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

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

Disadvantages

1. Expansion of railway lines
2. Sagging of overhead wire
3. Expansion of balance wheel or wrist-watch
4. Cracking of glass cup when hot water is poured into the glass cup
5. 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 40cm³ is heated from 30⁰C to 90⁰C, the increase in volume is

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

Answer

Volume expansivity = 3 x Linear expansivity = 3 x 2.0 x 10^-5K^-1 = 6.0 x 10^-5K^-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 cm³

Question 10

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

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

Answer

The cubic expansivity = 6.3 x 10^-6K-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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