# 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

**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}

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