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

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

Edited By Safeer PP | Updated on Apr 08, 2025 05:23 PM IST

NCERT Exemplar Class 11 Physics Solutions Chapter 11 discusses heat, temperature, and their impacts on various materials. Why does metal seem colder than wood at the same temperature? Physics Chapter 11 explains the phenomena such as thermal expansion, heat transfer and specific heat capacity. It also helps us to relate everyday applications, such as why there are expansion gaps in bridges or how thermos flasks keep the liquids warm!

This Story also Contains
  1. NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ I
  2. NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ II
  3. NCERT Exemplar Class 11 Physics Solutions Chapter 11: Very Short Answer
  4. NCERT Exemplar Class 11 Physics Solutions Chapter 11: Short Answer
  5. NCERT Exemplar Class 11 Physics Solutions Chapter 11: Long Answer
  6. Important Concepts and Formulas - NCERT Exemplar Class 11 Physics Solutions Chapter 11
  7. Topics in NCERT Exemplar Class 11 Physics Solutions Chapter 11
  8. NCERT Exemplar Class 11 Physics Solutions Chapter-Wise
  9. NCERT Exemplar Class 11th Solutions

NCERT Exemplar Class 11 Physics Solutions Chapter 11 helps the students to understand the difference between heat and temperature. It covers concepts such as thermal expansion heat capacity and laws such as Stefan's law and Newton's law of cooling. These solutions consist of elaborate descriptions and solution methods for improved exam preparation. It is crucial to grasp these topics in order to master thermal physics.

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NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ I

Question:11.1

A bimetallic strip is made of aluminium and steel (αAl>αsteel). 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 the option (d) Will bend with steel on 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 be increased due to heating. This leads to 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)tA180100=tB150
b)tA30150=tB100
c)tB180150=tA150
d)tB40100=tA180

Answer:

The answer is the option b)tA30150=tB100
Explanation: Now, from graph tA, we know that,
Lower fixed point (LFP) = 30
& upper fixed point (UFP) = 180
Similarly, in case of scale B,
UFP = 100
& LFP = 0
Thus, the formula,
tA(LFP)(UFP)A(LFP)A=tB(LFP)B(UFP)B(LFP)B
TA3018030=tB01000ortA30150=tB100

Question:11.4

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

Answer:

We know that,
Buoyant force (B.F.) =Vρlg
Where, V’ = volume of displaced liquid by dipped body
ρl = 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.Pl
Thus,
FAF0=ρAρ0,where,ρA>ρ0
FAF0>1or
FA>F0

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 the 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=LgorTL
Hence, the effective length and T increases on 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 the option (a) kinetic energy of random motion of molecules.
Explanation: Kinetic energy associated with random motion of molecules increases as vibration of molecules about their mean position increases on 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 α. The sphere is heated a little by a temperature ΔT so that its new Temperature is T+ΔT. The increase in the volume of the sphere is approximately
(a)2πRαΔT
(b)πR2αΔT
(c)πR3αΔT3
(d)4πR3αΔT

Answer:

The answer is the option
Explanation: Here, we know that,
α = coefficient of linear expansion
3α= ϒ = coefficient of cubical expansion
Now, V0=43πR3
γ=ΔVVΔT
Thus, ΔV=γVΔT
i.e.,ΔV=3α.43πR3ΔT
=4πR3ΔT
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 the option (c) Plate will cool fastest and sphere the slowest.
Explanation: On cooling, loss of heat is directly proportional to surface area exposed to the surrounding, material of object & temperature difference between the surrounding 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 circular plate is maximum, and it will cool fastest.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: MCQ II

Question:11.9

Mark the correct options:
a) A system X is in thermal equilibrium with Y but not with Z. System 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 system Y and Z may be in thermal equilibrium with each other

Answer:

