# Numerical Problems on Linear Expansion

#### 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? a 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:

l2 = l1 (1 + α (t2 – t1))

∴  l2 = 50 x (1 + 18 x 10-6  (25 – 0))

∴  l2 = 50 x (1 + 18 x 10-6  x 25)

∴  l2 = 50 x (1 + 450 x 10-6)

∴  l2 = 50 x (1 + 0. 000450)

∴  l2 = 50 x 1. 000450

∴  l2 = 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 = l2 = 64.576 cm.
• To Find: coefficient of linear expansion = α =?
• Solution:

l2 = l1 (1 + α (t2 – t1))

∴  l2 = l1 + α l1 (t2 – t1)

∴  l2 – l1 = 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: 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, a = 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:

l2 = l1 (1 + α (t2 – t1))

∴  l2 = 60 x (1 + 1.5 x 10-5  (110 – 10))

∴  l2 = 60 x (1 + 1.5 x 10-5  x 100)

∴  l2 = 60 x (1 + 1.5 x 10-3)

∴  l2 = 60 x (1 + 0. 0015)

∴  l2 = 60 x 1. 0015

∴  l2 = 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 = a = 2 x 10-5 /oC.
• Given: Difference in lengths = l2 – l1 = 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 = lo = ?
• Solution:

l1 = lo (1 + αt1)

∴  l1 = lo (1 + 2 x 10-5 x 10)

∴  l1 = lo (1 + 2 x 10-5)   ……….  (1)

l2 = lo (1 + αt2)

∴  l2 = lo (1 + 2 x 10-5 x 30)

∴  l2 = lo (1 + 60 x 10-5)   ……….  (2)

From equations (1) and (2)

l2 – l1 = lo (1 + 60 x 10-5) – lo (1 + 20 x 10-5)

∴  0.04  = lo (1 + 60 x 10-5 – 1 – 20 x 10-5)

∴  0.04  = lo x  40 x 10-5

∴  lo = 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 = l2 = 300.54 cm, Initial temperature = t1 = 100 oC, final temperature = t2 = 150 oC,
• To find: Initial length = lo = ? and coefficient of linear expansion = α =?
• Solution:

l1 = lo (1 + αt1)

∴  300.36 = lo (1 + 100α)   ……….  (1)

∴  l2 = lo (1 + αt2)

∴  300.54 = lo (1 + 150α)   ……….  (2)

Dividing equation (2) by (1)

∴  300.54/300.36 = lo (1 + 150α)/lo (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 = lo (1 + 100α)

∴  300.36 = lo (1 + 100 x 1.2 x 10-5)

∴  300.36 = lo (1 + 1.2 x 10-3)

∴  300.36 = lo (1 + 0.0012)

∴  300.36 = lo (1 .0012)

∴  lo = 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 = l2 – l1 =?
• 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

l2 = l1 (1 + α (t2 – t1))

∴  l2 = l1 + α l1 (t2 – t1)

∴  l2 – l1 = α l1 (t2 – t1)

∴  l2 – l1 = 1 x 10-5  x 1 x  (55 – 25)

∴  l2 – l1 = 10-5 x  30

∴  l2 – l1 = 3 x 10-4 m

∴  l2 – l1 = 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))

∴  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

∴  l A1  = 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 +  l A1α 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/ α= 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

∴ l B1 = 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 – l A1) + (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)

∴ l A1 αA=  l B1α B

∴ l A1 /l B1 =  α B/ α= 18 x 10-6/12 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) = 5

∴ ( (3/2)l B1 – l B1) = 5

∴ (1/2)l B1 = 5

∴ l B1 = 10 cm

l A1  = (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