Thermal Expansion

Fundamental Physics I by Dr. Abhijit Kar Gupta (email: [email protected]) 1

Thermal Expansion What is thermal expansion? When the temperature of a substance (solid, liquid or gas) is increased, the molecules or atoms in it vibrate faster and they tend to move away from each other, on average. This results in an expansion (increase of length, breadth and depth) of the substance as a whole. This is called thermal expansion. [Note: When the temperature is decreased (the substance is cooled), there is a contraction which can be considered as ‘negative’ expansion. A similar expansion or contraction is also possible with the application of a force. This has been discussed in the chapter dealing with elastic properties of matter.] Some facts:

• Generally the size of all material bodies increases with the increase of temperature. The exceptions are the water and some aqueous solutions in the range of 0 C0 to 4 C0 , the iodide of silver (resolidified) below 142 C0 and some nickel-steel alloys.

• Different substances expand at different rates. This depends on the physical

property of the substance. Different solid substances have different expansion rates like iron, glass, wood all have distinctly different rates of expansion. Expansions in some substances are not easily detectable with naked eye.

• The property of thermal expansion is utilized in various ways in our daily life and in scientific works.

Examples: Blacksmiths use to put red-hot iron rims on wooden cart wheels so that they cool and shrink tight. The property of thermal expansion of liquid mercury is used in clinical thermometers. • Liquids expand appreciably as compared to solids, in general.

Example: The expansion of mercury is greater than the expansion of glass. If the expansion of glass of a thermometer would be at the same rate as that of mercury, the mercury would not rise with the increase of temperature.


Thermal Expansion of Solids

The expansion of solids is of three types. Linear expansion of solids can be experimentally measured from the expansion of length of bars or rods of the solid. Surface or volume expansions of solids are generally difficult to measure directly and thus they are calculated from the linear expansion. In this chapter we consider isotropic solids only. Isotropic materials are those whose properties are the same in all directions. Thus the expansion is also the same in all directions for such materials. Examples: all amorphous solids, e.g., glass; the regular crystals, e.g., sodium chloride (table salt); metals. Demonstrations of Expansion of Solids:

• Bar and Gauge Experiment

A is a metal bar with a wooden rod attached to it. B is a metallic gauge as shown in the figure [Fig. #] . At normal room temperature, the bar A just fits into the gauge B. When the bar A is heated, it can not be fitted into the gauge B anymore. This is an indication that there has been a thermal expansion in the bar A. Now when the bar is cooled down, it fits into the gauge again. This means that the bar regains its original length by contraction.

• Ball and Ring Experiment

A metal ring B has been attached with a stand as shown in the adjacent figure [Fig. #]. A metal ball A is hanged from a hook above this fixed with the stand. At normal room temperature the metal ball just passes through the ring. Now if the ball is heated separately and then we attempt to insert this through the ring again, the ball gets stuck. From this it can be understood that the volume of the metal ball is increased due to heating and thus it is unable to pass through the ring. When it is allowed to cool down to previous temperature (room temperature), the ball gets back its original volume and then it can again pass through the ring. Thus we understand from this experiment that the volume of a solid material increases as the temperature is increased.

Fig. to be included

Fig. to be included


Experiments showing that the Expansions of Different Solids are Different:

• Ferguson’s Experiment

This experiment shows that the expansions for different materials due to the same increment of temperature are different. In this experiment, one end of a metal beam R is rigidly clamped at L and the other end is kept on a roller C as shown in the figure [Fig. #]. The roller can move on a glass plate G. A pointer P is attached to the roller which can rotate over a circular scale S. The beam R is heated from below by some Bunsen burners. As a result, the beam expands and the roller R rolls over the glass plate. Thus there is a deflection of the pointer. The amount of deflection depends on the amount of expansion of the beam. The deflection is seen to increase as the beam is heated more. Thus it is understood that the expansion of length of the beam is increased along with the increase of temperature. The same experiment is repeated with beams of different materials but of same length. Deflections in the scale are found to be different for different beams where the increase of temperature is kept fixed. Thus the expansions for different materials are found to be different for the same increment of temperature.

• Experiment with Bimetallic Strip When two thin bars of different metals, such as brass and iron, are welded or riveted together, the compound thin bar such made is called bimetallic strip. At the beginning, the bimetallic strip remains straight at normal room temperature. When it is heated, one side of the double strip becomes longer than the other as the expansion rate of one metal (brass) is more than that of other (iron). As a result, it is seen that the strip is bent into a curve with the brass on the outer side of the curve. Next, the strip is cooled below room temperature keeping it in ice. When the strip is cooled, it tends to bend in the opposite direction, because the metal that expands more (brass) also shrinks more.

Application of Bimetallic Strip: The movement of bimetallic strip is used to turn a pointer, regulate a valve or close a switch. A very important application is the thermostat. The back-and-forth

Fig. to be included

Fig. to be included


bending of this is applied to open and close an electrical circuit. Thermostats are used in refrigerators, air conditioning machines, electric toasters and various other devices. The idea is to control temperature as it is desired. Coefficient of Linear Expansion: Definition The thermal expansion of a solid in a particular direction that is the change in one dimension (length, breadth or thickness) is called linear expansion.

• For most of the solids, the change in size due to temperature change is very low*. • For small temperature changes, linear expansion is approximately proportional to

the change in temperature. Experimentally, it is seen that the amount of expansion of length of a rod is proportional to the (i) initial length of the rod and (ii) the increase of temperature. Let us suppose that the temperature is increased from 0T toT . Thus the change in temperature is )( 0TTT −=∆ . The length is changed from 0L to L . So the change in length is )( 0LLL −=∆ . Thus we have

L∆ ∝ 0L and L∆ ∝ T∆ . ∴ L∆ ∝ 0L T∆ Therefore, we can write L∆ = α 0L T∆ , (1) where the proportionality constant α is called the coefficient of linear expansion. The above equation (1) can also be written in the following form )( 0LL − = α 0L T∆ . ∴ L = 0L (1 + α T∆ ) Or L = 0L [1 + α )( 0TT − ]. (2) When the initial temperature, 0T = 0 and the final temperature is T , we can write TL = 0L (1 + α T ), (3) where TL is the length at temperature T . From equation (1) we get a definition of the coefficient of linear expansion:

* A typical metal expands only about 7% when its temperature rises from near 0 K to its melting point.


α = TL

L∆∆

0

. (4)

If we now put 0L = 1 and T∆ = 1, we have α = L∆ . Thus we can define α in the following way. Definition: The linear expansion of a solid material of unit length due to one unit of increase of temperature is called coefficient of linear expansion.

• The coefficient α is specific to a material. The magnitude of this depends on the atomic arrangements inside a matter.

• The coefficient of expansion may vary slightly for different temperature ranges. So it is not strictly a constant. However, this variation is negligible for most applications. Thus we considerα to be practically a constant and independent of temperature.

Dimension ofα : It can be analyzed from the relation (3) Dimension of α = =

• Thus we see, α depends only on temperature. Dimension of α is / 0C [also written as 10 )( −C ] or / 0F [also written as 10 )( −F ]; in Kelvin scale it is / K (also written as 1−K ).

As we have 1 0C = 59 0F , the coefficient of linear expansion for steel in two

scales can be related as

61012 −× / 0C = 95 × 61012 −× / 0F = 61067.6 −× / 0F .

• The value of α does not depend on the dimension of length. This is because it is essentially a ratio of two lengths. When we say the coefficient of linear expansion of steel is 61012 −× / 0C , we understand that if we take a steel rod of length 1 cm. and the temperature of it is increased by 1 0C then the increase of length will be 0.000012 cm. Likewise, when

Dimension of length Dimension of length × Dimension of temperature

1 Dimension of temperature


the length is taken to be 1 m or 1 ft, then the increments are 0.000012 m or 0.000012 ft.

Below is a list of mean coefficients of expansion of several solid substances between 0 C0 and 100 C0 .

Table # 1: Coefficient of linear expansion of different solids

Material α10 )( −C

Cadmium 32 610−× Lead 28 610−×

Magnesium 26 610−× Aluminium 24 610−×

Brass 20 610−× Copper 17 610−× Nickel 13 610−× Steel 12 610−× Glass 4-9 610−× Invar 0.9 610−×

Quartz (fused) 0.4 610−×

Numerical Examples with Solutions: Example #1 A steel scale 1 m long is calibrated at a temperature of 30 C0 . What is the length of this scale on a hot summer day when the temperature is 40 C0 ? Solution: From equation (1), L∆ = α 0L T∆ = 12 610−× ( 10 )( −C × (1 m)× )3040( 00 CC − = 0.12 310−× m = 0.00012 m. Thus the length at 40 C0 is 1.00012 m. Example #2 An Aluminium rod is expanded 0.5 cm when heated from 20 C0 to 80 C0 . What is its initial length? Assume α = 24 610−× / 0C .


Solution: L∆ = α 0L T∆

∴ 0L = ( )0TTL−∆

α.

Here, α = 24 610−× / 0C , 80=T 0C and 200 =T 0C

∴ 0L = )2080(1024

5.06 −×× − =

6024105.0 6

×× = 347.2 cm. Initial length

Example #3 The length of a copper rod is measured to be 200.166 cm and 200.664 cm at the temperatures 50 C0 and 200 C0 . What is the linear coefficient of thermal expansion for copper? [ H.S. (T)]

Solution: α = TL

L∆∆

0

= )(

)(

00

0

TTLLL−

−

Here, 0L = 200.166 cm, L = 200.664 cm, =0T 50 C0 and =T 200 C0 .

∴ α = )50200(166.200

166.200664.200−×

− = 150166.200

498.0×

= 16.6 0610 C−× .

Example #4 The length of a brass bar is 150 cm at 40 C0 . What will be its length at 100 C0 ? The coefficient of linear expansion of brass is 20 610−× / 0C . Solution: L = 0L (1 + α T∆ ) = 0L [1 + α )( 0TT − ] Here, =0L 150 cm, =0T 40 C0 , =T 100 C0 , α = 20 610−× / 0C . ∴ L = 150 [ ])40100(10201 6 −×+ − = 150 [ ]61012001 −×+ = 150 00120.1× = 150.18 cm. Increase in temperature generally causes increment in surface area and volume of a material. Therefore, there should be a coefficient associated with the surface expansion and also another with volume expansion. Coefficient of Surface Expansion If the area is increased from 0A to A due to the increase of temperature from an initial value 0T to T then we can write, the change in the area is A∆ = 0AA − and to the change in the temperature is T∆ = 0TT − .


As before, the change in area A∆ is proportional to the initial area 0A and the change in temperature T∆ : A∆ ∝ 0A and A∆ ∝ T∆ . ∴ A∆ ∝ 0A T∆ Or A∆ = β 0A T∆ , (5) where β is the coefficient of surface expansion (or area expansion). From above, can write 0AA − = β 0A T∆ . Or ( )TAA ∆+= β10 Or A = [ ])(1 00 TTA −+ β . (6) If the initial temperature is 00 =T and the final temperature is T then we can write, ( )TAA β+= 10 (7) From equation (5) we can write the expression for coefficient of surface expansion:

β = TA

A∆∆

0

. (8)

When we put 0A = 1 and T∆ = 1, we have β = A∆ . Thus the definition of β can be given as the following. Definition: The increase in surface area of a solid material of unit surface area due to one unit of increase of temperature is called coefficient of surface expansion. Nature of β :

• Like the coefficient of liner expansionα , the coefficient of surface expansion β is specific to a certain material (material property).

