Question

A brass wire 1.8 m long at 27 °C is held taut with little tension between two rigid supports. If the wire is cooled to a temperature of –39 °C, what is the tension developed in the wire, if its diameter is 2.0 mm ? Co-efficient of linear expansion of brass = 2.0 × 10¯⁵ K¯¹; Young’s modulus of brass = 0.91 × 10¹¹ Pa.

Answer 1

Given:

Length of the brass wire,

l=1.8 m

Change in temperature, ΔT=( −39−27 )=−66   o C

α=2.0× 10 −5   K -1

Young’s modulus, Y=0.91× 10 11  Pa

Diameter, D=2.0 mm=2× 10 −3  m

Area of the wire is,

A= π D 2 4 A= 22 7 × 1 4 × ( 2.0× 10 −3 ) 2 A=3.142× 10 −5   m 2

Now,

Y= Fl AΔl Δl= Fl AY αlΔT= Fl AY F=YAαΔT

Substitute the values

F=0.91× 10 11 ×3.142× 10 −5 ×2.0× 10 −5 ×( −66 ) F=−3774.2 N F≃−3 .78×10 2  N