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# How to solve questions on linear expansivity

Linear expansivity 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 linear expansity:

(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

Question 1

A wire of length 100.0m at 300C has a linear expansivity 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 2

Two metals P and Q are heated through the same temperature difference. If the ratio of the linear expansivities 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 =( linear expansivity * length)/increase in length

For metal P,

Qp = ( linear expansivity * length)p/increase in lengthp

For metal Q,

Qq = ( linear expansivity * lengthq)/increase in lengthq

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

increase in length of p:increase lengthq = ( linear expansivity * length)p / ( linear expansivity * lengthq)

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

A is the correct option

Question 3

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.
[Linear expansivity of the metal= 2.0 x 10-5K-1]

Solution

cubic expansivity = 3* linear expansivity = 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

Question 4

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* linear expansivity

0.0000063 = 3*linear expansivity

linear expansivity = 0.0000063/3 = 0.0000021

Area expansivity = 2*0.0000021 = 0.0000042

B is the correct option

Question 5

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 6

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? (Linear expansivity 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

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