 # HC Verma Solutions Class 12 Chapter 1 Heat and Temperature

HC Verma Solutions Class 12 Chapter 1 Heat and Temperature is provided here to help students understand the topics more clearly. Usually, before working on any exercise it is important to understand the key concepts and terminologies. Therefore the solutions are designed to help students learn more about the chapter topics and also tackle difficult questions. The chapter deals with the topic of heat which can be defined as the degree or level of hotness/ coldness of an object or environment. It is the transfer of energy from a hotter system to a cooler system due to a temperature difference between them. The SI unit of heat is joule (J). A systems temperature rises when the heat is transferred to it. Temperature is measured in degree Celsius, Kelvin or Fahrenheit.

In addition to understanding the concepts, the chapter also contains important exercises that students should solve and practice. These exercises are also extremely important as it gives students an idea to tackle complex problems based on the concepts related to Heat and Temperature. The HC Verma solutions further contain answers to all the exercises which students can refer and study effectively for their final exam and various competitive exams like JEE, NEET, etc.

Physics is a branch of science and is the basis of future technology. Students who aspire to excel in the developing field of science and technology should have strong basics in physics. These Solutions will help students master the concepts and enhance their problem-solving skills. Get solutions of HC Verma book for Class 12 below.

## Heat And Temperature Main Topics

Some of the exercises are based on topics such as;

• The relation between Heat and Temperature
• Thermal equilibrium
• The relation between Celsius and Fahrenheit, Kelvin
• Ice point
• Steam point
• Triple point
• Constant Volume gas thermometer
• Change of moment of inertia with temperature
• The coefficient of linear expansion

## Class 12 Important Questions In Chapter 1

1. Can two bodies be in thermal equilibrium if they are not in contact? Explain

2. When a glass bottle with a tightly closed metal lid is put in hot water for some time, it can be opened more easily. Why does this happen?

3. We take a metal sheet and make a circular hole in it. Then the sheet is heated. What will happen to the hole?

(a) The hole increases and becomes bigger (b) Becomes smaller (c) no changes in size (d) Is deformed.

4. Suppose the temperature at the surface of a lake is 4°C. The temperature of water at the bottom can be;

(a) 0°C (b) 4°C (c) 6°C (d) 9°C.

5. Mark the correct pair or pairs that may give equal numerical values for the temperature of a body?

(a) Platinum and Kelvin (b) Celsius and Kelvin (c) Kelvin and Fahrenheit

## HC Verma Solutions Vol 2 Heat and Temperature Chapter 1

Question 1: The steam point and the ice point of a mercury thermometer are marked at 80° and 20°. What will be the temperature in centigrade mercury scale when this thermometer reads 32°?

Solution:

Ice point = Lo = 20o and L1 = 32o

Steam point = L100 = 80 o

Let T be the temperature in centigrade scale corresponding to 32o

Using formula,

T = [L1 – L0]/[L100 – L0] x 100

= [32-20]/[80-20]

= 20oC

Question 2: A constant volume thermometer registers pressure of 1.500 × 104 Pa at the triple point of water and a pressure of 2.050 × 104 Pa at the normal boiling point. What is the temperature at the normal boiling point?

Solution:

The temperature at triple point pressure “Ptp” of water is standardized as, Ttp = 273.16 K.

We know, for constant volume gas themometer,

T = P/Ptr x 273.16 K

= [2.050×104]/[1.5 x104] x 273.16

= 373.31

Temperature at normal Boiling point is 373.31 K

Question 3: A gas thermometer measures the temperature from the variation of pressure of a sample of gas. If the pressure measured at the melting point of lead is 2.20 times the pressure measured at the triple point of water, find the melting point of lead.

Solution:

Pressure at Melting point= 2.20×Pressure at triple point of water (Given)

T = P/Ptr x 273.16 k

= [2.2 x ptr]/Ptr x 273.16 k

= 600.952 K

Question 4: The pressure measured by a constant volume gas thermometer is 40 kPa at the triple point of water. What will be the pressure measured at the boiling point of water (100°C)?

Solution:

Pressure at triple point of water = ptr = 40kPa = 40×103 Pa

Boiling point of water = T = 100°C = 373 K

Find the value of P:

We know, T = P/Ptr x 273.16 k

=> P = [TxPtr]/273.16 k

= [373x49x103]/273.16

= 54620 Pa

= 54.64 KPa

Question 5: The pressure of the gas in a constant volume gas thermometer is 70 kPa at the ice point. Find the pressure at the steam point.