• This coefficient of expansion also may vary slightly for different temperature ranges. Therefore, it is not strictly a constant. However, this is negligible for most practical applications and an average value within a certain range is always quoted.

Dimension:

• Dimension of β does not depend on the dimension of area as it is essentially the ratio of two areas and thus dimension of area cancels out.


• Dimension of β depends on temperature only and it is the inverse of temperature. The meaning of the coefficient of surface expansion for steel to be 061024 C−× (= 0.000024 /C )0 is the following. If we take a steel sheet of area 1 ft 2 or 1 cm 2 or 1 m 2 , then its area is increased by 0.000024 ft 2 or 0.000024 cm 2 or 0.000024 m 2 when the temperature is increased by 1 0C . Conversion from one scale to another is again the same thing as we have done in the case of linear expansion. We have to remember the relationship between the temperature differences in two scales

1 0C = 59 0F .

Thus the value of β in Fahrenheit scale is

61024 −× / 0C = 95 × 61024 −× / 0F = 61034.13 −× / 0F .

Coefficient of Volume Expansion: If the volume of a material is increased from 0V to V due to the increase of temperature

from an initial value 0T to T then we can write, the change in the volume is V∆ =

0VV − and the change in the temperature is T∆ = 0TT − . As before, the change in volume V∆ is proportional to the initial volume 0V and to the change in temperature T∆ : V∆ ∝ 0V and V∆ ∝ T∆ . ∴ V∆ ∝ 0V T∆ Or V∆ = γ 0V T∆ , (9) whereγ is the coefficient of volume expansion. From equation (9), we can write 0VV − = γ 0V T∆ . Or ( )TVV ∆+= γ10 Or V = [ ])(1 00 TTV −+ γ . (10) If the initial temperature is 00 =T and the final temperature is T then the formula (10) can be written as


( )TVV γ+= 10 (11) From equation (9) we can write the expression for coefficient of volume expansion:

γ = TV

V∆∆

0

. (12)

When in expression (12) we put 0V = 1 and T∆ = 1, we have γ = V∆ . Thus the definition of γ can be given as the following. Definition: The increase in volume of a solid material of unit volume due to one unit of increase of temperature is called the coefficient of volume expansion. Nature of γ :

• Like the coefficient of liner expansion and that of surface expansion, the coefficient of volume expansion γ is specific to a certain material (material property).

• This coefficient of expansion also may vary slightly for different temperature ranges. Therefore, it is not strictly a constant. However, this is negligible for most practical applications and an average value within a certain range is always quoted.

Dimension:

• Dimension of γ does not depend on the dimension of volume as it is essentially the ratio of two volumes and thus dimension of volume cancels out.

• Dimension of γ depends on temperature only and it is the inverse of temperature. The meaning of the coefficient of volume expansion for steel to be 061036 C−× (= 0.000036 /C )0 is the following. If we take a steel sheet of volume 1 ft 3 or 1 cm 3 or 1 m 3 , then its volume is increased by 0.000036 ft 3 or 0.000024 cm 3 or 0.000024 m 3 when the temperature is increased by 1 0C . Conversion from one scale to another is again the same thing as we have done in the case of linear expansion and surface expansion. We have to remember the relation between the temperature differences in two scales

1 0C = 59 0F .

Thus the value of γ in Fahrenheit scale is

61036 −× / 0C = 95 × 61036 −× / 0F = 61020 −× / 0F .


The Relation among the three Coefficients of Expansion: The relation between α and β

Suppose we take a square sheet of length 0L . The area is 200 LA = .

The temperature of the square sheet is increased from an initial temperature of 0T toT . The increase in temperature is 0TTT −=∆ . We call here Tt ∆= for the sake of calculation. The increase in temperature causes the increase in length which now becomes L . Thus we can write, L = 0L (1 + α T∆ ) = 0L (1 + α t ), where α is the coefficient of linear expansion. Therefore, the present area of the sheet is 2LA = = 0L 2 (1 + α t ) 2 = 0A (1 + 2α t + α 2 t 2 ). Now since α is usually much smaller than 1 and if the temperature difference =t T∆ is not very high then the term α t is still very small compared to 1. The square of this quantity α 2 t 2 will be even smaller than 1. Thus the term α 2 t 2 can be neglected as it is very small compared to the termα t . Hence we have an approximate relationship A = 0A (1 + 2α t ). (13) On the other hand, we can write from the definition of the coefficient of surface expansionβ , A = 0A (1 + β t ). (14) Comparing (13) and (14), we have β = 2α (15) The relation between β and γ


Suppose we take a cube of length 0L . The volume is 300 LV = .

The length is increased from 0L to L when the temperature is increased from 0T toT . We write the increase in temperature, 0TTTt −=∆= . If the linear expansion coefficient be α then L = 0L (1 + α T∆ ) = 0L (1 + α t ). The present volume of the cube is V = 3

0L (1 + α t ) 3

= 1(30V + α3 t + α3 2 t 2 + 3α 3t )

Since α t is small (less than 1), α 2 t 2 and 3α 3t are much smaller compared to α t . Therefore, we can write the approximate relation V = 1(3

0V + α3 t ). (16) On the other hand, we can write from the definition of the coefficient of volume expansionγ , V = 0V (1 + γ t ). (17) Comparing equations (16) and (17), αγ 3= . (18) Now from (15) and (18) we get

The above is the relation among the three coefficients of expansion. This can also be written in the following form: α : β :γ = 1 : 2 : 3

32γβα ==


Derivation of the relationship among α , β and γ by Calculus: The differential form of the expression of linear coefficient of expansion is

dTdL

L1

=α

The interpretation of the above is that if the temperature of a solid having coefficient of linear expansionα , is increased from T to dTT + , the length is increased from L to dLL + . Similarly, we can write the expressions of β and γ as

dTdA

A1

=β

dTdV

V1

=γ .

Suppose, we take a cube of length L whose each surface has an area 2LA = and the volume of the cube is 3LV = . Differentiating the area A = 2L with respect to temperatureT , we get

=dTdA

dTdLL2 .

Now,

dTdA

A1

=β = dTdLL

L2.1

2 = 2. dTdL

L1 = 2α . (19)

Again differentiating the volume 3LV = with respect to T , we get

Note: If a quantity, say x , is bigger than 1 then its successive powers become bigger and bigger than 1. If x is taken smaller than 1 then its successive powers become smaller and smaller than 1. Example: Suppose, =x 2, we then have 42 =x , 83 =x , 164 =x etc. If x is smaller than 1, say 2.0=x , then we have 04.02 =x , 008.03 =x , x 0016.04= . Therefore, when x is smaller than 1, the higher powers become negligible compared to x and thus we often neglect them for approximation.


=dTdV

dTdLL23 .

From the definition of γ ,

dTdV

V1

=γ = dTdLL

L2

3 3.1 = 3.dTdL

L1 = 3α . (20)

∴ From (19) and (20) we have

32γβα == .

---------------------Numerical Examples with Solutions-------------------------- Example #1 A brass disc has diameter 8 cm at 30 C0 . How much will be the increase in area of it when the temperature is raised to 80 C0 ? Given 06 /1018 C−×=α . Solution: The increase of area is β=∆A 0A T∆ = 2α 0A ( 0TT − ) as we know β = 2α . Here, the initial area ππ 1642

0 ==A sq. cm, 06 /1018 C−×=α , =0T 30 C0 and T = 80 C0 . ∴ A∆ = 2 )3080(1416.3161018 6 −××××× − = 36 610501416.316 −×××× = 0.0905 sq. cm. Example #2 A rectangular copper block measures 158 ′′×′′×′′ . How much of its volume will increase when its temperature is increased from 0 C0 to 800 C0 ? The coefficient of linear expansion for copper is = 04 /1016.0 C−× . [H.S.] Solution:

The initial volume of the block is 1580 ××=V = 40 cubic inch. The increase in volume is γ=∆V 0V T∆ = 3α 0V T∆ since the coefficient of volume expansionγ = 3α , where α is the coefficient of linear expansion. Here, we have =α 04 /1016.0 C−× , 400 =V cub. in, T∆ = 0TT − = )0800( − = 800 0C . ∴ V∆ = 3 800401016.0 4 ×××× − = 1.536 cub. in. Example #3 The volume of a lead bullet is 25 c.c. at 0 C0 . At 98 C0 , its volume increases by 0.021 c.c. What is the coefficient of linear expansion for lead? [H.S.(T)] Solution:

The volume expansion coefficient, TV

V∆∆

=0

γ


Here, 250 =V c.c., V∆ =0.021 c.c., 0TTT −=∆ = 98 – 0 = 98 0C .

∴ 9825

021.0×

=γ = 06 /1057.8 C−× .

The linear expansion coefficient, ==3γα

31057.8 6−× = 2.86 610−× / 0C .

Example #4 An aluminium sphere whose diameter is 20 cm, is heated from 0 C0 to 100 C0 . What will be its change in volume? The coefficient of linear expansion is = 23 610−× / 0C . Solution:

The initial volume of the sphere, 0V = 31034

×π c.c.

The change in volume, γ=∆V 0V T∆ = α3 0V T∆ . The coefficient of volume expansion, αγ 3= = 3×coefficient of linear expansion. The increase in temperature is 0TTT −=∆ , where 0T = 0 C0 is the initial temperature and T = 100 C0 is the final temperature.

∴ )0100(101416.33410233 36 −××××××=∆ −V = 28.9 c.c.

Example #5 A piece of metal weighs 46 gm in air. The same weighs 30 gm when that is immersed in a liquid of specific gravity 1.24 at temperature 27 C0 . When the temperature of the liquid is increased to 42 C0 , the metal piece weighs 30.5 gm when immersed. The specific gravity of the liquid is 1.20 at 42 C0 . What is the coefficient of linear expansion of the metal? [H.S. ’98] Solution:

At 27 C0 , the apparent decrease in weight of the metal is = 16)3046( =− gm. This should be the weight of the displaced liquid of same volume.

∴ The volume of the displaced liquid = 24.1

16 c.c.

Volume of metal piece = Volume of displaced liquid

Thus we can say, the initial volume of the metal piece is =0V24.1

16 c.c.

Similarly, the volume of the metal piece when immersed in the liquid at 42 C0 is

V = 20.1

5.1520.1

5.3046=

− c.c.

Therefore, the coefficient of volume expansion is

TV

V∆∆

=0

γ = )( 00

0

TTVVV−−

=

−

−11

00 VV

TT.

Here, the initial temperature is =0T 27 C0 and the final temperature is =T 42 C0 .


∴

−××

−= 1

1624.1

20.15.15

27421γ =

−× 1

2.1922.19

151

= 2.1915

02.0×

= 61044.69 −× 0/ C .

The coefficient of linear expansion, 61015.233

−×==γα 0/ C .

Change in Density of a Solid due to Change in Temperature:

We know, the density of a material VMD = =

VolumeMass .

When the temperature of a material is increased, its volume is increased but the mass remains unchanged. Thus the ratio of mass and volume changes which means the density changes. Density is increased when the volume is decreased and it is decreased when the volume is increased since the volume is in the denominator. Suppose, the density of a material is 0D and the volume is 0V at an initial temperature 0T . The density becomes D and the volume to V when the temperature is increased toT . The mass M remains unchanged.