The answer is the option (b) A system X is in thermal equilibrium with Y but not with Z. System 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 system Y and Z may be in thermal equilibrium with each other.
Explanation: (a) Here Y and Z are not in thermal equilibrium since,
Tx=Ty & Tx is not equal to Tz.
Thus, Tx=TyTz.hence, it is incorrect.
(b) System Y and Z are not in thermal equilibrium since,
Tx=Ty and TxTz
Thus, TyTz. Thus, (b) is correct.
(c) Y and Z are not in thermal equilibrium because,
TxTy and TxTz
Thus, TyTz. Hence, opt (c) is incorrect.
(d) Here,
TxTy & TxTz.
Here, Tz may be equal to Ty. 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 the 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 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 the option (a) The region AB represents ice and water in thermal equilibrium and (d) C to D all represents water and steam in equilibrium at boiling point.
Explanation: If the temperature of a substance does not change on applying heat continuously, then its state changes.
Here, AB = 0C & CD = 100C , 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 surrounding
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
d) all three phenomenon, 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:

  1. Loss of heat is proportional to temperature difference with surrounding and body. As milk cools, rate of cooling decreases with time. Thus, opt (a) is incorrect.
  2. The heat of milk falls exponentially by Newton’s law of cooling.
  3. As compared to heat lost by milk to surrounding while cooling it, a very small amount of heat also flows from surrounding to milk. Hence, it is correct.
  4. 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

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 ΔT and change in length Δl of a rod as follows:

S.No
l (m)
1
2
10
2
1
10
3
2
20
4
3
10
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 ‘α’, remains same.
Δt is same here.
Now, from observation no. 1
α=ΔLLΔt
α=4×1042×10=2×105/C0
Now, for observation no. 2,
ΔL=αLΔt
=2×105×1×10
=2×104m viz. not equal to 4×104m
Now, for observation no. 3,
ΔL=αLΔt
=2×105×2×20
=8×104 viz. not equal to =2×104m
Now, for the last observation, i.e., no. 4,
ΔL=αLΔt
=2×105×3×10
=6×104=6×104m.

Question:11.14

A student records the initial length l, change in temperature ΔT and change in length Δl of a rod as follows:

S.No
l (m)
1
2
10
2
1
10
3
2
20
4
3
10
If the first observation is correct, what can you say about observations 2, 3, and 4.

Answer:

Δt is same here.
Now, from observation no. 1
α=ΔLLΔt
4×1042×10=2×105C1
Now, for observation no. 2,
ΔL=αLΔt
=2×105×1×10
=2×104m viz. not equal to 4×104m
Now, for observation no. 3,
ΔL=αLΔt
=2×105×2×20
=8×104 viz. not equal to =2×104m
Now, for the last observation, i.e., no. 4,
ΔL=αLΔt
=2×105×3×10
=6×104=6×104m.

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 as compared to the wood.
Also, metal requires exceedingly small quantities of heat than wood to change temperature as the specific heat of metal is exceptionally low as compared to the 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,
xC=xF
Now, we know that,
C100=F32180
Thus,
x5=x329
Thus, 5x160=9x
-9x +5x = 160
-4x = 160x
Thus,x=40
Thus, 40F=40C

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 burner to food in utensil quickly and in a large amount when the base is copper.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Short Answer

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 ΔT.

Answer:

Now, we know that moment of inertia (I) of a rod when its axis is along perpendicular bisector it is =
112ML2
Now,ΔL=α.LΔT
Thus, I=112M(L+ΔL)2
=112M(L2+ΔL2+2LΔL)2
Since ΔL2 is a very small term, we will neglect it,
=M12(L2+2LΔL)
=ML212+MLΔL6×2L2L
=ML212+ML212.2ΔLL
I=I(1+2αΔT)
Thus, the new moment of inertia will increase by 2IαΔT.

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 0C and 1 atm takes water to ice state & increasing pressure from 0C to 1 atm takes water to ice state.
If we squeeze crushed ice, some parts of it melt into water at 0C 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 10C. 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 10C
Now, the heat required by -10C ice to 0C ice =msΔt
=100×1×[0(10)]
Thus, Q = 1000cal
Thus,
m=QL
=100080
= 12.5 gm
Thus, there is 12.5 gm of water and ice in the mixture and hence its temperature remains 0C.

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,
Rate of cooling(loss of heat) is proportional to Temperature difference of body & surrounding.
Thus, for option
  1. Since the temperature of the surrounding and water is small, water should be kept warmer, thus, the amount of heat energy lost here is less.
  2. Here a large amount of heat energy is lost since the temperature difference between water and surrounding is large.