Solution:

We know, T = P/Ptr x 273.16 k

The pressure at ice point = P1 = 70kPa

Now,

T1 = P1/Ptr x 273.16 k

=> 273 = [70×103]/Ptr x 273.16

=> P tr = [70×103x273.16]/273

Again, T2 = P2/P tr x 273.16 k

=> 373 = [P2x273]/[70×103x273.16]

=> P2 = [373x70x103]/273

=> P2 = 95.6 K Pa

Question 6: The pressures of the gas in a constant volume gas thermometer are 80 cm, 90 cm and 100 cm of mercury at the ice point, the steam point and in a heated wax bath respectively. Find the temperature of the wax bath.

Solution:

Pice = Pressure at ice point = 80cm of Hg

Tice = Ice point temperature = 0°C

Psteam = Pressure at steam point = 90cm of Hg

Tsteam = Steam point temperature = 100°C

Pressure of wax bath = P = 100cm of Hg

Let “T” be the temperature of the wax bath.

Find T.

Now, Question 7: In a Callender’s compensated constant pressure air thermometer, the volume of the bulb is 1800 cc. When the bulb is kept immersed in a vessel, 200 cc of mercury has to be poured out. Calculate the temperature of the vessel.

Solution:

v/[v-v’] = T/To

Where, T = temperature of vessel

V = Initial volume = 1800CC

V’= The poured out amount for compensation=200CC

T0 = Ice point temperature= 273K

=> T = 1800/(1800-200) x 273

= 307.125 K

Question 8: A platinum resistance thermometer reads 0° when its resistance is 80Ω and 100 o is when resistance is 90 Ω. Find the temperature at the platinum scale at which the resistance is 86 Ω.

Solution:

We know, T = [RT – Ro]/[R100-Ro] x 100

Where, RT = Resistance at which temperature is to be measured.

R0 = Resistance at 0°C

R100 = Resistance at 100°C

T = (86-80)/(90-80) x 100

= 60o C

Temperature at 86Ω is 60° C.

Question 9: A resistance thermometer reads R = 20.0Ω, 27.5Ω, and 50.0Ω at the ice point (0°C), the steam point (100°C) and the zinc point (420°C) respectively. Assuming that the resistance varies with temperature as Rθ = R0 (I + αθ + βθ2), find the values of R0, α and β. Here θ represents the temperature on the Celsius scale.

Solution:

Rθ = R0(1 + α θ + β θ2)

The values for 3 resistance ( Rθ ) are given at 3 temperatures (θ).

At temperature 0° C:

Rθ1 = Ro(1 + α x0 + β x 02) ….(1)

At temperature 100° C:

Rθ2 = Ro(1 + α x100 + β x 1002) ….(2)

At temperature 420° C:

Rθ3 = Ro(1 + α x420 + β x 4202) ….(3)

Solving these three equations simultaneously for three unknowns,

From (1), Rθ1 = Ro = 20 Ω

Put value of Ro in (2) and (3)

(2)=> 27.5 Ω = 20 Ω(1 + α x100 + β x 1002) and

(3)=> 50 Ω = 20 Ω (1 + α x420 + β x 4202)

from eqn.2 after re arranging, we get,

α = [0.375-1002x β]/100 …………(4) and

putting value of α in (3), we get

50 Ω = 20 Ω (1 + [[0.375-1002x β]/100 ] x420 + β x 4202)

Or

1.5 = 4.2 x 0.375 – 1002 x β + 4202 x β

Or

β = -5.58 x 10-7

Putting the value of β in (4) , we get

α = 3.806 x 10-3

Hence, the required values are

R0 = 20 Ω

α = 3.806 x 10-3

β = -5.58 x 10-7

Question 10: A concrete slab has a length of 10 m on a winter night when the temperature is 0°C. Find the length of the slab on a summer day when the temperature is 35°C. The coefficient of linear expansion of concrete is 1.0×10-5°C-1

Solution:

The length of the lab at 0° C = L0 = 10m

Change in temperature of slab from 0°C = θ = 35

Coefficient of linear expansion:

Using relation, Lt = Lo(1 + αθ) …(1)

Where,

L0 = Length at a reference temperature = 10m

α = coefficient of linear expansion = 1.0×10-5 °C-1

(1)=> Lt = 10(1 + 1 x 10-5 x 35) = 10.0035 m

Therefore, the length of the slab at 35°C is 10.0035m

Question 11: A metre scale made of steel is calibrated at 20°C to give a correct reading. Find the distance between 50 cm mark and 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10-5 oC-1.