∴ 0

0 VMD = and

VMD =

∴ The ratio of two densities, VV

DD 0

0

= .

If the coefficient of volume expansion of the material is γ , we can write γ+= 1(0VV T∆ ), where 0TTT −=∆ , the temperature difference.

∴ )1(0

0

0 TVV

DD

∆+=

γ = 1)1( −∆+ Tγ .

Thus we have )1(0 TDD ∆+= γ . We can also write, 1

0 )1( −∆+= TDD γ = )1(0 TD ∆− γ . [Expanding binomially and neglecting other terms after expanding as they are much smaller than T∆γ .] --------------------Numerical Examples with Solutions-------------------------


Example# 1 Density of glass is 2.6 gm/cc and 2.596 gm/cc at 10 C0 and 60 C0 respectively. Find the average coefficient of linear expansion of glass within this range of temperature. [H.S. ‘88] Solution:

We know =0D )1( TD ∆+ γ = [ ])(1 0TTD −+ γ , where γ is the coefficient of volume expansion. Here, 0D = 2.6 gm/cc, 596.2=D gm/cc, CT 0

0 10= and CT 060= . ∴ 2.6 = 2.596 [ ])1060(1 −+ γ

Or, 501 ×+ γ = 596.26.2 Or, 50γ =

596.2596.26.2 −

∴ γ = 50596.2

004.0×

= 30.8 610−× / 0C .

The coefficient of linear expansion, 3γα = = 10.27 610−× / 0C .

Thermal Stress: The length of a solid rod expands due to heating and it contracts due to cooling. But if the rod is rigidly fixed at two ends then this expansion or contraction can not happen. Thus there is an internal force developed inside the material of the rod. This internal force per unit area of the rod is called thermal stress. Demonstration of thermal stress Take an iron rod B which is placed inside a heavy metal frame Y as shown in the figure. One side of the rod is threaded and the other side is having two grooves 1P and 2P . The threaded side is fitted with a screw N. An iron pin is inserted into the groove 2P and this is made to sit tight with the frame by rotating the screw. Now if the rod is heated, there is a force developed due to thermal expansion for which the pin breaks down inside the groove. Now the pin is inserted into the groove 1P and this is made to sit tight with the outside of the frame. If the rod is cooled down, the pin 1P now breaks as the rod wants to contract. Thus we have a clear demonstration of thermal expansion or contraction.

Fig. to be included


Determination of thermal stress

Let a rod of length 0L be subjected to heating and the temperature of the rod is increased by T∆ . If the coefficient of linear expansion of the material of the rod be α then the expansion of the rod is L∆ =α 0L T∆ . Suppose, the same expansion in the rod be created due to a force F . If he cross-sectional area of the rod is A then the longitudinal stress is = AF / .

Longitudinal strain = 0LL∆ =

0

0

LTL ∆α

= α T∆

∴We can write the Young Modulus, TAFY

∆=α

/ .

∴ The thermal stress developed in the rod is = Y α T∆ . --------------------Numerical Examples with Solutions------------------------- Example# 1 A steel rod is clamped rigidly at two ends. The cross-sectional area of the rod is 4 sq. cm at 30 C0 . Find the force exerted by the rod at two ends when the temperature is increased to 60 C0 . [The Young Modulus for steel, 12101.2 ×=Y dyne/ cm 2 and the coefficient of linear expansion, 06 /1012 C−×=α .] Solution:

Here, the cross-sectional area 4=A sq. cm., 12101.2 ×=Y dyne/ cm 2 , 06 /1012 C−×=α , Temperature change, =∆T 0TT − = 3060 − = 30 0C . ∴ The force to be exerted due to thermal expansion is, AF = Y α T∆ = 301012101.24 612 ××××× − = 910024.3 × dyne. Example# 2 A wire is rigidly fixed at two ends. Determine the change in tension in the wire when the temperature is decreased by 10 C0 . The cross-section is = 0.01 sq. cm;

06 /1016 C−×=α , 111020×=Y dyne/ cm 2 [ J.E.E.] Solution:

Here, 01.0=A sq. cm. 111020×=Y dyne/ cm 2 , 06 /1016 C−×=α , =∆T 10 C0 . The change in tension in the wire = the force developed due to thermal contraction F . The force, AF = Y α T∆ = 101016102001.0 611 ××××× − = 32 510× dyne.


Practical applications of Thermal Expansion of Solids: The property of thermal expansion (and contraction) of solid is often utilized in a number of ways. Sometimes, trouble arises due to the expansion or contraction with temperature and our knowledge of this is used to avoid the trouble. So there are both advantages and disadvantages. Here are some examples. Advantages

• Metal lids: Metal lids on glass jars or metal caps of glass bottles are often loosened by heating under hot water. This is possible since the expansion of glass is negligible compared to a metal lid or cap.

• Iron rim: Blacksmiths put red-hot iron rims on wooden wheels of bullock carts. The radius of the rim is taken a little smaller than the outer radius of the wheel so that at normal temperature the rim can not be fitted onto the wheel. The iron rim expands when heated and thus can be easily put on the wheel. When it cools, the rim sits tight on the wheel.

• Metal strips: Two metallic strips are placed together and a hole is drilled for riveting. A bolt is inserted through the hole when heated and the two ends of it are beaten flat as shown in the figure [Fig.**] to hold the two metal strips tightly. When the rivet is cooled down, it shrinks and holds the two metal strips more strongly. This is a way by which big metal strips are joined to form structures of big steel bridges.

• Thermostat: We have already seen in the previous section (**), how a

bimetallic strip bends when heated. This property is used in thermostat. Thermostat is an automatic temperature control device which is used in Electric Micro oven, refrigerator, incubator, electric iron, electric heater, air cooling machine and in many other home appliances and instruments. It has been shown in the adjacent circuit how it works as an automatic switch. Here a bimetallic strip is shown to be made of invar and brass. The brass part is connected to the connecting point A of the thermostat. The bimetallic strip remains straight at ordinary temperature. Hence the thermostat circuit remains on. So the current flows through the connected heater and it keeps heating. The air around the heater gets heated and after a while the bimetallic strip also gets heated and bends. The coefficient of linear expansion for brass is more than that of invar. Thus the bimetallic strip bends in the opposite direction to the contact that is made. Because of this the circuit is disconnected and the current through the heater stops. As the heating stops, the bimetallic strip eventually comes back to normal temperature and becomes straight. Again the connection is established and the heater starts heating. This way the heating can be limited and thus the temperature can be controlled by controlling the current through the circuit.

Fig. to be included


• Fire alarm: Bimetallic strip is used in fire alarm circuit. Now the invar side is made to have contact with the thermostat. When there is fire, the bimetallic strip gets heated and it bends towards the invar side as the thermal expansion of invar will be much less than that of brass. Thus the contact is established and the current flows to start the alarm.

• Temperature measurement: A bimetallic strip can be used as a thermometer.

At an ordinary temperature it remains flat. Suppose we find out the temperature 0T at which the bimetallic strip is exactly flat or straight. As the temperature is increased more than this, it starts bending. As the temperature is raised higher, the amount of bending becomes greater which can be correctly measured by determining the curvature. The radius of curvature becomes smaller as the bending or curvature is more. Experimentally it is seen, the radius of curvature R is inversely proportional to the temperature

difference, )( 0TT − raised above 0T : 0

1TT

R−

∝ . Thus =− )( 0TTR Constant.

This constant can be determined by measuring R from a known temperature. However, this has not much practical importance.

Disadvantages

• Fishplate: Fishplates are used to join two iron rails of certain length to make

rail tracks. A gap is deliberately kept between two rail pieces and they are joined by fishplates fitted with nut and bolts. The gap is kept to allow thermal expansion due to excessive heating in the day time and that due to friction by running trains. Also it may be noticed that the holes in a fishplate are made elongated along the length of it so as to allow expansion or contraction along the length.

Such a gap or the arrangement of fishplate is not there for tram lines. Tram lines are implanted partly inside the ground and are surrounded by granite and concrete. The heat developed in the rails due to heating can be dissipated easily to the ground through the concrete and granite. Thus there is not much scope of bending of rails due to excessive heating.

• Bridge: Both the ends of a steel or iron bridge are generally not rigidly fixed to concrete foundations. One end is left free and placed on a rocker kept on the concrete slab (Fig, **). This allows the steel bridge to have free expansion or contraction due to temperature variation. There could be damage in the concrete foundation due to thermal expansion or contraction if both the sides would be rigidly fixed inside the concrete.

Fig. to be included

Fig. to be included


• Breaking of glass: A thick glass tumbler often breaks when hot water is

poured in it. Glass is a bad conductor of heat. Thus the inside of the glass tumbler expands more due to heating compared to that of outside where heat can not reach quickly. This uneven expansion (or contraction in the case of cooling) causes an unbalanced thermal stress on the wall of the glass and the glass breaks. For a thin glass wall this is less likely to happen as the heat can flow to all parts of it more efficiently. Pyrex is a special type of glass whose linear expansion coefficient is comparatively small (three times less than ordinary glass). The thermal expansion in such a glass material is much less and thus the possibility of breaking is also less likely. The beakers, test tubes etc. that are used in laboratories are made out of this kind of glass.

• Metal scale: A metal scale is standardized at a certain temperature. Thus the

marks in it are correct only at that temperature. Metal is good conductor of heat. The distance between two marks are expanded or contracted due to heating or cooling. Thus to have correct measure by such a scale we have to know the coefficient of expansion of the material of the scale and then to calculate the correct scale reading.

• Aluminium piston: The aluminium pistons of some car engines are made a

little bit smaller in diameter than the steel cylinders to allow for the greater expansion rate of alumimium.

• Cavity in Teeth: Dentists use filling material for the cavities of teeth which

has the same rate of expansion as teeth.

• Concrete Roadways: Concrete roadways and footpaths are intersected by gaps, sometimes filled with tar or some other material so that concrete can expand freely in summer and contract in winter.

• Electric Bulbs: The filament of an electric bulb is made of such a material

(tungsten, platinum) whose coefficient of thermal expansion is very close to that of the glass. A high vacuum is created inside the bulb and the filament is inserted and sealed at high temperature. If the glass and the filament are of different expansion rates, there can be gaps created in the seal after cooling.

Compensated Pendulum: The pointers in a pendulum clock are regulated by a pendulum. This pendulum is suspended by a metal rod. The metal rod expands in summer and contracts in winter. The time period of the pendulum depends on the effective length of the suspension rod which is the distance between the point of suspension and the centre of gravity of the pendulum

Fig. to be included


bob. In summer, this effective length increases due to increase in temperature. Hence the time period increases and the clock slows down. On the other hand in winter, due to decrease in temperature the effective length of the rod decreases. Thus the time period decreases and the clock runs faster. But the effective length of the suspension rod should be kept fixed to get correct timing in all seasons. Therefore, some arrangement is made to compensate the expansion and contraction of the suspension rod. The clock is then called compensated pendulum. The basic principle is demonstrated in the Fig. **. AB and CD are two different rods made of different metals. These two rods are joined by another rod BC. There is expansion in the two rods AB and CD due to increase in temperature. The lengths and the materials of the two rods are taken in such a way that the expansions in the two rods are same. Thus the distance between the suspension point O and the centre of gravity D of the bob remains unchanged.

