Science > Physics > Expansion of Solids > Numerical Problems on Linear Expansion of Solids
In this article, we shall study to solve problems to calculate the coefficient of linear expansion of solid, final length and the change in temperature.
Example – 01:
A metal scale is graduated at 0 oC. What would be the true length of an object which when measured with the scale at 25 oC, reads 50 cm? α for metal is 18 x 10-6 /oC.
Given: Initial temperature = t1 = 0 oC, final temperature = t2 = 25 oC, measured length = l1 = 50 cm, coefficient of linear expansion = α = 18 x 10-6 /oC.
To Find: Actual length = l2 =?
Solution:
l 2 = l 1 (1 + α (t2 – t1))
∴ l 2 = 50 x (1 + 18 x 10-6 (25 – 0))
∴ l 2 = 50 x (1 + 18 x 10-6 x 25)
∴ l 2 = 50 x (1 + 450 x 10-6)
∴ l 2 = 50 x (1 + 0. 000450)
∴ l 2 = 50 x 1. 000450
∴ l 2 = 50.225 cm
Ans: the true length of object is 50.225 cm
Example – 02:
A metal rod is 64.522 cm long at 12 oC and 64.576 cm at 90 oC. Find the coefficient of linear expansion of its material.
Given: Initial temperature = t1 = 12 oC, final temperature = t2 = 90 oC, initial length = l1 = 64.522 cm, final length = l 2 = 64.576 cm.
To Find: coefficient of linear expansion = α =?
Solution:
l 2 = l 1 (1 + α (t2 – t1))
∴ l 2 = l 1 + α l1 (t2 – t1)
∴ l 2 – l 1 = α l1 (t2 – t1)
∴ 64.576 – 64.522 = α x 64.522 x (90 – 12)
∴ 0.054 = α x 64.522 x 78
∴ α = (0.054)/(64.522 x 78) = 1.073 x 10-5 /oC.
Ans: The coefficient of linear expansion is 1.073 x 10-5 /oC.
Example – 03:
A metal bar measures 60 cm at 10 oC. What would be its length at 110 oC, α = 1.5 x 10-5 /oC.
Given: Initial temperature = t1 = 10 oC, final temperature = t2 = 110 oC, initial length = l1 = 60 cm, coefficient of linear expansion = α = 1.5 x 10-5 /oC.
To Find: final length = l2 =?
Solution:
l 2 = l 1 (1 + α (t2 – t1))
∴ l 2 = 60 x (1 + 1.5 x 10-5 (110 – 10))
∴ l 2 = 60 x (1 + 1.5 x 10-5 x 100)
∴ l 2 = 60 x (1 + 1.5 x 10-3)
∴ l 2 = 60 x (1 + 0. 0015)
∴ l 2 = 60 x 1. 0015
∴ l 2 = 60.09 cm
Ans: the length at 110 oC will be 60.09 cm
Example – 04:
A rod is found to be 0.04 cm longer at 30 oC than it is at 10 oC. Calculate its length at 0 oC if coefficient of linear expansion = α = 2 x 10-5 /oC.
Given: Difference in lengths = l2 – l 1 = 0.04 cm, Initial temperature = t1 = 10 oC, final temperature = t2 = 30 oC, coefficient of linear expansion = α = 2 x 10-5 /oC.
To find: Initial length = l o = ?
Solution:
l 1 = l o (1 + αt1)
∴ l 1 = l o (1 + 2 x 10-5 x 10)
∴ l 1 = l o (1 + 2 x 10-5) ………. (1)
Now, l 2 = l o (1 + αt2)
∴ l 2 = l o (1 + 2 x 10-5 x 30)
∴ l 2 = ll o (1 + 60 x 10-5) ………. (2)
From equations (1) and (2)
l 2 – l 1 = l o (1 + 60 x 10-5) – l o (1 + 20 x 10-5)
∴ 0.04 = l o (1 + 60 x 10-5 – 1 – 20 x 10-5)
∴ 0.04 = lo x 40 x 10-5
∴ l o = 0.04 / (40 x 10-5) = 100 cm
Ans: The length of rod at 0 oC is 100 cm
Example – 05:
The length of iron rod at 100 oC is 300.36 cm and at 150 oC is 300.54 cm. Calculate its length at 0 oC and coefficient of linear expansion of iron.