Solution:

Using relation, Lt = Lo(1 + αθ) …(1)

Where,

Lo = Length at a reference temperature

α = coefficient of linear expansion and

θ = Change in temperature.

Here, 20°C can be taken as reference temperature and Lo =length between adjacent centimeter markings = 1 cm or 0.01 m

Also given, θ= -10° C and α = 1.1 × 10-5 °C-1,

(1) => Lt = 0.01(1 + 1.1 × 10-5 x (-10)) = 0.99989cm

Thus, final length between 50cm and 51cm marks will be 0.99989cm.

Question 12: A railway track (made of iron) is laid in winter when the average temperature is 18°C. The track consists of sections of 12.0 m placed one after the other. How much gap should be left between two such sections so that there is no compression during summer when the maximum temperature goes to 48°C? Coefficient of linear expansion of iron = 11 ×10-6 °C-1

Solution:

Using relation, Lt = Lo (1 + αθ) …(1)

Where,

Lo = Length at a reference temperature

α = coefficient of linear expansion and

θ = Change in temperature.

Here, Lo = 12 m and α = 11 × 10-6 °C-1

At temp 18o C

(1) => L18 = 12(1 + 11 × 10-6 x 18) = 12.00237 m

At temp 48o C

(1) => L48 = 12(1 + 11 × 10-6 x 48) = 12.006336 m

Now, ΔL = L48 – L18 = 12.006336 – 12.00237 = 0.4 cm

So, then there must be a gap of 0.4 cm between the rails.

Question 13: A circular hole of diameter 2.00 cm is made in an aluminium plate at 0°C. What will be the diameter at 100°C? α for Aluminium = 2.3 x 10-5 oC-1.

Solution:

d2 = d1( 1 + α Δt) …(1)

Here,

d1 = 2m = 2 x 10-2 cm

t1 = 0 oC and t2 = 100 oC

α = 2.3 x 10-5 0C

(1) => d2 = (2 x 10-2) (1 + (2.3 x 10-5) 102)

= 0.020046 m = 2.0046 cm

Question 14: Two-metre scales, one of steel and the other of aluminium, agree at 20oC . Calculate the ratio aluminium -centimetre/steel-centimetre at (a) 0°C, (b) 40°C and (c) 100°C.

α for steel = 1.1 x 10-5 oC-1 and for aluminium 2.3 x 10-5 oC-1

Solution:

Using relation, Lt = Lo(1 + αθ) …(1)

Where,

L0 = Length at a reference temperature

α = Coefficient of linear expansion and

θ = Change in temperature.

Here, L0 = 12 m and α = 11 × 10-6 °C-1

Let ‘Al’ and ‘S’ will be used as subscripts to denote Aluminium and Steel respectively.

(a) At temp 0oC

LAl/LS = [Lo(1 + αAl (-20))] /[Lo(1 + αS(-20))]

= [1 – 2.3×10-5x20)] /[1 – 1.1×10-5 x20]

= 0.99977

(b) At temp 40oC

LAl/LS = [Lo(1 + αAl (20))] /[Lo(1 + αS(20))]

= [1 + 2.3×10-5x20)] /[1 + 1.1×10-5 x20]

= 1.000249

(c) At temp 100oC

LAl/LS = [Lo(1 + αAl (80))] /[Lo(1 + αS(80))]

= [1 + 2.3×10-5x80)] /[1 + 1.1×10-5 x80]

= 1.000954

Question 15: A meter scale is made up of steel and measures correct length at 16°C. What will be the percentage error if this scale is used (a) on a summer day when the temperature is 46°C and (b) on a winter say when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10–6 °C–1.

Solution:

T0 = 16° = 289 K and α = 11× 10-6 C–1

(a) The temperature at which the scale measures during a summer day = Ts = 46°= 319 K.

Therefore, ΔT = 319 – 289 = 30K

The change in length due to linear expansion, ΔL = lα ΔT

Where l is length of the meter scale.

