An iron wire AB has diameter of 0 .6 mm and length 3 m at 0 ∘ C .The wire is now stretched between the opposite walls of a brass casing at 0 ∘ C . What is the extra tension that will be set up in the wire when the temperature of the system is raised to 40 ∘ C ? Given α b r a s s = 18 × 10 – 6 / K α i r o n = 12 × 10 – 6 / K Y i r o n = 21 × 10 10 N/m 2

Step-by-step solution

Increase in length of the iron wire = L α Δ t = L × 12 × 10 - 6 × 40 = 48L × 10 - 5 m Increase in length of the brass tube = L α Δ t = L × 18 × 10 - 6 × 40 = 72L × 10 - 5 m Since the increase in length of the brass tube is greater than that of the iron wire, the latter will be tightened. Let T be the tension in the wire. Then Stress = T ⁡ π r ⁡ 2 = T ⁡ π 3 × 1 0 - 4 2 = T ⁡ 9 π × 1 0 - 8 Strain = 7 2 L ⁡ - 4 8 L ⁡ × 1 0 - 5 L ⁡ = 2 4 × 1 0 - 5 Since stress = Y × strain , T ⁡ 9 π × 1 0 - 8 = 2 1 × 1 0 1 0 × 2 4 × 1