Given: initial length = l1 = 300.36 cm, final length = l 2 = 300.54 cm, Initial temperature = t1 = 100 oC, final temperature = t2 = 150 oC,
To find: Initial length = l o = ? and coefficient of linear expansion = α =?
Solution:
l 1 = l o (1 + αt1)
∴ 300.36 = l o (1 + 100α) ………. (1)
l 2 = l o (1 + αt2)
∴ 300.54 = l o (1 + 150α) ………. (2)
Dividing equation (2) by (1)
∴ 300.54/300.36 = l o (1 + 150α)/ l o (1 + 100α)
∴ 1.0006 = (1 + 150α)/(1 + 100α)
∴ 1.0006(1 + 100α) = (1 + 150α)
∴ 1.0006 + 100.06α = 1 + 150α
∴ 1.0006 – 1 = 150α – 100.06α
∴ 0.0006 = 49.94α
∴ α = 0.0006/49.94
∴ α = 1.2 x 10-5 /oC
From equation (1)
300.36 = l o (1 + 100α)
∴ 300.36 = l o (1 + 100 x 1.2 x 10-5)
∴ 300.36 = l o (1 + 1.2 x 10-3)
∴ 300.36 = l o (1 + 0.0012)
∴ 300.36 = l o (1 .0012)
∴ l o = 300.36/1 .0012 = 300 cm
Ans: The length of rod at 0 oC is 300 cm and coefficient of linear expansion is 1.2 x 10-5 /oC
Example – 06:
By how much will a steel rod 1 m long expand when heated from 25 oC to 55 oC? The coefficient of volume expansion of steel is 3 x 10-5 /oC.
Given: Initial temperature = t1 = 25 oC, final temperature = t2 = 55 oC, initial length = l1 = 1m, coefficient of volume expansion = γ = 3 x 10-5 /oC.
To Find: Increase in length = l 2 – l 1 =?
Solution:
coefficient of volume expansion (γ) = 3 x coefficient of linear expansion (α)
3 x 10-5 = 3 x α
∴ α = 1 x 10-5 /oC
l 2 = l 1 (1 + α (t2 – t1))
∴ l 2 = l 1 + α l 1 (t2 – t1)
∴ l 2 – l 1 = α l 1 (t2 – t1)
∴ l 2 – l 1 = 1 x 10-5 x 1 x (55 – 25)
∴ l 2 – l 1 = 10-5 x 30
∴ l 2 – l 1 = 3 x 10-4 m
∴ l 2 – l 1 = 0.3 x 10-3 m = 0.3 mm
Ans: The rod will expand by 0.3 mm.
Example – 07:
A brass rod and an iron rod are each 1m in length at 0 oC. Find the difference in their lengths at 110 oC. α for brass is 19 x 10-6 /oC and α for iron is 10 x 10-6 /oC.
Given: Initial temperature = t1 = 0 oC, final temperature = t2 = 110 oC, initial length = lBr1 = lFe1 =1m, coefficient of linear expansion for brass = αBr = 19 x 10-6 /oC. Coefficient of linear expansion for iron = αFe = 10 x 10-6 /oC.
To Find: Difference in length = lBr2 – lFe2 =?
Solution:
For brass rod
lBr2 = l Br1 (1 + α Br (t2 -t1))
For iron rod
lFe2 = l Fe1 (1 + α Fe (t2 -t1)) ………. (2)
Subtracting equation (2) from (1)
∴ lBr2 – lFe2 = l Br1 (1 + α Br (t2 -t1)) – l Fe1 (1 + α Fe (t2 -t1))
∴ lBr2 – lFe2 = 1 (1 + 19 x 10-6 (110 – 10)) – 1 (1 + 19 x 10-6 (110 – 10))
∴ lBr2 – lFe2 = 1 + 19 x 10-6 x 100 – 1 – 10 x 10-6 x 100
∴ lBr2 – lFe2 = 9 x 10-6 x 100 = 900 x 10-6 m
∴ lBr2 – lFe2 = 0.9 x 10-3 m = 0.9 mm
Ans: The difference in their lengths is 0.9 mm
Example – 08:
A brass rod and an iron rod are each 1m in length at 20 oC. At what temperature the difference in their lengths is 1.4 mm. α for brass is 18.92 x 10-6 /oC and α for iron is 11.92 x 10-6 /oC.