=> ΔL/l = α ΔT

Now, percentage error = ((ΔL/l) x 100)%

=> α ΔT x 100 = 11 x 10-6 x 30 x 100 = 0.033%

(b) Tw = 6° = 279 K

and Δ T = 289 – 279 = 10K

Now, percentage error = ((ΔL/l) x 100)%

> α ΔT x 100 = 11 x 10-6 x 10 x 100 = 0.011%

Question 16: A metre scale made of steel reads accurately at 20°C. In a sensitive experiment, distances up to 0.055 mm in 1 m are required. Find the range of temperature in which the experiment can be performed with this metre scale. Coefficient of linear expansion of steel = 11 × 10–6 °C–1.

Solution:

We know, ΔL = Lo α ΔT

Where Δ L is the change in length and Δ T is the change in temperature.

0.055 x 10-3 = 11 x 10-6 x 1 x (20 ± T2)

=> 5 = 20 ± T2

=> T2 = 15oC

Either T2 = 15oC or T2 = 25oC

The experiment can be performed with the given metre scale is 15° C to 25° C.

Question 17: The density of water at 0°C is 0.998 g cm-3 and at 4°C is 1.000g cm-3. Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4°C.

Solution:

Density of water at 0°C = ρ0 = 0.998 g cm-3

Density of water at 4°C: ρ4 = 1.000 g cm-3

Temperature Range= θ = 4°C

We know, f4 = f0(1 + γΔt)

=> fo = f4/(1 + γΔt)

=> 0.998 = 1/(1 + γ x 4)

=> γ = 0.0005 = 5 x 10-4 oC-1

As the density decreases,

γ = -5 x 10-4 oC-1

Hence the average Calculate the average coefficient of volume expansion of water in the temperature range 0 to 4°C is -5 x 10-4 oC-1 .

Question 18: Find the ratio of the lengths of an iron rod and an aluminium rod for which the difference in the lengths is independent of temperature. Coefficients of linear expansion of iron and aluminium are 12 × 10–6 °C–1 and 23 × 10–6 °C–1 respectively.

Solution:

Let the original length of iron rod be LFe and

original length of aluminium rods be LAl

Let, L’Fe and L’Al be the changed lengths (ΔL) when temperature is changed by ΔT.

Given:

Coefficient of Linear Expansion of iron rod = α Fe = 12 x 10-6 °C–1

Coefficient of Linear Expansion of Aluminium Rod = αAl = 20 x 10-6 °C–1

Now,

L’Fe = L Fe (1 + α Fe ΔT) and L’Al = L Al (1 + α Al ΔT)

Difference in lengths is independent of temperature it shows that their difference is constant.

=> L’Fe – L’Al= LFe – L Al ………(A)

Therefore, L’Fe – L’Al = L Fe (1 + α Fe ΔT) – L Al (1 + α Al ΔT)

= L Fe – L Al + L Fe α Fe ΔT – L Al α Al ΔT ……(1)

Using equation (A)

(1)=> L Fe α Fe ΔT = L Al α Al ΔT

=> L Fe / L Al = 23/12

Ratio of the lengths of an iron rod and an aluminium rod is 23:12.

Question 19: A pendulum clock gives correct time at 20°C at a place where g = 9.800 m s–2. The pendulum consists of a light steel rod connected to a heavy ball. It is taken to a different place where g = 9.788 m s–2. At what temperature will it give correct time? Coefficient of linear expansion of steel = 12 × 10–6 °C–1.

Solution:

From question, we are given

Temperature at which the pendulum shows correct time = T1 = 20°C

Value of gravitational acceleration at a place where T1 is 20°C = g1 = 9.8 m s–2

Value of g at different places = g2 = 9.788 m s–2

Coefficient of linear expansion of steel = α = 12 x 10-6 °C–1

Let, T2 be the temperature at a place where value of g = 9.788 m s–2.

We know, t = 2π √(l/g)

Where, l is the length of the rod.

Now, t1 = 2π √(l1/g1) and t2 = 2π √(l2/g2)

For obtaining correct time both the time periods should be same. This implies: t1 = t2.

=>2π √(l1/g1) = 2π √(l2/g2)

=>√(l1/g1) = √(l2/g2) …(1)

Also, we know, changes in length due to linear expansion, l2 = l1(1 + αΔT)

Substituting this result in equation (1), we get (g2/g1) = 1 +12 x 10-6 oC-1 x ΔT

=>[9.788/9.8] = 1 +12 x 10-6 oC-1 x ΔT

=> ΔT = -102.04 oC

Because: ΔT = T2 – T1

=> T2 – T1 = -102.04

=> T2 = -102.04 + 20 = -82.04 oC

Question 20: An aluminium plate fixed in a horizontal position has a hole of diameter 2.000 cm. A steel sphere of diameter 2.005 cm rests on this hole. All the lengths refer to a temperature of 10°C. The temperature of the entire system is slowly increased. At what temperature will the ball fall down? Coefficient of linear expansion of aluminium is 23 × 10–6 °C–1 and that of steel is 11 × 10–6 °C–1.