Suppose the initial lengths of the two rods AB and CD are 1L and 2L respectively and the linear coefficients of expansion of the materials of the two rods are 1α and 2α . The lengths of the rods become α1L T∆ and α2L T∆ respectively, due to the increase in temperature T∆ . ∴ The condition for the effective length of suspension to remain unchanged is α1L T∆ = α2L T∆

Or, 1

2

2

1

αα

=LL .

Thus the lengths of the rods are to be selected such that they are inversely proportional to their thermal expansion coefficients. In this picture, 1L > 2L and thus 2α > 1α . Here the coefficient of thermal expansion for CD rod is greater than that of AB rod. Compensated Balance Wheel: There is a balance wheel inside a wrist watch or small clocks instead of pendulums. The rotational oscillation of such a wheel regulates the clock. The time period of oscillation of such a wheel depends on the radius of it. The radius is increased due to increase in temperature so the time period increases. Thus the watch goes slow. On the other hand, when the radius is decreased due to decrease in temperature the time period is decreased. Thus the watch runs faster. The compensated balance wheel is used in order to keep the radius unchanged in all seasons. The compensated balance wheel is divided into three parts as shown in the Fig. **. Each part is made of a bimetallic strip. The outer side of the bimetallic strip is brass and the inner side is of stainless steel. One end of each part is attached to the centre of the wheel with a spoke and the other part carries a heavy screw. The lengths of the spokes are increased as the temperature is raised. As a result the radius of the balance wheel is

Fig. to be included


increased and the screws go away from the centre. The effective radius of the wheel is thus increased. At the same time the bimetallic strips bend inwards and the screws move towards the centre. As a result the effective wheel radius decreases. The two effects compensate and finally the effective radius remains unchanged due to temperature variation. Thus the time period does not change and the clock gives correct time in all seasons. --------------------Numerical Examples with Solutions------------------------- Example# 1. The gap between two segments of a rail line is 0.5 inch at 10 C0 . If the length of each segment is 66 ft then at what temperature the gap will be closed?

C06 /1011 −×=α . [H.S.] Solution: The gap will be closed when the increase in length of each segment will be equal to the size of the gap.

The increase in length, L∆ = 0LL − = 0.5 in = 12

5.0 ft, and the increase in temperature,

T∆ = 0TT − , =0L 66 ft, =0T 10 C0 , ?=T We know, L∆ = α 0L )( 0TT −

Or, 12

5.0 = 66 61011 −×× )( 0TT −

Or, 0TT − = 116612

105.0 6

××× = 57.39 Or, T = 10 +57.39 = 67.39

∴The gap will be closed at 67.39 C0 . Example# 2. The difference of lengths of two metal rods is always 25 cm. The coefficients of linear expansion of them are C05 /1028.1 −× and C05 /1092.1 −× . Find their lengths at 0 C0 . [H.S. ‘01] Solution:

Let us assume, the lengths of the two rods at 0 C0 are 1L and 2L . We also assume 1L > 2L . ∴According to question, we have 21 LL − = 25. The coefficients of linear expansion for the materials of the two rods are

C051 /1028.1 −×=α and C05

2 /1092.1 −×=α respectively. If the temperature is increased by T∆ , then the expansion in two rods has to be the same so as to keep the difference of lengths unchanged.

Fig. to be included


∴ TL ∆××× −51 1028.1 = TL ∆××× −5

2 1092.1

Or, 23

28.192.1

2

1 ==LL 21 2

3 LL =⇒

∴ 2523

22 =− LL 502 =⇒ L cm. , 5023

1 ×=L = 75 cm.

∴ The lengths of the two rods at C0 are 75 cm and 50 cm, respectively. Example# 3. The length difference between an iron and a copper rod at 50 C0 is 2 cm. The same difference is observed when the two rods are at a temperature, 450 C0 . What will be the length difference at 0 C0 ? Given, C06 /1012 −×=α (iron) and

C06 /1017 −×=α (copper). [J.E.E.] Solution:

Suppose the lengths of iron rod and copper rod at 50 C0 are x cm and y cm, respectively. Since the difference in lengths of the two rods remains unchanged when the temperature is raised to 450 C0 , the expansions in the two rods should be equal. Expansion in iron rod = )50450(1012 6 −××× −x = 4001012 6 ××× −x cm. Expansion in copper rod = )50450(1017 6 −××× −y = 4001017 6 ××× −y cm. ∴ According to question, 4001012 6 ××× −x = 4001017 6 ××× −y

Or, yx1217

= .

Looking at the values of the linear expansion coefficients, we can say that the iron rod is bigger than the copper rod. Thus we have, x > y .

∴ 2=− yx Or, 21217

=− yy

Or, 8.4524

==y cm ∴ =x 2 + 4.8 = 6.8 cm.

Suppose, the lengths are 0x and 0y respectively, at 0 C0 . ∴ 6.8 = { }5010121 6

0 ××+ −x Or, 796.60 =x cm ∴ 796.42796.60 =−=y cm. Example# 4. A steel scale gives correct reading at 15 C0 . A distance measured by this scale at 30 C0 is 2000 ft. What is the possible error in measuring the distance? The coefficient of linear expansion for steel is 0.000011 / C0 . [H.S. ‘94] Solution:

If the steel scale measures 1 ft at 30 C0 then the actual reading should be = [ ])]1530(000011.011 −×+× = 15000011.01 ×+ = 1.000165 ft.


∴ The actual reading for 2000 ft would be = 2000 000165.1× = 2000.33 ft. ∴ The scale now measures )200033.2000( − ft. = 0.33 ft less; this is the error involved. Example# 5. A steel scale gives correct reading at 0 C0 . The length of a rod is measured to be 50 cm by this scale at 50 C0 . What will be the correct length of the rod? [The coefficient of linear expansion for steel, C06 /1012 −×=α ] [H.S. ’92, ‘95] Solution:

If the steel scale measures 1 cm at 15 C0 then the actual reading should be = [ ])]015(000012.011 −×+× = 15000012.01 ×+ = 1.00018 cm.

∴ The actual reading corresponding to 50 cm is = 50 00018.1× = 50.009 cm. Example# 6. A metal scale gives correct reading at 25 C0 . If the length of a rod is measured to be 80 cm by this scale at 15 C0 , find the correct length of the rod at this temperature. Given, C06 /1015 −×=α [H.S. ‘97] Solution:

If the reading is 1 cm. by the metal scale at 15 C0 , then the actual reading should be = [ ])]2515(000015.011 −×+× = 10000015.01 ×− = 0.99985 cm.

∴ The correct length corresponding to a reading of 80 cm is = 0.99985 80× = 79.988 cm. Example# 7. A steel scale is errorless at 50 F0 . The length of a brass rod is measured to be 1.5 m by the scale at 50 C0 . What will be the actual length of the rod at 100 C0 ? [ The coefficient of linear expansion for steel is = C06 /102.11 −× and that for brass is =

C06 /1018 −× .] [H.S. ‘00] Solution:

The relation between the Fahrenheit and the Celsius scale is 9

325

−=

FC .

∴ 9

32505

−=

C Or, 10=C .

∴ The temperature of 50 F0 measured in Fahrenheit scale is 10 C0 in Celsius scale. The steel scale is thus errorless at 10 C0 . If the reading by this scale is 1 m at 50 C0 , the actual reading should be

[ ])]1050(0000112.011 −×+× = 400000112.01 ×+ = 1.000448 m. ∴ The actual reading corresponding to 1.5 m is = 000448.15.1 × = 1.5007 m. ∴ The actual length of the brass rod at 50 C0 is = 1.5007 m. ∴ The actual length of the brass rod at 100 C0 is = )]50100(000018.01[5007.1 −×+× =

)50000018.01(5007.1 ×+× = 1.502 m. Example# 8. A brass scale attached to a barometer is errorless at 0 C0 . The coefficient, α for brass is = 0.00002 / C0 . The barometer gives reading 75 cm at 27 C0 . What should be the actual reading? [I.I.T.] Solution:


The barometer scale is errorless at 0 C0 . The scale will expand at 27 C0 . The reading of 75 cm of length at 27 C0 will be = [ ])027(00002.0175 −×+× = 00054.175× = 75.0405 cm. ∴ The actual reading of the barometer is = 75.0405 cm. Example# 9. A thin steel ring is heated up to 95 C0 . The ring is now exactly fitted on to a steel cylinder at this temperature. The system is now cooled down to 20 C0 . The diameter of the cylinder is 10 cm at 20 C0 . Find the thermal stress developed inside the ring. The Young modulus of steel, 25 /1021 cmkgY ×= and the coefficient of linear expansion, ./1012 06 C−×=α [J.E.E. ‘88] Solution:

The thermal stress = Y α T∆ = )2095(10121021 65 −×××× − = 1890 kg/cm 2 = 1890 9801000×× dyne/cm 2 = 1.8522 910× dyne/cm 2 . Example# 10. A pendulum clock gives correct time at 20 C0 . Sometimes, the clock slows down to 30 seconds a day. What is the temperature during that time? The coefficient of linear expansion of the material of the pendulum bar is = ./1018 06 C−× [H.S. ‘89] Solution: The pendulum slows down means the time period of oscillation is increased. This happens when the length of the bar (suspension rod) is increased. This suggests that the temperature is increased. Let us suppose, the length of the pendulum bar is 1L at 20 C0 and it is 2L when the temperature is increased by T∆ . ∴We can write, 12 LL = [1 + α T∆ ] When the clock gives correct time, it strikes off each second for half a swing. The full time period for each swing is then 1t = 2 s.

∴ gL122 π= (i)

1 day = 606024 ×× = 86400 s. When the clock goes slow by 30 seconds, it actually strikes 86370)3086400( =− times a day which means, 86370 half swings in 86400 seconds. Thus the actual time required for

one half swing now is = 8637086400 s.

∴ The time period (= one full swing), 2t = 8637086400 2× s.

We can write, [ ]

gTL

gLt ∆+

==α

π12 12

2 (ii)

∴ (ii) ÷ (i):


[ ]

gLg

TL

1

1

2

12

2

28637086400

π

απ

∆+

=×

Or, 8637086400 = [ ] 2/11 T∆+α = [1 + 0.000018× T∆ ] 2/1

Or, 8637086400 = 1 + T∆×× 000018.0

21

[Binomially expanding and neglecting other terms as they are very small]

Or, 000018.0

218637086400

×

−=∆T =

000018.086370230

××

Or, 6.38=∆T ∴ The required temperature is = 20 + 38.6 = 58.6 C0 . Example# 11. A pendulum clock, made of iron, gives correct time at 20 C0 . Find out how much time the clock runs fast or slow when the temperature is changed to 40 C0 . Given, the coefficient of volume expansion of iron = C06 /1036 −× . [H.S. ’92, ’03; I.I.T.] Solution:

Let us suppose, the length of the pendulum bar is 1L at 20 C0 and 2L at 40 C0 , and the corresponding time periods are 1t and 2t respectively.

We can write, gLt 1

1 2π= and gLt 2

2 2π= .

∴ 1

2

1

2

LL

tt

=

Now, we have =1t 2 s, since the time period at 20 C0 is equal to 2 strikes by the pendulum. Also, we have [ ])2040(112 −+= αLL = ( )α2011 +L .