Given: Initial temperature = t1 = 20 oC, initial length = lBr1 = lFe1 =1m, coefficient of linear expansion for brass = αBr = 18.92 x 10-6 /oC, coefficient of linear expansion for iron = αFe = 11.92 x 10-6 /oC. Difference in length = lBr2 – lFe2 = 1.4 mm = 1.4 x 10-3 m.
To Find: final temperature = t2 =?
Solution:
For brass rod
lBr2 = l Br1 (1 + α Br (t2 -t1))
For iron rod
lFe2 = l Fe1 (1 + α Fe (t2 -t1)) ………. (2)
Subtracting equation (2) from (1)
∴ lBr2 – lFe2 = l Br1 (1 + α Br (t2 -t1)) – l Fe1 (1 + αFe (t2 -t1))
∴ 1.4 x 10-3 = 1 (1 + 18.92 x 10-6 (t2 -20)) – 1 (1 + 11.92 x 10-6 (t2 -20))
∴ 1.4 x 10-3 = 1 + 18.92 x 10-6 (t2 -20) – 1 – 11.92 x 10-6 (t2 -20)
∴ 1.4 x 10-3 = 7 x 10-6 (t2 -20)
∴ (t2 -20) =1.4 x 10-3/7 x 10-6 = 200
∴ t2 = 200 + 20 = 220 oC
Ans: At 220 oC the difference in their lengths is 1.4 mm.
Example – 09:
A rod A and a rod B are of equal length at 0 oC. If at 100 oC they differ by 1mm find their lengths at 0 oc, αA = 8 x 10-6 /oC, αB = 12 x 10-6 /oC.
Given: Initial temperature = t1 = 0 oC, final temperature = t2 = 100 oC, initial length = lA1 = lB1, coefficient of linear expansion for rod A = αA = 8 x 10-6 /oC, coefficient of linear expansion for rod B = αB = 12 x 10-6 /oC. Difference in length = lB2 – lA2 = 1 mm = 1 x 10-3 m.
To Find: Initial lengths of rod at 0 oC
Solution:
For rod A
lA2 = l A1 (1 + α A (t2 -t1))
For rod B
lB2 = l B1 (1 + α B (t2 -t1)) ………. (2)
Subtracting equation (2) from (1)
∴ lB2 – lA2 = l B1 (1 + α B (t2 -t1)) – l A1 (1 + αA (t2 -t1))
∴ 1 x 10-3 = l B1 (1 + 12 x 10-6 (100 – 0)) – l A1 (1 + 8 x 10-6 (100 – 0))
∴ 1 x 10-3 = l B1 (1 + 12 x 10-6 (100)) – l A1 (1 + 8 x 10-6 (100))
Now, given lA1 = lB1
∴ 1 x 10-3 = l A1 (1 + 12 x 10-4 – l A1 (1 + 8 x 10-4)
∴ 1 x 10-3 = l A1 (1 + 12 x 10-4 – 1 – 8 x 10-4)
∴ 1 x 10-3 = l A1 x 4 x 10-4
∴ l A1 = 1 x 10-3 / 4 x 10-4= 2.5 m
∴ lA1 = lB1 = 2.5 m
Ans: At 0 oC the lengths of the two rods is 2.5 m.
Example – 10:
A brass rod and an iron rod are 4 m and 4.01 m respectively at 20 oC. At what temperature the two rods have the same length. α for brass is 18.92 x 10-6 /oC and α for iron is 11.92 x 10-6 /oC.
Given: Initial temperature = t1 = 20 oC, initial length = lBr1 = lFe1 =1m, coefficient of linear expansion for brass = αBr = 18 x 10-6 /oC, coefficient of linear expansion for iron = αFe = 12 x 10-6 /oC, lBr2 = lFe2
To Find: final temperature = t2 =?
