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 ‘thermal stress’ developed at the junction? The ends of the rod are free to expand (Co-efficient of linear expansion of brass = 2.0 × 10–5 K–1, steel = 1.2 × 10–5 K–1).​

Answer 1

Answer:

heyyy!

Step-by-step explanation:

Initial temperature, T1 = 40°C

Final temperature, T2 = 250°C

Change in temperature, ΔT = T2 – T1 = 210°C

Length of the brass rod at T1, l1 = 50 cm

Diameter of the brass rod at T1, d1 = 3.0 mm

Length of the steel rod at T2, l2 = 50 cm

Diameter of the steel rod at T2, d2 = 3.0 mm

Coefficient of linear expansion of brass, α1 = 2.0 × 10–5K–1

Coefficient of linear expansion of steel, α2 = 1.2 × 10–5K–1

For the expansion in the brass rod, we have:

Change in length (∆l1) / Original length (l1) = α1ΔT

∴ ∆l1 = 50 × (2.1 × 10-5) × 210

= 0.2205 cm

For the expansion in the steel rod, we have:

Change in length (∆l2) / Original length (l2) = α2ΔT

∴ ∆l1 = 50 × (1.2 × 10-5) × 210

= 0.126 cm

Total change in the lengths of brass and steel,

Δl = Δl1 + Δl2

= 0.2205 + 0.126

= 0.346 cm

Total change in the length of the combined rod = 0.346 cm

wid luv ❤

Answer 2

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Refer the attachment.