∴ ( ) 2/1

1

12 2012

)201(2 α

α+×=

+=

LL

t

Or, ( ) 2/162 10122012 −××+×=t [as

3γα = , γ being the coefficient of volume

expansion]

Or, )101220211(2 6

2−×××+×=t [Binomially expanding and neglecting other terms]

Or, 00024.21012202 62 =××+= −t second.

Now the time period of oscillation of the pendulum becomes 2.00024 s instead of 2 s. Thus the clock now goes 00024.0)200024.2( =− s slow in every 2 s. 1 day = 606024 ×× = 86400 s.

∴In a day, that is in 86400 s, the clock goes slow by 2

8640000024.0 × = 10.37 s.


Example# 12. There are 3 steel and 2 brass rods in a compensated pendulum. If the average length of each steel rod is 90 cm, what is the average length of each brass rod? Given, the coefficient of linear expansion for steel = C06 /1012 −× and that for brass =

C06 /1019 −× . Solution: According to the working principle of compensated pendulum, we ca say The increment of length for 3 steel rods = the increment of length for 2 brass rods. Or, 66 101921012903 −− ×××=××× L [ =L length of each brass rod]

Or, 192

12903×××

=L Or, L = 85.26 cm.

Example# 13. A meter scale has to be made out of steel so that the maximum error in measuring1 mm can be 0.0005 mm at any temperature. What can be maximum change in temperature that can be tolerated? Given, the coefficient of linear expansion for steel =

C06 /1022.13 −× [I.I.T.] Solution: Suppose, the maximum change in temperature is Tδ for which there can be the maximum error 0.0005 mm in measuring 1 mm. For the increase in temperature of Tδ , the reading for 1 mm will be = 1.0005 mm. ∴ 1.0005 = 1 + α Tδ Or, 1.0005 = 1 + Tδ×× −61022.13

Or, Tδ = 61022.130005.0

−× = 37.82 C0 .

Example# 14. A equilateral triangle is formed by three rods at 0 C0 . One of the three rods is made of invar (the thermal expansion of invar is negligible) and the rest two are made of some other material. When the triangle is heated up to 100 C0 , the two rods, made of same material, subtend an angle of ( )3/ θπ − between them. Show that the coefficient of linear expansion of the material of the two rods is C0200/3θ . [J.E.E.; H.S. ‘99] Solution:

Suppose, the length of each rod is l at 0 C0 and the coefficient of linear expansion of the material of two rods (AD and BD, shown in the figure**) = α .

∴The length of each of the rods (AD and BD) at 100 C0 will be ).1001(1 α+= ll It is assumed that the base AB is made of invar and thus it does not expand appreciably. Now a perpendicular is drawn from the vertex of the triangle on to the base which bisects the base (see figure). Now from the triangle ADO, we can write

2/sin)2/6/sin(2/ 1

πθπll

=−

Or, 1

)1001()2/6/sin(2

αθπ

+=

−ll

Or, )1001(2

1)2/6/sin(α

θπ+

=−

Fig. to be included


Or, =−2

sin6

cos2

cos6

sin θπθπ)1001(2

1α+

Or, )1001(2

12

.231.

21

αθ

+=− [ For θ small,

22sin θθ

→ and 12

cos →θ ]

Or, )1001(

1123

αθ

+−=

Or, 1

1001001

10023 α

ααθ =

+= [ α100 is neglected, being small compared to 1]

∴ θα200

3= / C0 .

Example# 15. A uniform pressure P is applied on all sides of a solid cube. How much of its temperature has to be raised so that the volume of the cube will be unchanged? Given, the coefficient of volume expansion = γ and the bulk modulus = B . [I.I.T.] Solution: If the initial volume of the cube is = V and the decrease in volume due to applied pressure P is = V∆ then the bulk modulus,

VVPB∆

= Or, B

PVV =∆ .

Suppose, the increase of temperature has to be T∆ to keep the volume unchanged. Then we have, γ=∆V V T∆

∴ γ V T∆ = B

PV Or, T∆ = γB

P .

Example# 16. A 100 cm long rod is clamped at one end. A screw is attached to the free end of the rod. The pitch of the screw is 0.5 mm. The screw can go back and forth along the length of the rod when rotated. A circular scale with 100 divisions is attached with this. The screw moves one smallest division on the linear pitch scale for one full rotation of the circular scale. One smallest division on the linear scale is 0.5 mm. At 20 C0 , the linear scale reads something little more than 0 and the circular scale reads 92. If the

Note here that the original triangle was a equilateral triangle and thus the

value of each angle = 3π . When two sides expand equally, the triangle

becomes an isosceles triangle where the angle between two equal sides

now has decreased to a value ( )3

θπ− , usually only a little less than the

previous value. One can thus easily think that θ is small.


temperature of the rod is raised to 100 C0 , the linear scale shows a little above 4 and the circular scale reads 72. Find the coefficient of linear expansion for the material of the rod. [I.I.T.] Solution: Screw pitch = 0.5 mm. Total divisions of the circular scale = 100.

The least count of the screw = 100

5.0 = 0.005 mm. One division in the circular scale is

equal to 0.005 mm in the linear scale. Thus the total scale reading for 20 C0 is = 92005.05.00 ×+× = 0.46 mm. The total scale reading for 100 C0 is = 72005.05.04 ×+× = 2.36 mm. ∴ The increase in length = 46.036.2 − = 1.9 mm = 0.19 cm. Thus we can write, 0.19 = )20100(100 −××α

∴ 80100

19.0×

=α = C06 /1075.23 −× .

Example# 17. Two rods of equal cross-section are joined fact to face. The joint rod is 1 m long at 25 C0 and one in the composition is a 30 cm long copper rod. The joint rod is expanded by 1.91 mm at 125 C0 . If the joint rod is kept between two rigid walls, it does not expand even when the temperature is increased. What is the Young modulus and the thermal coefficient of linear expansion for the material of the other rod? For copper, the coefficient of linear expansion C05 /107.1 −×= and Young modulus, 211 /103.1 mNY ×= . [I.I.T.; J.E.E.] Solution:

At 25 C0 , the length of the copper rod = 30 cm and the length of the other rod = 7030100 =− cm.

At 125 C0 , the increase in length of the copper rod = )25125(107.130 5 −××× − = 0.051 cm = 0.51 mm. ∴ The increase in length of the other rod = 51.091.1 − = 1.4 mm. If the coefficient of linear expansion of the material of the other rod isα ,

)25125(7014.0 −××= α Or, C05 /100.21007014.0 −×=×

=α

Thermal stress due to change in temperature T∆ is = Y α T∆ . Thermal stress in copper rod = T∆×××× −511 107.1103.1 Thermal stress in the other rod = TY ∆××× −5102 Since the cross-sectional area of the two rods is same, the thermal stress in two rods is also same. ∴ TY ∆××× −5102 = T∆×××× −511 107.1103.1

Or, 115

511

10105.1102

107.1103.1×=

××××

= −

−

Y N/m 2 .

Example# 18. An aluminium sphere is immersed in oil at 20 C0 and at 1 atm pressure. How much has the pressure to be increased at 35 C0 so that the sphere will not expand?


Given, the coefficient of linear expansion of aluminium = C06 /1023 −× and the bulk modulus (coefficient of volume deformation), K = 210 /107.7 mN× . Solution:

Suppose, the volume of sphere at 20 C0 = 0V and the volume at 35 C0 = V . ∴ [ ])2035(10 −+= γVV = [ ]15310 ×+ αV = [ ]15102331 6

0 ×××+ −V

Or, 6

0

1010351 −×+=VV Or, 6

0

1010351 −×=−VV

Or, 6

00

0 101035 −×=∆

=−

VV

VVV

If we now assume that the pressure is increased to P , then

0VVPK∆

= .

∴0VVKP ∆

= = 610 101035107.7 −××× N/m 2

= 101300

1010357.7 4×× atm = 786.7 atm. [ 3103.101 × N/m 2 = 1 atm]

∴The pressure is to be increased by 785.7 atm above the standard atmospheric pressure of 1 atm. Example# 19. A solid brass cylinder, having mass 500 gm and radius 3 cm, is placed on a frictionless bearing. At 20 C0 , the angular speed of this cylinder around its axis is 60 radian/sec. Will there be any change in the angular speed, angular momentum and rotational kinetic energy of the cylinder when the temperature of it is increased to 100 C0 ? If so, then determine the percentage change in the above mentioned quantities. Given, α for brass = C05 /102 −× . [J.E.E.] Solution:

The moment of inertia of the solid cylinder, 2

21 MRI = .

The radius, R at 20 C0 is 3 cm.

∴The moment of inertial at 20 C0 is 21 )3(

21 MI = = 9

21

×M gm cm 2 .

When the temperature is raised to 100 C0 , the radius = )]20100(1021[3 5 −××+× − = 3.0048 cm.

∴The moment of inertial at 100 C0 is 22 )0048.3(

21 MI = = 029.9

21

×M gm cm 2 .

The angular momentum remains constant. The angular speed changes.

∴ 2211 ωω II = ⇒ 2

112 I

I ωω = =

029.921

60921

×

××

M

M =

029.9609× = 59.808 radian/sec.


The percentage change in angular speed is = 1001

12 ×−ωωω

= %32.010060

60808.59−=×

− . [Negative sign is due to the decrease in angular speed]

The rotational kinetic energy at 20 C0 , 2111 2

1 ωIE = .

The rotational kinetic energy at 100 C0 , 2222 2

1 ωIE = .

∴ The percentage change in rotational kinetic energy is = 1001

12 ×−

EEE =

100)60(9

)60(9)808.59(029.9100 2

22

211

211

222 ×

××−×

=×−ω

ωωI

II = -0.32%.

Example# 20. Two rods of different materials are joined end to end and the two other ends of the rods are clamped at two opposite and parallel walls. The area of cross-section of both the rods is A . The length, the coefficient of linear expansion and the Young modulus of one rod are 1l , 1α and 1Y whereas for the other rod they are 2l , 2α and 2Y . The temperature of the rods are increased by CT 0 . Determine the aggregate force developed and the lengths of the rods in this situation. [I.I.T.] Solution:

Total thermal expansion in the two rods is = TlTl 2211 αα + . Due to the clamping at two opposite walls, there will be a thermal stress developed inside the two rods. The contraction in the two rods will then balance the thermal expansion. Let us say, the force developed inside each rod be F . If the contraction in the first rod is 1l∆ and that in the second rod is 2l∆ , we can write from the definition of Young modulus,

111 /

/llAFY

∆= Or,

1

11 AY

Fll =∆ and

222 /

/llAFY

∆= Or,

2

22 AY

Fll =∆ .

The total contraction in the two rods is = =∆+∆ 21 ll1

1

AYFl +

2

2

AYFl .

∴ TlTl 2211 αα + = 1

1

AYFl +

2

2

AYFl Or,

+=+

2

2

1

12211 )(

Yl

Yl

FllAT αα

Or, 1221

221121 )(YlYl

llYATYF+

+=

αα .


The length of the first rod when the temperature is increased by CT 0 = Initial length +

expansion – contraction = 1l + Tl 11α1

1

AYFl

− = 1l + Tl 11α ×−1

1

AYl

1221

221121 )(YlYl

llYATY+

+ αα

= 1l + Tl 11α1221

221121 )(YlYlllTYl

++

−αα = +1l

1

1

2

2

2211 )(

lY

lY

YYT

+

− αα

Similarly, the present length of the second rod = +2l

1

1

2

2

1122 )(

lY

lY

YYT

+

− αα .

