# How to Solve Questions on Linear Expansivity

In this article, i will solve questions on linear expansivity, define area and cubic expansion.

**What is Linear expansion: **It is defined as the increase in length, of the unit length of the material for one degree temperature rise.

In order to solve any question on L.E:

(i) Write down the formula

(ii) Write down the data given

(iii) Understand what you are asked to find

(iv) Make proper substitution

(v) Evaluate

Table of Contents

**Advantages and disadvantages of Linear expansivity**

**Advantages **

- Used for making bimetallic strip which is used in thermostat
- Removal of tight glass stopper
- Red-hot rivets in ship building
- Expansion of metals is used in bimetallic thermometer

**Disadvantages**

- Expansion of railway lines
- Sagging of overhead wire
- Expansion of balance wheel or wrist-watch
- Cracking of glass cup when hot water is poured into the glass cup
- Expansion of metals or concrete bridges

## Area Expansivity

The increase in area of an object with temperature change is called superficial expansivity or area expansivity.

The superficial expansivity of a material is defined as,

Its value is twice that of L.E.

## Cubic Expansivity

Increasing temperature usually causes increase in volume for both solid and liquid materials.

The cubic expansivity of a solid is defined as

Its value is thrice that of L.E.

**Steps on how to solve questions on linear expansivity**

Question 1

What is meant by the statement: The linear expansivity of a solid is 1.0 x 10-5 k-1

Solution

The statement means A unit length of the solid will expand in length by a fraction 1.o x 10-5 of the original per kelvin rise in temperature.

Question 2

A wire of length 100.0m at 300C has a L.E of 2 x 10-5K-1. Calculate the length of the wire at a temperature of -100C {UTME 2013}

A. 99.92m B. 100.08m C. 100.04m D. 99.96m

Solution

l1 = new length

l0 = original length

Q = temperature change

0.00002 = change in length / 100(30-(-10))

0.00002 = change in length / 100*40

0.00002*4000 = change in length

change in length = 0.08

change in length = 100 – x

x = 100 – 0.08 = 99.92m

A is the correct option

Question 3

Two metals P and Q are heated through the same temperature difference. If the ratio of the L.E of P to Q is 2: 3 and the ratio of their lengths is 3:4 respectively, the ratio of the increase in lengths of P to Q is {UTME 2012}

A. 1 : 2 B. 2 : 1 C. 8 : 9 D. 9 : 8

Solution

Temperature difference =( L.E * length)/increase in length

For metal P,

Qp = (L.E * length)p/increase in lengthp

For metal Q,

Qq = ( L.E * lengthq)/increase in lengthq

( linear expansivity * length)p/increase in lengthp = ( L.E * lengthq)/increase in lengthq

increase in length of p:increase lengthq = ( L.E * length)p / ( L.E * lengthq)

increase in length of p:increase lengthq = 2/3 *3 /4 = 6 / 12 = 1:2

A is the correct option

Question 4

A metal of volume 40cm3 is heated from 300C to 900C, the increase in volume is {UTME 2011}

A. 0.40cm3 B. 0.14cm3 C 4.00cm3 D. 1.20cm3.

[L.E of the metal= 2.0 x 10-5K-1]

Solution

cubic expansivity = 3* L.E = 3 * 0.00002 = 0.00006

0.00006 = increase in volume / 40*(90 – 30)

0.00006 = increase in volume / 40*60

increase in volume = 0.00006*2400 = 0.144cm3

B is the correct answer

Question 5

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10-6K-1. The area expansivity is {UTME 2007}

A. 6.3 x 10-6K-1 B. 4.2 x 10-6K-1 C. 2.1 x 10-6K-1 D. 2.0 x 10-6K-1

Solution

cubic expansivity = 3* L.E

0.0000063 = 3* L.E

L.E = 0.0000063/3 = 0.0000021

Area expansivity = 2*0.0000021 = 0.0000042

B is the correct option

Question 6

Calculate the length which corresponds to a temperature of 20^0C if the ice and steam points of an ungraduated

thermometer are 400 mm apart

A. 80mm

B. 20mm

C. 30mm

D. 60mm

Solution

ice point = lower fixed point = 0^oC

steam point = upper fixed point = 100^oC

(100 -0)/(20 -0) = 400/(x – 0)

100/20 = 400 / x

20*400 = 100x

x = 8000/100

x = 80mm

A is the correct option

Question 7

A steel bridge is built in the summer when its temperature is 35.0°C. At the time of construction, its length is 80.00m. What is the length of the bridge on a cold winter day when its temperature is -12.0°C? (L.E of steel is 1.2 x10^-5)

Solution

initial length = 80 m

initial temperature = 35.0°C

final length = ?

final temperature = -12.0°C

Change in temperature = final – initial = -12-35 = -47.0°C

let change in length = x

o.oooo12 = x/80(-47)

x = 0.000012*80*(-47) = -0.04512 m

note

x = final length – initial length

-0.04512 = final length – 80

final length = 80 -0.04512 = 79.95488 m

The length of the bridge on a cold winter day = 79.95488 m

Question 8

A copper rod whose linear expansivity =1.70×10^-5°c is 20cm longer than an aluminum rod whose L.E =2.20×10^-5°c. How long should the copper rod be if the difference in their length is to be independent of temperature.