NCERT Exemplar Class 11 Physics Solutions Chapter 11: Long Answer

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 αiron=1.2×105K and αbrass=1.8×105Kwhat 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-
αiron=1.2×105K &
αbrass=1.8×105K
Now, L11L1B=10cm …… (i)
Thus, α=ΔLL0ΔTorL2L1L1ΔT
Now, L2=L1+L1αΔT
=L1(1+αBΔT)
If the rod is heated then, the length will become,
L21andL2B
Now, L21L2B=10cm
L11(1+α1ΔT)L1B(1+αBΔT)=10
L11+α1L11ΔTL1BL1BαBΔT=10
Thus, from (i),
10+(α1L11αBL1B)ΔT=10orα1L11αBL1B=0
α1L11=αBL1B
L11L1B=αBα1
=1.8×1051.2×105
=32
Now, letL11=3x and L1B=2x
L11L1=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 (βbrass=6×106Kandβiron=3.55×105K)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 V1i& V1b as the initial volumes of the iron and brass container, i.e., V1iV1b=100cc.
After heating by ΔTK the difference will be the same, but the new volumes will be V2i &V2b.
Now, V2iV2b=100cc
γ=ΔVVΔT
Therefore, V2V1=γV1ΔT
V2=V1+γV1ΔT
=V1(1+γΔT)
V2i=V1i(1+γiΔT)
V2b=V1b(1+γbΔT)
V1i+V1iγiΔT(Vib+VibγbΔT)=100cc.
100+(V1iγiV1bγb)ΔT=100
V1iγiV1bγb=0
V1iV1b=γbγi
=6×1053.55×105
=63.55
=12071
Now, let V1i=120x,V1b=71x
& V1iV1b=100
Thus, 120x – 71x = 100
49x = 100
x=10049
= 2.04
Thus, V1i=120×2.04=245cc
V1b=71×2.04=145cc

Question:11.24

Calculate the stress developed inside a tooth cavity filled with copper when hot tea at temperature of 57C is drunk. You can take body temperature to be 37C and 1.7 x 10-5 0C, bulk modulus for copper =140×109N/m2.

Answer:

Let ΔT be the change in temperature, ΔT=5737=20C
Let α be linear expansion of body, α=1.7×105K
& ϒ be the cubical expansion =3α=3×1.7×105
=5.1×105K
Let V be the volume of the cavity and ΔV be the increase in its volume which is a result of increase in temperature by ΔT.
ΔV=γV.ΔT
ΔVV=γΔT
Now, we know that,
Thermal stress production = B x Volumetric strain
=B.ΔVV
=B.γΔT
=140×109×5.1×105×20
=1.428×108Nm2
Thus, the stress is 1.01×105Nm2, viz., 103 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 20C, it is deformed as shown in the figure. Find x if αsteel=1.2×105C.
Numerical on strain

Answer:

Given data:
αsteel=1.2×105C
L0 = 10m
& ΔT=20C
Now, by using Pythagoras theorem,
x2=[(L+ΔL)12]2(L2)2
=14[L2+ΔL2+2LΔL]L24
Thus, x=L24+ΔL24+2LΔL4L24 ……… (neglecting ΔL2sinceΔL2<<L)
x2=2LΔL4
Thus, x=122LΔL
ΔL=Lcα.ΔT
=10×1.2×105×20
=240×105
Thus, x=122×10×240×105
=402×0.3×102
=10.8×102m
Thus, x=10.8cm

Question:11.26

A thin rod having length Lo at 0C and coefficient of linear expansion α has its two ends maintained at temperatures θ1and θ2, respectively. Find its new length.

Answer:

From one end to another end the temperature of rod varies from θ1 to θ2
Mean temperature of rod= θ1+θ22C

A thin rod
Here, the rate of flow of heat from A to C to B is equal
Thus, θ1>θ>θ2
Thus, dθdt=KA(θ1θ)L02
=KA(θθ2)L02
Here K is the coefficient of thermal conductivity.
Therefore, θ1θ=θθ2
θ=θ1+θ22
Thus, L=L0(1+αθ)
=L0[1+α(θ1+θ22)]

Question:11.27

According to Stefan’s law of radiation, a black body radiates energy σT4 from its unit surface area every second where T is the surface temperature of the black body and σ=5.67×108W/m2K4 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 30C, 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=Uc. How much momentum per unit time does it impart on unit area at a distance of 1 km?