Example# 21. Two rods of same coefficient of linear expansion 2α and of equal length

2l and another of coefficient 1α and length 1l are freely jointed to form an isosceles triangle. The base of the triangle 1l is nailed at the middle. What can be the relation between 1l and 2l so that the distance between the vertex and the mid point of the base remains unchanged even if the temperature is changed? Solution: ABC is an isosceles triangle and D is the mid point of the base BC which is nailed (as shown in the Figure**).

According to the figure, AD = 2

122 2

−

ll = x (say,)

∴2

122

2

2

−=

llx

As the temperature is changed by T∆ = t , the lengths of the rods are changed to

)1( 222 tll α+=′ and )1( 111 tll α+=′ . If now the length AD = y , then we can write

2

12

22

2

′−′=

lly = 2

12

12

22

2 )1(41)1( tltl αα +−+

= )21(41)21( 1

212

22 tltl αα +−+ [neglecting square terms, as they are small]

According to question, 22 yx =

Or, =

−

212

2 2ll )21(

41)21( 1

212

22 tltl αα +−+

Fig. to be included


Or, tl 22

22 α = 21 tl 1

21 α Or,

2

1

1

2

21

αα

=ll This is the required relation.

Example# 22. Two metal strips of length 0l and width x are riveted at a temperature t such that the ends of them coincide. The coefficients of linear expansion of the materials of the rods are 1α and 2α ( 1α > 2α ). This bimetallic strip bends like an arc of a circle when the temperature is raised to )( tt δ+ . Find the radius of curvature of the bent strip. Solution:

Let us suppose, the lengths of the strips, AB and CD at higher temperature are 1l and 2l and their radii of curvatures are 1r and 2r where θ is the angle subtended at the centre as shown in the Fig.**. ∴ θδα 1101 )1( rtll =+= ……………(i) and θδα 2202 )1( rtll =+= ……………..(ii) Subtracting (ii) from (i),

tlrr δααθ )()( 21021 −=− ∴21

210 )(rr

tl−−

=δαα

θ = x

tl δαα )( 210 − [ xrr =− 21Q ]

Now, adding up (i) and (ii),

θδαα tll

rr)(2 2100

21++

=+ = θ

δαα tll )(2 2100 ++

= xtl

tll×

−++δαα

δαα)(

)(2

210

2100 = tl

xtllδααδαα

)(})(2{

210

2100

−++

.

Example# 23. A 600 cm steel rod is clamped firmly at one end and the other end of it is kept at 10 cm away from the suspension point of a lever. If the temperature of the rod is increased by 50 C0 , the lever is rotated by an angle of 2 0 . Find the increment in length of the rod and the coefficient of linear expansion. Solution: The increment in length of the rod,

θrl =∆ [Here, =r 10 cm, 180

2 πθ ×= radian.]

= 9180

210 ππ=×× = 0.349 cm.

The coefficient of linear expansion of the material of the rod,

Tll∆×

∆=

0

α [Here, 0l = 600 cm (initial length), T∆ = 50 C0 ]

= 50600

349.0×

= C06 /106.11 −× .

Fig. to be included


Example# 24. Three rods of equal length at 0 C0 form an equilateral triangle, ABC (as shown in Figure**). The coefficient of linear expansion for the material of the rod AB is α and that for other two rods isβ . How much will be the increase in angle C at Ct 0 ? [J.E.E. ‘98] Solution:

At 0 C0 , the angle C∠ = 60 C0 = 3π .

Let us now assume, C∠ = φ2 , where θπφ −=6

(θ very small)

[Note that the angle C may decrease or increase according to the amount of expansions of the rods. Here, in the figure, it is shown to decrease.] If l be the length of each rod at 0 C0 , we can write for the extended triangle,

090sin)1(

sin)1(2/ tltl β

φα +

=+ Or,

)1()1(2/sin

tltl

βαφ

++

=

Or, )1(2

16

sint

tβαθπ++

=

− Or, =− θπθπ sin

6coscos

6sin

)1(21

ttβα++

Or,

++

=−tt

βαθθ

11

21sin

23cos

21

Or, θ31− = tt

βα

++

11 [Q θ small ∴ θθ =sin , 1cos =θ ]

Or,

++

−=tt

βαθ

111

31 =

)1(3)(ttβαβ+−

∴ The total increase in the angle C at Ct 0 is = =θ2)1(3

)(2tt

βαβ+− .

--------------------Discussions on some Questions------------------------- Q.1 Is it possible for the difference in length of a brass rod and a steel rod to remain same at all temperatures? [I.I.T.] Ans. This is possible only when the amount of expansion (or contraction) of the brass rod is equal to that of the steel rod for the same increase (or decrease) in temperature. Let us suppose, the coefficient of linear expansion for brass = Bα , the coefficient for steel = Sα and at a particular temperature, the length of the brass rod = BL , the length of the steel rod = SL . According to question, the increment in length in the two rods will be the same for an increase in temperature, T∆ .

Fig. to be included


∴ =∆TL BBα TL SS ∆α Or, B

S

S

B

LL

αα

=

Thus the ratio of lengths of the two rods has to be equal to the inverse of the ratio of the coefficients of their materials. Q.2 A brass disc sits tight inside a hole in steel. Will you heat up this to remove the disc out of the hole? Given, α for brass = C06 /1019 −× and α for steel = C06 /1012 −× . [I.I.T.] Ans. We have to cool this place to remove the brass disc out of the steel hole. The diameter of the brass disc is equal to the inner diameter of the steel hole at the initial stage. The value of the coefficient of linear expansion of brass is bigger than the coefficient of linear expansion of steel. Thus for the same amount of heating, the expansion of the diameter of brass disc will be more than the expansion of the inner diameter of the steel hole. For this the brass disc will sit there more tightly. On the other hand, when we cool this place, the decrease in the diameter of brass disc will be more than the decrease in the diameter of the steel hole. Hence the brass disc will get loose and come out. Q.3 An isosceles triangle is formed by three zinc rods. Will the base angles of the triangle change due to a change in temperature? Explain. [H.S. ‘88] Ans. Let us suppose, the initial length of two equal arms of the triangle is = a and the initial length of the base is = b . Thus the ratio of the lengths of the arms = baa :: . If the coefficient of linear expansion for zinc isα , then due to increase in temperature t , the ratio of the lengths of the arms is now )1(:)1(:)1( tbtata ααα +++ = baa :: . Hence the ratio of the lengths of the arms of the triangle remains the same and so we can conclude that the angles also remain unchanged. Q.4 A copper strip and an iron strip of same size are riveted together. Explain what happens when such a combination is heated. [J.E.E. ‘87] Ans. The bimetallic strip made of copper and iron bends when it is heated. The coefficient of linear expansion for iron and the coefficient for copper are not same. Otherwise it would remain straight. The coefficient of linear expansion for copper is more than that for iron. Thus the expansion in copper strip will be more than that in iron strip. The combination will thus be bent because of unequal expansion. The copper strip will be on the outer side as it expands more and the iron strip will be on the inner side when bent. Q.5 An iron rod is inserted in a circular iron ring along its diameter. Explain whether the ring to remain circular when the combination is heated. [J.E.E. ‘85] Ans. The ring remains circular when the combination is heated. Let us suppose, the initial diameter of the ring is d and the coefficient of linear expansion isα .


∴ The perimeter of the ring = π d and the length of the rod = d .

∴ The ratio of the perimeter of the ring and the length of the rod = ππ=

dd .

If now the temperature of the combination is increased by T∆ , the perimeter of the ring = )1( Td ∆+απ and the length of the rod = )1( Td ∆+α .

∴ The ratio of the perimeter of the ring and the length of the rod = πααπ

=∆+∆+

)1()1(

TdTd .

Thus it shows that the ring will remain circular. Q.6 The area of cross-section and the length of two rods are same. But the materials of the rods are not same; the elastic and thermal properties are different. The two rods are fixed in two rigid walls along their lengths. What can be the relation between their Young modulus and coefficient of linear expansion so that the meeting point of the two rods may not change even when they are heated? [J.E.E. ‘89] Ans. Two rods are fixed rigidly. If the thermal stresses in them are equal, when heated, the meeting point of the rods does not change. Suppose, the Young modulus and the coefficient of linear expansion for the first rod are

1Y and 1α and those for the second rod are 2Y and 2α respectively. If the temperature is increased by T∆ , the thermal stress developed in first rod = TY ∆11α and the thermal stress developed in the second rod = TY ∆22α . Therefore, the condition for the meeting point of the rods to remain unchanged is

TY ∆11α = TY ∆22α Or, 1

2

2

1

αα

=YY .

Q.7 Does the volume expansion of a solid body due to heating differ if it contains a hole in it? Justify your answer. [J.E.E.] Ans. The change in volume of a solid body due to the change in temperature does not depend on whether the body contains a hole in it. If the hole would be filled up with solid of same material, it would expand (or contract) at the same rate. Thus the volume of the hole will expand (or contract) with the same rate as that of the solid body. There will be no effect. Q.8 A solid and a hollow cylinder of same size and made of same material are taken. Explain if the expansions in them be same when they are heated. [H.S. ‘90] Ans. The volume expansions in the solid and in the hollow cylinders of same size and same material will be the same if their temperatures are raised to same amount. Suppose, the length and radius of the two cylinders be 1l and 1r . The initial volume of both the cylinders, 1V = 1

21 lrπ .

As the temperature is increased by T∆ , we consider the changes in the hollow cylinder: The length, )1(12 Tll ∆+= α and the radius, )1(12 Trr ∆+= α .


∴ The new volume of the hollow cylinder, )1()1( 1

2212

222 TlTrlrV ∆+×∆+== ααππ = 3

12

1 )1( Tlr ∆+απ = )31(1 TV ∆+ α [As T∆α is small, other terms are neglected.] = )1(1 TV ∆+ γ [Since, αγ 3= , the coefficient of volume expansion] For the solid cylinder we can write, the new volume,

)1(12 TVV ∆+=′ γ [In this case we consider the volume expansion altogether.]

∴ 22 VV =′ , the volume expansion in both the cylinders is same. Q.9 A pendulum has length l , time period T and the coefficient of linear expansion α .

Show that the change in time period due to the change in temperature t∆ is = tT∆α21

and the rate of change of time period is = Tα21 . [J.E.E. ‘94]

Ans.

The time period of the pendulum at the initial stage is glT π2= .

If the temperature is increased by t∆ , the length is changed to )1( tl ∆+α .

∴The new time period, g

tlT )1(2 ∆+=′

απ = ( ) 2/112 tgl

∆+απ

=

∆+ tT α

211 [Neglecting other terms of the Binomial expansion]

∴The increase in time period in the time interval t∆ is

= TT −′ = TtT −

∆+ α

211 = tT∆α

21 .

∴The rate of increase in time period = Tt

tTα

α

212

1

=∆

∆.