Solution:

From question, we are given

Let dAl be diameter of the hole in the aluminium plate and dst be diameter of the Steel sphere resting on the hole.

Initial temperature = T1 = 10 ° C

αAl and αst be the Coefficient of linear expansion of aluminium and Coefficient of linear expansion of Steel respectively.

Here,

dAl = 2.000 cm, dst = 2.005 cm, αAl =23 × 10–6 °C–1 and αst = 11 × 10–6 °C–1

Now, change in diameter of Steel sphere and aluminium plate:

d’st = dst(1 + αst ΔT) and d’Al = d Al (1 + α Al ΔT)

Where d’st and d’Al changed diameters.

For the steel ball to fall through the hole the changed diameter of the aluminum plate and steel ball should be equal.

=> d’st = d’Al

=> dst(1 + αst ΔT) = d Al (1 + α Al ΔT)

=> 2.005(1 + 11×10-6 x ΔT) = 2 (1 + 23×10-6 x ΔT)

=>-5 x 10-3 = -23.945 x 10-6 x ΔT

=> ΔT = 208.811

Because: ΔT = T2 – T1

=> T2 – T1 = 208.811

=> T2 = 208.811 + 10 = 218.811 oC

Question 21: A glass window is to be fit in an aluminium frame. The temperature on the working day is 40°C and the glass window measures exactly 20 cm × 30 cm. What should be the size of the aluminium frame so that there is no stress on the glass in winter even if the temperature drops to 0°C? Coefficients of linear expansion for glass and aluminium are 9.0 × 10–6 °C–1 and 24×10–6 °C–1 respectively.

Solution:

From question, we are given

Let IGl and bGl be length and breadth of the Glass window.

αGl and αAl be the Coefficient of linear expansion for glass and aluminium respectively.

Here,

IGl = 20 cm, bGl = 30 cm, αGl =9.0 × 10–6 °C–1 and αAl = 24 × 10–6 °C–1

Now, l’ = l(1 – α ΔT)

[l’ is the changed length and the value of α is negative]

Now,Let the changed length of aluminium frame = Changed length of glass window.

=> l’AL = l’Gl

=> lAl(1 – αAl ΔT) = lGl (1 – αGl ΔT) => lAl = 20.01 cm

Similarly, changed breadth of aluminium frame = Changed breadth of Glass window

=> b’AL = b’Gl

=> bAl(1 – αAl ΔT) = bGl (1 – αGl ΔT) => bAl = 30.018 cm

Therefore, dimensions are: 20.01 cm x 30.018 cm.

Question 22: The volume of a glass vessel is 1000 cc at 20°C. What volume of mercury should be poured into it at this temperature so that the volume of the remaining space does not change with temperature? Coefficients of cubical expansion of mercury and glass are 1.8 × 10–4 °C–1 and 9.0 × 10–6 °C–1 respectively.

Solution:

Volume of remaining space = V’g – V’Hg

Here Vg = 1000 CC, T1 = 20oC, γHg = 1.8×10-4 oC and γg = 9.0 × 10–6 °C–1.

VHg = ?

The remaining space should not change with temperature which means: Initial = Final

V’g – V’Hg = Vg – VHg …(1)

Now,

V’g = vg(1 + γg ΔT)

V’Hg = vHg(1 + γHg ΔT)

=> V’g – V’Hg = vg(1 + γg ΔT) – vHg(1 + γHg ΔT)

Using (1),

=> vg/vHg = γHgg

=> (γg vg)/γHg = vHg

=> vHg = [1000x9x10-6]/[1.8×10-4] = 50 CC

Question 23: An aluminium can of cylindrical shape contains 500 cm3 of water. The area of the inner cross section of the can is 125 cm2. All measurements refer to 10°C. Find the rise in the water level if the temperature increases to 80°C. The coefficient of linear expansion of aluminium = 23 × 10–6 °C–1 and the average coefficient of volume expansion of water = 3.2 × 10–4 °C–1 respectively.

