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

Initial temperature, T₁ = 27°C 
Length of the brass wire at T1, l = 1.8 m
Final temperature, T₂ = –39°C
Diameter of the wire, d = 2.0 mm = 2 × 10^–3 m
Tension developed in the wire = F
Coefficient of linear expansion of brass, α = 2.0 × 10^–5 K^–1
Young’s modulus of brass, Y = 0.91 × 10¹¹ Pa
Young’s modulus is given by the relation:
Y = \(\frac{Sress}{Strain}\) = \(\frac{\frac{F}{A}}{\frac{\Delta L}{L}}\)
\(\Delta\)L = \(\frac{F \times L}{A \times Y}\) ...… (i)
Where,
F = Tension developed in the wire 
A = Area of cross-section of the wire.
ΔL = Change in the length, given by the relation:
ΔL = αL(T₂ – T₁) ...… (ii)
Equating equations (i) and (ii), we get:
αL(T₂-T₁) = \(\frac{FL}{\pi}\)(\(\frac{d}{2}\))²xY
F = α(T₂-T₁)πY(\(\frac{d}{2}\))²
F = 2 x 10^-5 x (-39-27) x 3.14 x 0.91 x 10¹¹ x (\(\frac{2 \times 10^{-3}}{2}\))
F = -3.8 x 10² N
(The negative sign indicates that the tension is directed inward.)
Hence, the tension developed in the wire is 3.8 ×10² N.