Solution

For copper,

L.E of copper = x

change in length of cooper = y

original length of copper = c

temperature change = t

x = y/(c*t)

For aluminium,

linear expansivity of aluminium = a

change in length of aluminium = b

original length of aluminium = d

temperature change = r

a = b/(d*r)

According to the question, copper is 20cm longer than an aluminum rod i.e. c = d + 0.2

y/t = x*c = x*(d+0.2) = 1.70×10^-5*(d+0.2)

b/r = a*d = 2.20×10^-5*d

note, y/t = b/r since the difference in length is independent on temperature

2.20×10^-5*d = 1.70×10^-5*(d+0.2)

d = 1.7/2.2 * (d+0.2)

d = 0.7727d + 0.1545

d-0.7727d = 0.15

0.2273d = 0.1545

d = 0.68m

c = d + 0.2 = 0.68 +0.2 = 0.88m

Question 9

A metal of volume 40cm^{3} is heated from 30^{0}C to 90^{0}C, the increase in volume is

A. 0.40cm^{3} B. 0.14cm^{3} C 4.00cm^{3} D. 1.20cm^{3}.

[L.E of the metal = 2.0 x 10^{-5}K^{-1}]

Answer

Volume expansivity = 3 x Linear expansivity = 3 x 2.0 x 10^{-5}K^{-1} = 6.0 x 10^{-5}K^{-1}

Increase in volume = x

V.E = (x) / (initial volume x temp rise)

6.0 x 10^{-5} = x / (40 x (90 – 30))

X = 6.0 x 10^{-5} x 40 x 60 = 14400 x 10^{-5} = 0.144 cm^{3}

Question 10

A blacksmith heated a metal whose cubic expansivity is 6.3 x 10^{-6}K^{-1}. The area expansivity is

A. 6.3 x 10^{-6}K^{-1} B. 4.2 x 10^{-6}K^{-1} C. 2.1 x 10^{-6}K^{-1} D. 2.0 x 10^{-6}K^{-1}

Answer

The cubic expansivity = 6.3 x 10^{-6}K-1

Recall that

cubic expansivity = 3 x L.E

Therefore,

L.E = (6.3 x 10-6)/3 = 2.1 x 10^{-6}

Area expansivity = 2 x L.E = 2 x 2.1 x 10^{-6} = 4.2 x 10^{-6} K^{-1}

Recommended: Solved questions on other topics in physics

The linear expansivity of a metal p is twice that of another metal Q. When the material are heated through the same temperature change, their increase in length is the same calculate the ratio of the original length of P to that of Q. Please help me solve this

liner expansivity of p = xp

linear expansivity of q =xq

xp = 2Xq

same temperature= y

same increas in length = l0

lp is the original length of p

lq is the original length of q

xp = l0/(lp.y)

xq = l0/(lq.y)

l0/(lp.y) = 2 (l0/(lq.y))

l0/(lp.y) = 2l0/(lq.y)

1/lp = 2/lq

lp/lq = 1/2

a copper rod whose linear expansivity =1.70×10^-5°c is 20cm longer than an aluminum rod whose linear expansivity=2.20×10^-5°c. How long should the copper rod be if the difference in their length is to be independent of temperature.

Check question 8, i have provided the solution to it.

A metal of length 17.01m is heated until it’s temperature rises by 70^oc if it’s new length is 18.02m .calculate it’s linear expansivity?

L.E = (18.02 – 17.01) / (17.01 x 70) = 1.01 /(17.01 x 70) = 8.48 x 10^-4

18.02_17.01/17.01*70=3.445*10^-4c

The ratio of a linear expansivity of a copper to that of iron is approximately 1:5 A specimen of iron and specimen of copper expand by the same amount per unit raise in temperature. What is the ratio of the length of iron to that of copper.

