absorbed/released without temperature change during solid-liquid-gas transitions.
Formula:

$
Q=m L
$

where $L=$ latent heat.

5. Expansion of Solids, Liquids, and Gases:

Substances expand on heating:

Linear Expansion:

$
\Delta L=L_0 \alpha \Delta T
$

Area Expansion:

$
\Delta A=A_0 \beta \Delta T
$

Volume Expansion:

$
\Delta V=V_0 \gamma \Delta T
$

6. Newton’s Law of Cooling:

The rate of cooling is directly proportional to the temperature difference between the body and the surroundings.
Formula:

$
\frac{d T}{d t} \propto\left(T-T_s\right)
$

7. Thermal Conductivity (K):

Describes heat flow through a material.
Formula:

$
\frac{Q}{t}=K A \frac{\Delta T}{L}
$

where $Q=$ heat, $A=$ area, $L=$ thickness.

8. Stefan–Boltzmann Law:

Hot bodies radiate heat energy.
Formula:

$
E=\sigma T^4
$

where $\sigma$ is the Stefan constant.

9. Emissivity and Absorptivity:

Black bodies absorb all radiation; polished surfaces reflect more.
Kirchhoff’s Law: Good absorbers are good emitters.

10. Thermal Equilibrium:

When bodies in contact attain the same temperature, no net heat flow occurs.

Advantages of NCERT Exemplar Class 11 Physics Solutions Chapter 11 Thermal Properties of Matter

Through thermal behaviour, one can easily understand the Physics and NCERT Exemplar Class 11 Solutions to this chapter provide an understanding of the chapter and be able to pass the exam. These solutions comprise a broad range of conceptual and numerical problems that assist in strengthening the problem-solving abilities and the insight into the concepts of heat, temperature, expansion, and cooling.

These model solutions facilitate clarity of thought by providing clear accounts of how thermal expansion, heat flow, specific heat and other thermal effects work in a way that is easy to understand, which helps you solidify your theoretical knowledge.
The questions provided in the NCERT exemplar assist in enhancing the number-solving skills since they involve application-based questions that involve thermal properties and real-life situations, which makes the learning meaningful.
The solutions are very useful in preparation for exams since they contain the pattern of most frequently asked questions and step-by-step detailed answers, which make one faster and accurate in competitive and school examinations.
These Exemplar solutions serve as a great revision, and they motivate students to repeatedly solve meaningful questions, thus helping them to be more confident when taking the test or board exams.
They are also useful in developing stronger analytical skills as most of the questions are theoretical and need reasoning and application of laws like the laws of thermal physics, like Stefan's law of cooling and Newton's law of cooling, which are fundamental in learning thermal physics.

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

The solutions of NCERT Exemplar Class 11 Physics prove very helpful in passing the examinations because they contain conceptual questions, numericals, and application-based problems of all the chapters. These solutions can be used by students not only to prepare for CBSE but also for such competitive exams as JEE or NEET. Chapter links enable the students to obtain systematic content of the material in a single place.

Chapter 2 Units and Measurement

Chapter 3 Motion in a Straight Line

Chapter 4 Motion in a Plane

Chapter 5 Laws of Motion

Chapter 6 Work, Energy, and Power

Chapter 7 System of Particles and Rotational Motion

Chapter 8 Gravitation

Chapter 9 Mechanical Properties of Solids

Chapter 10 Mechanical Properties of Fluids

Chapter 11 Thermal Properties of Matter

Chapter 12 Thermodynamics

Chapter 13 Kinetic Theory

Chapter 14 Oscillations

Chapter 15 Waves

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

NCERT solutions for Class 11 Physics Chapter 1 Units and Measurement

NCERT solutions for Class 11 Physics Chapter 2 Motion in a straight line

NCERT solutions for Class 11 Physics Chapter 3 Motion in a Plane

NCERT solutions for Class 11 Physics Chapter 4 Laws of Motion

NCERT solutions for Class 11 Physics Chapter 5 Work, Energy and Power

NCERT solutions for Class 11 Physics Chapter 6 System of Particles and Rotational Motion

NCERT solutions for Class 11 Physics Chapter 7 Gravitation

NCERT solutions for Class 11 Physics Chapter 8 Mechanical Properties of Solids

NCERT solutions for Class 11 Physics Chapter 9 Mechanical Properties of Fluids

NCERT solutions for Class 11 Physics Chapter 10 Thermal Properties of Matter

NCERT solutions for Class 11 Physics Chapter 11 Thermodynamics

NCERT solutions for Class 11 Physics Chapter 12 Kinetic Theory

NCERT solutions for Class 11 Physics Chapter 13 Oscillations

NCERT solutions for Class 11 Physics Chapter 14 Waves

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also, Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

Q: Are thermal properties crucial for JEE mains?

Yes, this chapter is one of the most crucial physics chapters as, every year, questions are asked from these topics.

Q: How are these questions solved?

The questions are solved in the most detailed way for the students. We have made sure to add both theory and numerical in a way that is both CBSE accepted and is easy to retain.

Q: Are these solutions helpful?

Yes, these solutions are highly helpful to clear both theoretical understandings of the chapter, and also get an idea of how to solve numerical questions in an exam.

Q: What are daily life applications of thermal expansion?

It is used in railway tracks, bridges, hot water pipes, and thermostats to stop the damage due to expansion.

Q: Difference between heat capacity and specific heat capacity?

Heat capacity is total heat needed to change temperature, while specific heat is heat required to raise 1 kg of a substance by 1°C.

Q: Why is the anomalous expansion of water important?

Water expands below 4°C, making ice float, which helps aquatic life survive in winter.

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