Answer:

Given: E = σT4 per sec per sq. m
Thus, total E is equal to radiated from all surface area A per second will be power radiated by nuclear weapon
Thus, P = σAT4
Now, σ=5.67×108W/m2K4 &
P=5.67×108×(4×πR2)(106)4
=5.67×4×3.14××0.5×0.5×108×1024 …. (R = 0.5m, T = 106K)
=5.67×3.14×10248×1
Thus, P18×1016Watt
=1.8×1017J/s …… (i)
(b) P=18×1016Watt from (i)
Now, for evaporation of water 10% of this power is required.
Thus, E=10100×18×1016Watt
=1.8×1017J/s
Now, energy required by m kg of water at 30C to evaporate at 100C = E required to heat up water from 30C+100C+ E required to evaporate water into vapour
=mSw(T2T1)+mL
=m[Sw(T2T1)+L]
Thus,
18×1016=m[4180(10030)+22.6×105]
m=18×101625.5×105
7×109kg
(c) Now, we know that,
Momentum per unit time p=Uc
=18×10163×108

Thus,
p=6×108kgms1
Required momentum per unit time
=6×1084×3.14×(103)2
6×1084×3.14×106=47.77kgms1m2

We here provide our students with the best NCERT Exemplar Class 11 Physics solutions chapter 11 PDF download that they can get for the NCERT questions. NCERT is the major part of entrance exam preparation. Therefore, being well acquainted with the NCERT questions and their solutions can give a major boost to the chapter's prep. We have here detailed NCERT Exemplar Class 11 Physics chapter 11 solutions, which can be studied both online and offline by downloading the same in PDF format.

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

1. Temperature and Heat:

  • Heat is a form of energy transferred due to a temperature difference.
  • Temperature measures the degree of hotness or coldness of a body.
  • Formula: Q=mcΔT (Heat energy, mass, specific heat, temperature change)
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2. Measurement of Heat:

  • Heat is measured in joules (J) or calories (cal).
  • Specific Heat Capacity (c): The heat required to raise the temperature of 1 kg of a substance by 1C.

3. Ideal Gas Equation and Absolute Temperature:

  • Equation: PV=nRT ( P= Pressure, V= Volume, n= number of moles, R= Universal Gas Constant, T= Temperature in Kelvin)
  • Absolute temperature is measured in Kelvin (K) and follows gas laws.

4. Thermal Expansion:

  • Solids, liquids, and gases expand when heated.
  • Linear Expansion Formula: ΔL=αLΔT
  • Volume Expansion Formula: ΔV=βVΔT

5. Specific Heat Capacity:

  • Different materials require different amounts of heat to change temperature.
  • Formula: c=QmΔT

6. Calorimetry:

  • Used to measure heat exchange.
  • Principle: Heat lost = Heat gained
  • Formula: m1c1ΔT1=m2c2ΔT2

7. Change of State:

  • Phase change occurs at constant temperature (melting, boiling, freezing).
  • Latent Heat Formula: Q=mL ( L= Latent Heat)

8. Heat Transfer:

  • Conduction: Heat transfer in solids. Formula: Q=kAΔTd
  • Convection: Heat transfer in fluids.
  • Radiation: Heat transfer without a medium. Stefan's Law: E=σT4

9. Newton's Law of Cooling:

  • Formula: dTdt(TTs)
  • The rate of cooling depends on the temperature difference with the surroundings.

Topics in NCERT Exemplar Class 11 Physics Solutions Chapter 11

NCERT Exemplar Class 11 Physics Solutions Chapter-Wise

NCERT Exemplar Class 11th Solutions

Check Class 11 Physics Chapter-wise Solutions

Also, Read NCERT Solution subject-wise -

Check NCERT Notes subject-wise -

Also Check NCERT Books and NCERT Syllabus here

Frequently Asked Questions (FAQs)

1. 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.

2. 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.

3. 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.

4. 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.

5. 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.

6. 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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A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times.  Assume that the potential energy lost each time he lowers the mass is dissipated.  How much fat will he use up considering the work done only when the weight is lifted up ?  Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate.  Take g = 9.8 ms−2 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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