Solution:
For brass rod
lBr2 = l Br1 (1 + α Br (t2 -t1))
For iron rod
lFe2 = l Fe1 (1 + α Fe (t2 -t1)) ………. (2)
Given lBr2 = lFe2
∴ l Br1 (1 + α Br (t2 -t1)) = l Fe1 (1 + αFe (t2 -t1))
∴ 4 (1 + 18 x 10-6 (t2 -20)) = 4.01(1 + 12 x 10-6 (t2 -20))
∴ 4 + 72 x 10-6 (t2 -20) = 4.01 + 48.12 x 10-6 (t2 -20)
∴ 72 x 10-6 (t2 -20) – 48.12 x 10-6 (t2 -20) = 4.01 – 4
∴ 23.88 x 10-6(t2 -20) = 0.01
∴ (t2 -20) = 0.01/(23.88 x 10-6)
∴ t2 – 20 = 418.8
t2 = 418.8 + 20 = 438.8 oC
Ans: At 438.8 oC the rods will have the same length
Example – 11:
The difference in lengths of rod A and a rod B is 60 cm at all temperatures. Find their lengths at 0 oC, αA = 18 x 10-6 /oC, αB = 27 x 10-6 /oC.
Given: Initial temperature = t1 = 0 oC, final temperature = t2 oC, coefficient of linear expansion for rod A = αA = 18 x 10-6 /oC, coefficient of linear expansion for rod B = αB = 27 x 10-6 /oC. Difference in length = lB2 – lA2 = 60 cm = 60 x 10-2 m, (t2 -t1) is same for both rod.
To Find:Initial lengths of rod at 0 oC
Solution:
For rod A
lA2 = l A1 (1 + α A (t2 -t1))
lA2 = l A1 + lA1α A (t2 -t1))
lA2 – l A1 = l A1α A (t2 -t1)) ………. (1)
For rod B
lB2 – l B1 = l B1α A (t2 -t1)) ………. (1)
As the difference between the two rods is always 60 cm their expansion should be equal
lA2 – l A1 = lB2 – l B1
l A1α A (t2 -t1)) = l B1α A (t2 -t1))
∴ l A1 αA= l B1α B
∴ l A1 /l B1 = α B/ αA = 27 x 10-6/18 x 10-6 = 3/2
∴ l A1 = (3/2) l B1
Length of rod A is greater than rod B
Now (l A1 – l B1) = 60
∴ ( (3/2) l B1 – l B1) = 60
∴ (1/2) l B1 = 60
∴ lB1 = 120 cm
l A1 = (3/2)l x 120 = 180 cm
Ans: At 0 oC the length of rod A is 180 cm and that of rod B is 120 cm
Example – 12:
The difference in lengths of rod A and a rod B is 5 cm at all temperatures. Find their lengths at 0 oC, αA = 12 x 10-6 /oC, αB = 18 x 10-6 /oC.
Given: Initial temperature = t1 = 0 oC, final temperature = t2 oC, coefficient of linear expansion for rod A = αA = 12 x 10-6 /oC, coefficient of linear expansion for rod B = αB = 18 x 10-6 /oC. Difference in length = lB2 – lA2 = 5 cm = 5 x 10-2 m, (t2 -t1) is same for both rod.
To Find:Initial lengths of rod at 0 oC
Solution:
Assuming length of rod B is greater than rod B
For rod A
lA2 = l A1 (1 + α A (t2 -t1))
For rod B
lB2 = l B1 (1 + α B (t2 -t1)) ………. (2)
Subtracting equation (2) from (1)
∴ lB2 – lA2 = l B1 (1 + α B (t2 -t1)) – l A1 (1 + αA (t2 -t1))
∴ lB2 – lA2 = l B1 + l B1α B (t2 -t1) – l A1 – l A1 αA (t2 -t1)
∴ lB2 – lA2 = (l B1 – lA1) + (l B1α B – l A1 αA)(t2 -t1)
∴ 60 x 10-2= 60 x 10-2 + (l B1α B – l A1 αA)(t2 -t1)
∴ 0= (l B1α B – l A1 αA)(t2 -t1)
∴ 0= (l B1α B – l A1 αA)
∴ lA1 αA= lB1α B
∴ lA1 /l B1 = α B/ αA = 18 x 10-6/12 x 10-6 = 3/2
∴ lA1 = (3/2)l B1
Length of rod A is greater than rod B
Now (l A1 – l B1) = 5
∴ ((3/2) l B1 – l B1) = 5
∴ (1/2) l B1 = 5
∴ lB1 = 10 cm
lA1 = (3/2)l x 10 = 15 cm
Ans: At 0 oC the length of rod A is 15 cm and that of rod B is 10 cm
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