-----------------------------Questionnaire--------------------------------

Very brief questions: Marks: 1

Answer in a few words

1. What is the unit of the coefficient of linear expansion? [/ 0C or often written as / C0 or ]10 −C

2. What is the relation between the coefficients of linear expansion and volume expansion? [ ]3αγ =


3. A bimetallic strip made of brass and iron remains straight at 20 C0 . If the temperature brought down to 0 C0 the strip bends,. Which metal will be on the convex side of the bent strip? [iron]

4. In Celsius scale, the coefficient of linear expansion of platinum is C06 /109 −× . What is this value in Fahrenheit scale? [ F06 /105 −× ]

Fill in the blanks

1. Due to increase in temperature, each side of a copper cube is increased by 0.1%.

The volume of the cube is increased by ------------. [0.3%] 2. If the temperature is increased, the density of a solid -------------. [decreases] 3. A pendulum clock keeps correct time at 20 C0 . If the room temperature is 40 C0 ,

the clock goes -----------. [slow] 4. The thermal expansion of invar is -------- than that of other metals or alloys. [less]

Multiple answer type

1. When a hollow metal sphere is heated, the volume of the cavity inside

(a) remains same, (b) decreases, (c) increases [(c)] 2. Which of the following has highest percentage increase when a solid metal sphere

is heated? (a) length, (b) area, (c) volume, (d) density [(c)]

3. A steel scale gives correct reading at 10 C0 . When this is used to measure a rod at 30 C0 , the measured length will be (a) equal to correct length, (b) less than the correct length, (c) more than the correct length [(b)]

4. Platinum wire can be sealed with a glass wall because platinum and glass have (a) equal density, (b) equal melting point, (c) equal specific heat, (d) equal coefficient of linear expansion [(d)]

5. The ratio of lengths of two iron rods is 1:2 and the ratio of cross-section of them is 2:3. The ratio of volume expansion of the two rods due to same temperature increase is (a) 1:2, (b) 1:3, (c) 2:3, (d) 1:6 [(b)]

6. A brass disc is tightly sitting inside an iron hole. What is the most useful technique to pull the disc out of the hole? [The coefficient of linear expansion of brass is more than that of iron.] (a) The combination has to be heated, (b) The combination has to be cooled, (c) Hitting is required than cooling or heating, (d) Has to be heated first and then to immerse in water for cooling. [(b)]

7. An isosceles triangle is formed by three iron rods. What will be the change in angles when the triangle is heated? (a) No change, (b) The base angles increase but the vertex angle decreases, (c) The base angles decrease but the vertex angle increases, (d) Nothing can be said [(a)]


8. At 0 C0 , an aluminium rod (coefficient of linear expansion 1α ) of length 1l and an iron rod (coefficient of linear expansion 2α ) of length 2l are joined together to form a single rod of length 21 ll + . If the expansion in the two rods is same for the

temperature increase of t C0 , the value of 21

1

lll+

will be

(a) 2

1

αα , (b)

1

2

αα , (c)

21

1

ααα+

, (d) 21

2

ααα+

[I.I.T.-J.E.E. ‘03]

[(d)]

9. The coefficient of linear expansion for brass is C06 /1019 −× . A brass circular disc has area 25 cm 2 at 0 C0 . The area at 80 C0 is (a) 25.038 cm 2 , (b) 25.076 cm 2 , (c) 25.114 cm 2 , (d) 25.38 cm 2 [(b)]

10. At 10 C0 , the length of each rail segment is 25 m. How much gap has to be kept between two such segments such that there will not be any deformation up to a temperature, 50 C0 ? [For steel, α = C06 /1011 −× .] (a) 5.5 mm, (b) 8.25 mm, (c) 1.1 mm, (d) 1.65 mm [(c)]

11. Two plain strips of equal area made of iron and copper are kept one on top of the other and riveted. What will be the nature of the curvature of this bimetallic strip at 0 C0 and at 100 C0 ? [The coefficient of linear expansion for copper is more than that of iron.] (a) The copper strip will be on the convex side at both the temperatures, (b) The iron strip will be on the convex side at both the temperatures, (c) The copper strip will be on the convex side at 0 C0 , and on the concave side at 100 C0 , (d) The iron strip will be on the convex side at 0 C0 , and on the concave side at 100 C0 [(d)]

12. In which of the following the bimetallic strip is not used? (a) A metal wire is sealed inside a glass, (b) Thermostat, (c) Fire alarm, (d) Compensated balance wheel of a watch [(a)]

13. The length of a brass rod has to be measured at different temperatures. Which of the following materials should be chosen to make a scale? (a) wood, (b) steel, (c) brass, (d) platinum [(a)]

14. A pendulum clock keeps correct time at Ct 0 . When the temperature is increased (a) The clock always keeps correct time. (b) The clock runs fast, (c) The clock goes slow, (d) Whether the clock goes fast or slow depends on the material of the pendulum bar. [(c)]

15. The material of a rod has Young modulus Y and the coefficient of linear expansion,α . If such a rod of length l and cross-section A is fixed firmly between two rigid walls and then temperature is increased by Ct 0 , what will be thermal stress developed? (a) αlAY t , (b) αAY t , (c) αlY t , (d) αY t [(b)]

16. Two rods of same length are kept rigidly fixed between two unmovable walls and the temperatures of them are increased by the same amount. What will be the


relation between the Young modulus and the coefficient of linear expansion of the materials of the two rods when the thermal stress in them are equal?

(a) 2

1

2

1

αα

=YY , (b)

2

1

2

1

αα

=YY , (c)

1

2

2

1

αα

=YY , (d)

1

2

2

1

αα

=YY [(c)]

17. A bimetallic strip is made of two exactly similar strips of copper and brass. The coefficients of linear expansion for the two materials are Cα and Bα . If now the temperature is raised by T∆ , the bimetallic strip bends like an arc for which the radius of curvature is R . Which of the following is correct?

(a) TR ∆∝ , (b) 2)(1T

R∆

∝ , (c) CBR αα −∝ , (d) CB

Rαα −

∝1 [I.I.T. ‘99]

[(d)]

Short questions: Marks: 2 1. There is an expansion in solid due to heating – how can it be shown by an

experiment? 2. Describe an experiment to show that different materials undergo different expansions

due to same increase in temperature. 3. Define coefficient of linear expansion for solid. [H.S.’97] 4. ‘All solids expand in length due to increase in temperature. In some cases this

expansion has advantages and in some cases this expansion causes disadvantages.’ – Establish this with examples. [H.S.’91]

5. ‘Metal scale does not give correct reading at all temperatures’.- Explain this. [H.S.’97] 6. What is bimetallic strip? How is it used to measure temperature? [H.S. ‘03] 7. The coefficient of linear expansion of iron is C06 /1012 −× - Explain this. [H.S. ‘03] 8. Is the value of α same in Celsius and Fahrenheit scales? 9. The coefficient of linear expansion for brass is C06 /1019 −× . What is this value in

Fahrenheit scale? [H.S.’95] 10. Why does a bimetallic strip bend due to heating or cooling? 11. How does a bimetallic strip work as an automatic switch in an electrical circuit? [H.S. ‘93] 12. A bimetallic strip can be used as an alarm – explain. 13. What is invar? Write any usage of this. 14. An isosceles triangle is formed by three zinc rods. Do the base angles change due to

the rise in temperature? Explain with logic. 15. It is claimed that the difference in length of a brass rod and a steel rod is same at all

temperatures. Is it possible? 16. An iron rod is fixed along the diameter of a circular iron ring. Explain if the ring

remains circular even when the combination is heated. 17. A copper strip and an iron strip of same size are riveted. Explain what happens when

this is heated.


18. A brass disc is struck inside a steel hole. What will you do to pull the disc out of the hole? Will you heat this place? Given, α for brass = C06 /1019 −× and α for steel = C06 /1012 −× .

19. Does the volume change due to temperature depend on whether there are pores inside a material (when other things remain the same)? Explain logically.

20. A hollow cylinder and a solid cylinder of same size are heated to same degree. Do they have same expansion? Answer with reasons. [H.S.’90]

21. Why iron rods and not other rods are used for building construction? [H.S. ‘95] 22. Platinum wire is easily sealed inside a glass and not copper wire. Why? [H.S. ‘92] 23. Why does a thick glass tumbler break when hot water is poured in it? 24. Why thin glass beaked is used to heat water in laboratory? 25. The tight lid of glass bottle opens up due to little heating – why? 26. A gap is kept between two rail pieces – why? 27. How do the shapes change for an equilateral triangle and a square made of metal rods

when they are heated? 28. A metal pipe is heated. What will be the changes in (i) volume, (ii) density, (iii)

internal diameter and (iv) external diameter of this pipe? 29. Why both the ends of a big steel bridge are not inserted into cement foundation? 30. Why are the holes made in elliptical shape in a fishplate for inserting the bolts

through it while joining two rail segments? 31. Give three examples where thermal expansion of solid is utilized. 32. Two rods of length l and width d are riveted. What will be the mean radius of

curvature when the temperature of this riveted combination is raised by C0θ ? The materials of two rods have coefficient of linear expansion 1α and 2α . [H.S. ‘05]

33. Find out the expression for thermal stress. [H.S. ‘05] 34. Two rods have same cross-section and same length but they are made of different

materials. The rods are fixed in two unmovable walls along their lengths. The elastic constant and the thermal coefficients of the materials of the two rods are different. What will be the relation between the coefficient of linear expansion and Young modulus of the two rods so that the joining position of the two rods remains unchanged when heated.

Medium questions: Marks: 4 1. Define coefficient of linear expansion. Show that the unit of coefficient of linear

expansion does not depend on the unit of length but it depends on the unit of temperature. [2+2] [H.S. ’96, ‘99]

2. Establish the relation between the coefficient of linear expansion and the coefficient of surface expansion of solid. [4] [H.S. ‘99]

3. Establish the relation between the coefficient of linear expansion and the coefficient of volume expansion of solid. [4]

[H.S. ’91, ’95, ’97, ’01, ’03, ’05; J.E.E. ‘98] 4. Find a relation of how density of solid changes with temp change and discuss. [2+2] [H.S. ’94, ’98, ‘01] 5. Explain the working principles of (i) thermostat, (ii) compensated balance wheel.


[2+2] 6. Why does a common pendulum clock run fast in winter and go slow in summer?

What can be the arrangement so that it may keep correct time in all seasons? [2+2] 7. What is compensated pendulum? Explain its principle. [2+2] 8. Prove by an experiment where force is developed due to thermal expansion. [4] 9. What is thermal stress? Determine thermal stress inside a rod. [2+2] 10. Define coefficient of linear expansion. Show that the unit of it does not depend on the

unit of length. A metal rod is fixed at two ends inside a circular metal ring along its diameter. Show that the ring remains circular when the combination is heated. [2+2]

[H.S.(XI) ‘06] Short problems: Marks: 2 1. The coefficient of linear expansion for iron is 0.0000067/ F0 . What is the value of the

coefficient in per degree Celsius? [Ans. 0.000012/ C0 ] 2. Temperature of a 2 m long rod is increased by 100 C0 . What will be the increase in

length? Given, C0/000012.0=α . [Ans. 0.24 cm] 3. The temperature of a 3 m long iron rod is raised from 0 C0 to 200 C0 . The length

increases by 7.2 mm. Calculate the coefficient of linear expansion of iron. [Ans. 0.000012/ C0 ] 4. The length of a brass rod is increased by 0.5 cm when the temperature is increased by

100 C0 . What was the initial length of the rod? For brass, C06 /1019 −×=α . [Ans. 263.157 cm] 5. A zinc rod is 2 m at 30 C0 . What will be its length at 60 C0 ? For zinc,

C0/00003.0=α . [Ans. 200.18 cm] 6. The lengths of an iron rod and a zinc rod at 0 C0 are 25.55 cm and 25.5 cm,

respectively. At what temperature the lengths of them will be equal? The coefficients of linear expansion for iron and zinc are 61010 −× and 61030 −× per C0 , respectively.