Solution:

Volume of water in Aluminium Can : Vw = 500 cm3

Area of the inner cross section of the can : A = 125 cm2.

Initial Temperature while measuring: T1 = 10 ° C.

Increased temperature : T2 = 80 °C

Thus, Change in Temperature : ΔT = T2-T1 = 80-10= 70 ° C.

The coefficient of linear expansion of aluminium : α = 23×10–6 °C–1.

The Average coefficient of volume expansion of water:

Γ = 3.2×10–4 °C–1.

Now,

increased volume of water= v’w = vw(1 + γΔT)

= 500(1 + 3.2 x 10-4 x 70)

= 511.2 cm3

The aluminium can will be expanded too due to temperature increase.

Rise in volume of water = increased volume of water – Volume of water in Aluminium Can

= 511.2 – 500

= 11.2 cm3

Now,

Rise in water level = [Rise in volume]/[Ares of cross section]

= 11.2 cm3/125cm2

= 0.0896 cm

Question 24: A glass vessel measures exactly 10 cm × 10 cm × 10 cm at 0°C. It is filled completely with mercury at this temperature. When the temperature is raised to 10°C, 1.6 cm3 of mercury overflows. Calculate the coefficient of volume expansion of mercury. Coefficient of linear expansion of glass = 6.5 × 10–6 °C–1.

Solution:

Vg = 10cm x 10cm x 10cm = 1000 cm3.

VHg = 1000 cm3.

T1 = 0 °C and T2 = 10 ° C.

vg = volume of a glass vessel and vHg = Volume of mercury in the vessel

T1 = Temperature at which glass is filled completely with mercury:

T2 = Increased temperature

Change in temperature: ΔT = T2-T1 = 10 ° C.

Coefficient of Volume Expansion of Glass: γg: 3× αg = 3× 6.5× 10-6 °C–1

Now,

Volume of mercury overflown is equal to the difference in changed volumes of mercury and glass respectively.

v’Hg = v’g = 1.6 cm3

vHg (1 + γHg ΔT) – vg(1 + γg ΔT) = 1.6 cm3

1000(1 + γHg x 10) – 1000(1 + 3× 6.5× 10-6 x10) = 1.6 cm3

10000 x γHg – 0.196 = 1.6

=> γHg = [1.6+0.195]/10000 = 1.795 x 10-4 oC-1

Question 25: The densities of wood and benzene at 0°C are 880 kg m3 and 900 kg m–3 respectively. The coefficients of volume expansion are 1.2 × 10–3 °C–1 for wood and 1.5 × 10–3 °C–1 for benzene. At what temperature will a piece of wood just sink in benzene?

Solution:

Let V’ is the changed volume due to change in temperature ΔT.

For wood: v’w = vw(1 + γw ΔT)

For Benzene: v’b = v b (1 + γ b ΔT)

Using Density-Volume Relation,

Changed Density of wood: ρ’w = M w /v’ w

Changed Density of benzene: ρ’b = M b /v’ b

Let’s equate the final densities of wood and benzene to obtain the required condition, i.e. ρ’ w = ρ’ b

=> M w /v’w = M b /v’ b Solving above equations for ΔT, we get

ΔT = 20/0.24

or ΔT = T2 – T1 = 20/0.24

=> T2 = 83.33 + 0oC = 83.33

Therefore, At 83.33 °C the piece of wood will just sink in benzene.

Question 26: A steel rod of length 1 m rests on a smooth horizontal base. If it is heated from 0°C to 100°C, what is the longitudinal strain developed?

Solution:

When the steel rod is heated up to 100 °C , due to thermal expansion it’s length will increase with temperature.

A longitudinal strain develops when there exists an opposing force to the expansion of the length. Hence, there will be zero longitudinal strain as there is no opposite in this case.

Question 27: A steel rod is clamped at its two ends and rests on a fixed horizontal base. The rod is unstrained at 20°C. Find the longitudinal strain developed in the rod if the temperature rises to 50°C. Coefficient of linear expansion of steel = 1.2 × 10–5 °C–1.

Solution:

for linear thermal expansion of a body, ΔL = LαΔT

Where ΔL is the Change in Length and L is the initial length at T1.

