Question

A brass rod of length 50 cm and diameter 3.0 mm is
joined to a steel rod of the same length and diameter.
What is the change in length of the combined rod at
250°C, if the original lengths are at 40.0°C? Is there a​

Answer 1

Here,

intial temperature ( T1) = 40°C

Final temperature ( T2) = 250°C

so, increase in temperature (∆T) = 250 - 40 =210°C

For brass rod,

Intial length ( Lo) = 50 cm

Let Final length = L

Coefficient of linear expansion (a) = 2 × 10^-5 /°C

L = Lo(1 + a∆T)

= 50( 1 + 2 × 10^-5 × 210)

=50( 1 + 420 × 10^-5)

= 50× 1.00420

= 50.21 cm

Increase in length ( ∆L) = L2 - L1

= 50.21 - 50

= 0.21 cm

For steel rod,

Intial length ( Lo') = 50 cm

Coefficient of linear expansion ( a') = 1.2 × 10^-5/K

Final length ( L') = Lo'(1 + a'∆T)

= 50( 1 + 1.2 × 10^-5 × 210)

= 50 × 1.00252

= 50.126 cm

Increase in length ( ∆L') = 50.126 cm - 50 cm = 0.126 cm

Total inches in length = ∆L + ∆L'

= 0.21 + 0.126

= 0.336 cm

Answer 2

Answer:

0.336 cm

Explanation:

Here,

intial temperature ( T1) = 40°C

Final temperature ( T2) = 250°C

so, increase in temperature (∆T) = 250 - 40 =210°C

For brass rod,

Intial length ( Lo) = 50 cm

Let Final length = L

Coefficient of linear expansion (a) = 2 × 10^-5 /°C

L = Lo(1 + a∆T)

= 50( 1 + 2 × 10^-5 × 210)

=50( 1 + 420 × 10^-5)

= 50× 1.00420

= 50.21 cm

Increase in length ( ∆L) = L2 - L1

= 50.21 - 50

= 0.21 cm

For steel rod,

Intial length ( Lo') = 50 cm

Coefficient of linear expansion ( a') = 1.2 × 10^-5/K

Final length ( L') = Lo'(1 + a'∆T)

= 50( 1 + 1.2 × 10^-5 × 210)

= 50 × 1.00252

= 50.126 cm

Increase in length ( ∆L') = 50.126 cm - 50 cm = 0.126 cm

Total inches in length = ∆L + ∆L'

= 0.21 + 0.126

= 0.336 cm