[98 C0 ] 7. A railway line is made by a 15 m long steel rail segment. How much gap has to be

kept between two such segments so as to tolerate maximum 40 C0 temperature increase? For steel, C06 /1011 −×=α . [0.66 cm]

8. At 10 C0 , there are gaps of 0.5 inches between two rail segments. At what temperature will the gaps vanish when each segment is 66 ft long? C06 /1011 −×=α

[Ans. 67.4 C0 ] 9. The height of Eiffel tower is 335 m. The minimum temperature in winter is 0 F0 and

the maximum temperature is 100 F0 . The tower is made of steel. The coefficient of linear expansion of steel is C06 /1012 −× . What will be the maximum change in height in two seasons? [Ans. 22.3 cm]

10. The distance between Allahabad and Delhi is 390 mile by railroad. The mean temperature in winter is 38 F0 and the mean temperature in summer is 119 F0 . How much total gap has to be kept in the rail lines? For iron, C0/000012.0=α . [Ans. 0.2106 mile]


11. The diameter of a wooden wheel is 50 inches. An iron rim of inner perimeter 156.1 inches has to be fitted on this. How much temperature of the rim has to be increased for this purpose? For iron, C06 /1012 −×=α . [Ans. 522.97 C0 ]

12. At 30 C0 , a metal ball has diameter 4.02 cm and a circular hole in a brass sheet has diameter 4 cm. How much temperature of the brass sheet has to be increased so that the ball at 30 C0 just passes through this hole? For brass, C06 /1018 −×=α . [Ans. 307.8 C0 ]

13. A steel spherical bob has to be inserted through a brass ring. The bob has diameter 25 cm and the ring has inner diameter 24.9 cm at 20 C0 . If the brass and the ring are heated together, find the temperature at which the bob just passes through the ring. For steel, C06 /1012 −×=α and for brass, C05 /102 −×=α . [Ans. 525 C0 ]

14. A steel made scale is errorless at 15 C0 . Find out the error when a distance is measured to be 200 cm at 30 C0 ? The coefficient of linear expansion for steel = 0.000012/ C0 . [Ans. 0.036 cm less]

15. The diameter of a brass disc is 10 cm at 30 C0 . What will be the increase in area at 70 C0 ? For brass, C06 /1018 −×=α . [Ans. 0.113 sq. cm]

16. There is a to and fro motion of a piston of diameter 4 inches. How much gap has to be kept between the cylinder and piston so that they just touch? The coefficients of linear expansion for the materials of piston and cylinder are F05 /1036.1 −× and

F05 /1062.0 −× . [Ans. 0.01296 inches] 17. The difference in length of two rods is 25 cm at all times irrespective of temperature.

If the coefficient of linear expansion for the rod A is C05 /1028.1 −× and that for the rod B is C05 /1092.1 −× , find the lengths of the rods at 0 C0 . [Ans. 75 cm, 50 cm]

18. A brass scale is correct at 20 C0 . A length is measured to be 50 cm at 40 C0 . What is the correct length? For brass, C06 /1018 −×=α . [Ans. 50.018 cm]

19. The length of a brass rod is 1.5 m at 50 C0 , measured by a steel scale which gives correct reading at 68 F0 . Find the correct length of the rod at 50 C0 . For steel,

C06 /102.11 −×=α [Ans. 1.5005 m] 20. What is the density of brass at 100 C0 if the density is 7.8 gm/cc at 0 C0 ? For brass,

C06 /1018 −×=α . [Ans. 7.76 gm/cc] 21. The density of glass is 2.6 gm/cc and 2.596 gm/cc at 10 C0 and 60 C0 , respectively.

What is the mean value of coefficient of linear expansion for glass in this temperature range? [Ans. 10.27 C06 /10−× ]

22. Each side of a square plate is 100 cm at 0 C0 and there is a circular hole of diameter 40 cm at the centre of the square. At what temperature, each side of the square plate will be 101 cm and what will be the diameter of the hole at that temperature? Given, α for the metal = C06 /105.12 −× [Ans. 800 C0 , 40.4 cm]

23. The length of a steel structure of a bridge is 0.5 km. The maximum and minimum temperatures can be 116 F0 and 44 F0 , respectively. What is the gap that has to be kept to allow thermal expansion? [For steel, C05 /101 −×=α ] [J.E.E. ‘92]

[Ans. 20 cm]


24. An isosceles triangle is formed by three brass rods. The base of the triangle is 3 inches and the height 12 inches. What will be change in base angles when the temperature rise is by 50 C0 ? Given, C06 /1019 −×=α . [Ans. no change]

Medium level problems: Marks: 4 1. A steel scale gives correct result at 10 C0 . What is the distance between two cm

marks on that scale at 30 C0 and at 60 F0 ? A brass rod is found to be 5 m measured by that scale. What will be the correct length of the rod at 10 C0 ? (For brass, C0/000018.0=α , for steel, C0/000012.0=α ) [Ans. 1.00024 cm, 1.000067 cm, 4.9994 m]

2. A pendulum clock keeps correct time at 25 C0 . How much fast or slow the clock will run when the temperature is 0 C0 ? The coefficient of linear expansion for brass is

C06 /1019 −× . [Ans. 20.52 second fast] 3. A pendulum clock keeps correct time at 20 C0 . Sometimes, the clock goes 10 seconds

fast in a day. What is the temperature of the day? For the material of the pendulum bar, =α C06 /1018 −× . [Ans. 7.3 C0 ]

4. A pendulum clock completes 86405 half oscillations. The clock goes 15 s slow at the end of the day although it runs fast at the beginning of the day. Find the maximum and minimum temperature of the day. For the material of the pendulum bar, =α C06 /1016 −× . [Ans. 28.94 C0 ]

5. There are three iron rods and two brass rods in a compensated pendulum. The coefficients of linear expansion for iron and brass are C06 /1012 −× and C06 /1019 −× , respectively and the length of each iron rod is 50 cm. Find the average length of brass rods. [Ans. 63.16 cm]

6. If the density of mercury is 13.56 gm/cc at 15 C0 , what is the mass of 600 cc mercury at 130 C0 ? What will be the volume of 600 gm mercury at that temperature? [For mercury, C05 /1018 −×=γ ] [Ans. 7.968 kg, 45.18 cc]

7. The difference between the lengths of an iron and a copper rod is 2 cm at 50 C0 and at 450 C0 . What are the lengths of the rods at 0 C0 ? [ CFE

06 /1012 −×=α and CCu

06 /1017 −×=α ] [Ans. 6.796 cm and 4.796 cm] 8. Volume can be measured accurately at 10 C0 by a measuring cylinder made of glass.

Find the percentage error when this is used at 70 C0 . The coefficient of linear expansion for glass = C06 /109 −× . [Ans. 0.162%]

9. Prove that if the temperature change is T∆ , the change in density is T∆−=∆ αρρ 3 . =α Coefficient of linear expansion.

10. Prove that the expansion in volume V due to rise in temperature, T∆ is TVV ∆⋅=∆ α3 , where =α coefficient of linear expansion.


Advanced level problems: 1. The area of cross-section of a 25 cm long steel rod is 0.8 sq. cm. What is the tension

required to increase an amount of length which can be found from increasing the temperature of the rod by 10 C0 ? For steel, 12102×=Y dyne/cm 2 and C05 /10−=α .

[Ans. 71016× dyne] 2. The temperature of a steel rod is increased by 15 C0 . What is the thermal stress

required to stop the expansion along the length? For steel, 12102×=Y dyne/cm 2 and C05 /102.1 −×=α . [Ans. 71036× dyne/cm 2 ]

3. Four rods of equal length are used to form a square ABCD. The coefficient of linear expansion for AB and CD rods is α and the coefficient of linear expansion for the other rod is β . Prove that , when the temperature of the rods is increased by 0t , the

angles between the diagonals are changed by .)(21 tβα −

4. The temperature of a brass weight is raised from 20 C0 to 100 C0 . The density of brass is 8.4 gm/cc at 20 C0 . How much will be the increase in volume due to the rise in temperature? [ C06 /1019 −×=α ] [Ans. 0.54 cc]

5. The internal perimeter of a steel wheel is 157 cm. This has to be fitted on a ring whose diameter is 50 cm. How much increase in temperature is required for this purpose? What will be force per unit area on the ring due to the wheel when cooled? For steel, 12101.2 ×=Y dyne/cm 2 and C06 /1012 −×=α .

[Ans. 50.03 C0 , 1.261 910× dyne.] 6. Two strips of different materials are riveted to form a single strip of width 0.1 cm.

What will be the radius of curvature when the temperature is increased by 100 C0 ? Given, α for the two metals, C06 /1018 −× and C06 /1012 −× . [Ans. 167.2 cm]

7. Two equal rods of copper and an aluminium rod are joined to form an isosceles triangle ABC. Another rod AD of some unknown material is joined from the vertex A to the mid point of base (made of aluminium). For a small change in temperature, the rods don’t bend. In this situation, find the coefficient of linear expansion for material of the rod AD. For copper, C06 /1016 −×=α ; for aluminum, C06 /1026 −×=α .

[Ans. C06 /107.12 −× ] 8. The temperature of a brass sphere whose radius is10 cm is increased at the rate of

2.5 C0 in a second. Find out the rate of increase of (i) diameter, (ii) surface area, (iii) volume. The coefficient of expansion for brass, C06 /1018 −×=α .

[Ans. (i) 0.0009 cm/s, (ii) 0.1131 cm 2 /s, (iii) 0.5655 cm 3 /s] 9. A metal rod weighs 46 gm in air. It weighs 30 gm when immersed in a liquid of

specific gravity 1.24 at 127 C0 . If the temperature of the liquid is 42 C0 , the metal rod weighs 30.5 gm. If the specific gravity of the liquid is 1.20 at 42 C0 , find the coefficient of linear expansion of the metal. [Ans. C05 /1031.2 −× ]

10. The lengths of two rods, made of brass and iron, are 50 cm and one end of each is joined with the other. On the other ends, a needle of diameter 1 mm is kept between them. An index attached with this needle indicates its rotation. When the rods are


heated, the index rotates by 10 0 . Calculate the temperature change. For brass, C05 /108.1 −×=α ; for iron, C05 /102.1 −×=α . [Ans. 29.1 C0 ]

11. A thin rod of negligible mass and of area of cross-section, 26104 m−× is hanging vertically from the top. The length of the rod is 0.5 m at 100 C0 . The rod is now cooled down to 0 C0 . But a mass is attached to resist the length contraction due to cooling. Find the (i) mass that is attached, (ii) the energy stored in the rod. Give, the coefficient of linear expansion for the material of the rod, K/10 5−=α and 1110=Y N/m 2 ; 10=g m/s 2 . [I.I.T. ‘97]

[Ans. (i) 400 kg, (ii) 2 joule] =============================================================

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Thermal Expansion