Given: α = 1.2 × 10–5 °C–1 and Δ T = T1 – T2 = 30 °C

=> ΔL/L = α ΔT

= 1.2 × 10–5 x 30

= 3.6 x 10-4

But, Strain is Change in length divided by original length:

S = ΔL/L

So, S = 3.6 x 10-4

Hence, Longitudinal strain developed in the rod is 3.6 x 10-4

Question 28: A steel wire of cross-sectional area 0.5 mm2 is held between two fixed supports. If the wire is just taut at 20°C, determine the tension when the temperature falls to 0°C. Coefficient of linear expansion of steel is 1.2 × 10–5 °C–1 and its Young’s modulus is 2.0 × 1011 N m–2.

Solution:

A= 0.5 mm2= 0.5×10-6 m2.

T1 = 20 ° C.

T2 = 0 °C

Δ T = 20 ° C.

αs = 1.2 × 10–5 °C–1

and Y= 2.0 × 1011 N m–2.

We know, Young’s Modulus: Y = Stress/Strain = (F/A)/(ΔL/L)

=> Y = [FL]/[AΔL] …(1)

Where, F is the force or tension between the wire and the fixed supports and

A is the cross -section area.

For Linear Expansion: ΔL = Lα ΔT ….(2)

Δ L is the Change in Length due to decrease in temperature and L is the original length at T1.

Using (2) in (1), we have

Y = [FL]/[A α L ΔT]

or F = YA L αs ΔT

F = YAαs ΔT

= 2 x 1011 x 0.5 x 10-6 x 1.2 x 10-5 x 20

= 24

=> F = 24 N, the tension in the wire when the temperature falls to 0°C is 24 N.

Question 29: A steel rod is rigidly clamped at its two ends. The rod is under zero tension at 20°C. If the temperature rises to 100°C, what force will be rod exert on one of the clamp? Area of cross section of the rad = 2.00 mm2. Coefficient of linear expansion of steel = 12.0 × 10–6 °C–1 and Young’s modulus of steel = 2.00 × 1011 N m–2.

Solution:

T1 = 20 ° C, and T2 = 100 ° C.

So, Δ T= T2 – T1 = 100 – 20 = 80 ° C.

A = 2.00 mm2 = 2.00 × 10-6 m2.

α = 12.0 × 10–6 °C–1

Y = 2.00 × 1011 N m–2.

We know, Young’s Modulus: Y = Stress/Strain = (F/A)/(ΔL/L)

=> Y = [FL]/[AΔL] …(1)

Where, F is the force or tension between the wire and the fixed supports and

A is the cross -section area.

For Linear Expansion: ΔL = Lα ΔT ….(2)

Using (2) in (1), we have

Y = F/[A α ΔT]

or F = YA L α ΔT

= 2 x 1011 x 2 x 10-6 x 12 x 10-6 x 80

= 384

=> F = 384 N

when the temperature is increased to 100° C , the rod will exert a force of 384N on one of the champ.

Question 30: Two steel rods and an aluminium rod of equal length l0, and equal cross-section are joined rigidly at their ends as shown in the figure bellow. All the rods are in a state of zero tension at 0°C. Find the length of the system when the temperature is raised to θ. Coefficient of linear expansion of aluminium and steel are αa and αs respectively. Young’s modulus of aluminium is Ya and steel is Ys.

 Steel Aluminium Steel

Solution:

At T1 = 0° C

Length of two steel rods and one aluminium rod joined rigidly at T1= l

Coefficient of linear expansion of aluminium : αa

Coefficient of linear expansion of steel are : αs

Young’s modulus of aluminium : Ya

Young’s modulus of steel : Ys

Increased Temperature : T2 = θ

Change in temperature : Δ T= θ – 0 = θ

We know, Young’s Modulus: Y = Stress/Strain = (F/A)/(Δl/l)

=> Y = [Fl]/[AΔl] …(1)

For Linear Expansion: Δl = lα ΔT ….(2)

Using (2) in (1), we have

Y = F/[A α ΔT]

or F/A = stress = Y α ΔT …(1)

The Total Young’s Modulus of whole system:

Total Y = [Total Stress]/[Total Strain]

Now, Total stress = stress due to two steel rods + stress due to one aluminium rod

= (2 Ys x αs x θ) + Ya x αa x θ) ….(2)

[using (1)]

Total Young’s Modulus = (2 x Ys) + Ya …..(3)

Total Strain = [Total Stress]/Total Y] Which is the length of the system when temperature is increase to θ.