The equation to use to solve this linear expansivity question is xcu/xfe = Lfe/Lcu (linear expansivity of copper/L.E of iron = Initial length of fe/ initial length of copper)

xcu/xfe = 1:5 according to your question

i.e 1/5 = Lfe/Lcu, this means the ration of the length of iron to copper is still 1:5

Describe briefly how you can experimentally determine the linear expansivity A. of a solid, stating all precautions needed for accurate result. B what will be the new area of 400cm2 of a solid, if it’s temperature is raise by 10k?

An iron rob is 1.58m long at0c.what i must be the length of the brass rob at 0c if the difference between the length of the two robs is to remain the same at all temperature. Linear expansivity of iron= 1.2*10^_5k-1 linear expansivity of brass=1.9*10^_5k-1

if i get ur question right, what is means is that change in length of iron – change in length of brass = 0 at all temperature.

let assume the temperature change is 10.

1.2 x 10^-5 = (change in length of iron) / (1.58 x 10)

1.2 x 10^-4 x 1.58 = change in length of iron —- eq 1

1.9 x 10^-5 = (change in length of brass) / (y x 10) : y is the length of brass at 0 degree

1.9 x 10^-4 x y = change in length of brass ——eq 2

equating eq 1 to eq 2

1.2 x 10^-4 x 1.58 = 1.9 x 10^-4 x y

y = (1.2 x 10^-4 x 1.58) / (1.9 x 10^-4) = 0.998m

how do i attempt questions that has both linear expansivity and change in volume

Thanks Mr. Olajire…

This has been really helpful…

May your knowledge continue to increase.

You are welcome Joel

Thank you very much, may the Almighty continue to increase you in knowledge. This was 100% helpful. Found my question and it’s solution. ???

You are welcome

a metal of length 10cm is heated until it temperature rises by 80 degree Celsius,if the new length 20m calculate the linear expansivity?

The diameter of a rod is 5mm at 35°c what is the temperature when the diameter is 4.99mm (linear expansivity =1.2×10^-5 k-1)

In this case, use area expansivity to solve the question.

at 5mm diameter, the area = 550/28 mm2

at 4.99mm diameter, = 547.8022/28 mm2

area expansivity = 2 x L.E = 2.4 x 10^-5

A.E = (A1 – A0)/A0 (t-35)

An aluminum rod of length 1.8cm at 10`c is heated, produced a difference in length 40.007m. Calculate temperature to which it is heated ( linear expansivity of aluminium=0.000023 . show full working

Chidi, check the question again. An initial length of 1.8 cm to produce a difference in length of 40.007 m. This looks unreal.

A metal length of 16.01 is heated until it’s temperature rises by 60 degrees. If its new length is 15.05meters calculate its linear expansivity.

L.E = (15.05 – 16.01)/ 16.01 x (-60) = -0.96/(-960.6) = 9.993 x 10^-4

If the linear expansitivity of a rod is 4×10-5 per degree celcius,what will be the new length of the rod if itz heated from 15degree to 95 degree celcius from its original length of 20cm

The volume of a small piece of metal is 2.000 cm3 at 30 oC and 7.000 cm3 at 70 oC. Determine the cubic expansivity of the metal?

Cubical expansion = change in volume / original * change in temperature

C.E.=7cm^3 – 2cm^3 / 2cm^3 * {70″c – 30″c}

c.e.=5cm^3/2cm^3 * 40″c.

c.e.=500/80=6.25*10^(-2)

A cube with side 100cm at 0’c is heated to 100’c.if the side become 101cm long find linear expandability

A metal of length 50.0m at 20c, if the length of the metal increase to 50.0006m when heated to a temperature of 70c. Calculate the linear expansivity of the metal.

What is the name of the matal???

Pls i need calculation of beta and gramma

Calculate the cubic expansivity of glass

(linear expansivity = 8.5 x 10-6)

Please help me with this

cubic expansivity = 3 x linear expansivity. Therefore, you will multiply the value of L.E by 3 to get C.E

Pls Need help ASAP

an iron rod is 1.58m long at 0 degree celsius what must be the length of a brass rod at 0 degree celsius if the difference between the two rods is to remain the same at all temperature linear expansivity of iron is 0.00002 per kelvin linear expansivity of brass is 0.000019 per Kelvin

Let’s make y as the length of brass at 0 degree

let’s make x as differences in length

let’s use 20 degree at change in temp for both iron and brass rod

What this statement ” if the difference between the two rods is to remain the same at all temperature” means is that the change in change all all temperature is the same.

For iron,

0.00002 = x/(1.58)*20

x = 1.58*20*0.00002

for brass,

0.000019 = x/(y)*20

x = y*20*0.000019

therefore,

1.58*20*0.00002 = y*20*0.000019

y = 0.000632/0.00038 = 1.66m

1.66m is the length of brass at 0 degree.