Question 31: A steel ball initially at a pressure of 1.0 × 105 Pa is heated from 20°C to 120°C keeping its volume constant. Find the pressure inside the ball. Coefficient of linear expansion of steel = 12 × 10–6 °C–1 and bulk modulus of steel = 1.6 × 1011 N m–2.

Solution:

Pressure = P = BγΔθ

Where P is the pressure inside the ball.

B = Bulk modulus of steel = 1.6 x 1011 Nm-2

Δθ = Change in temp = 100oC

(Here, γ = 3α; given α = 12 x 10-6 oC)

=> P = 1.6 x 1011 x 3 x 12 x 10-6 x 100

= 5.8 x 108 Pa

Question 32: Show that moment of inertia of a solid body of any shape changes with temperature as I = I0(1 + 2αθ), where I0 is the moment of inertia at 0°C and α is the coefficient of linear expansion of the solid.

Solution:

Moment of inertia at 0° C = I0

Change in temperature : Δ T = θ – 0 = θ

We know, I0 = MRo2 …(1)

Where M is the mass of the body, I0 is the initial moment of inertia at 0° and R0 is the radius of gyration at 0° C.

When temperature of the solid body increases, the radius of gyration also changes due to thermal expansion.

Let I be the changed moment of inertia when temperature is θ.

I = MR’2 ….(2)

Where R’ = Ro(1 + α ΔT); R’ is the changed radius of gyration due to expansion.

(2)=> I = M[Ro(1 + α ΔT)] 2

= MRo2 (1 + 2αθ + α2 θ2)

[Given: ΔT = θ]

α2 θ2 is neglected, we get

I = I0 (1 + 2αθ)

[Using (1)]

Hence Proved.

Question 33: A torsional pendulum consists of a solid disc connected to a thin wire (α= 2.4 × 10–5 °C–1) at its centre. Find the percentage change in the time period between peak winter (5°C) and peak summer (45°C).

Solution:

Temperature change in winter: Δ T1= 5° C

Temperature change in summer : Δ T2 = 45° C

and α= 2.4 × 10–5 °C–1

Due to change in temperature the moment of inertia will also change.

So,

I’ = Io(1 + 2α ΔT)

Where, I’ is the changed moment of inertia and I0 is the initial moment of inertia at 0 ° C

We know, time period for torsional pendulum: T = 2π√[I/K] …(1)

Here, k is torque constant of the wire.

During winter:

I’ = Io(1 + 2α ΔT1)

(1)=> t1 = 2π√[I’/K]

= 2π√[(I0 (1 + 2α ΔT1)/K] During Summer:

I’ = I0 (1 + 2α ΔT2)

(1)=> t1 = 2π√[I’/K] Now, Let us find percentage change in the time period between peak winter (5°C) and peak summer (45°C) :

% change = [change in quantity]/[original quantity] = [t2 – t1]/t1

or % change = (t2/t1) – 1) x 100

Using values of t1 and t2, we get => % change = 9.59 x 10-4 x 100 = 0.0959 %

Question 34: A circular disc made of iron is rotated about its axis at a constant velocity ω. Calculate the percentage change in the linear speed of a particle of the rim as the disc is slowly heated from 20°C to 50°C keeping the angular velocity constant. Coefficient of linear expansion of iron = 1.2 × 10-5 °C–1.

Solution:

We know, Angular Velocity: ω = v/r

Where v and r be the velocity and radius of the particle.

When temperature increased, the disc undergoes thermal expansion.

Let v be the velocity of particle at T1 and v’ be the velocity of particle at T2.

and r be the radius of particle at T1 and r’ be the changed radius of particle at T2.

From question, we have

T1 = 20 ° C

T2 = 50 ° C

Δ T = 50-20= 30° C

α = 1.2 × 10–5 °C–1

Now,

Angular Velocity at T1: ω1 = v/r

Angular Velocity at T2 : ω2 = v’/r’

Also we know, Thermal linear expansion of radius is r’ = r(1 + α ΔT)

ω is constant even after heating the disc, so

ω = v/r = v’/r’

=> v/r = v’/[r(1 + α ΔT)]

=> v’ = v(1 + α ΔT)

= v(1 + 1.2×10-5x30) = 1.00036 v

=> v’ = 1.00036 v

Let us find the percentage change in the linear speed of a particle of the rim when the disc is slowly heated from 20°C to 50°C :

% change = [change in quantity]/[original quantity] = [(v’ – v)/v] x 100

= [(1.00036v – v)/v] x 100

= (1.00036 -1) x 100

= 0.036