### Q.9:- 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 × 10^{11} Pa.

**Answer:-**Initial temperature, *T*_{1} = 27°C

Length of the brass wire at *T*_{1}, *l* = 1.8 m

Final temperature, *T*_{2} = –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 × 10^{11} Pa

Young’s modulus is given by the relation:

γ = Stress / Strain = (*F*/*A*) / (∆*L*/*L*)

∆*L* = *F* X *L* / (*A* X *Y*) ……**(i)**

Where,

*F *= Tension developed in the wire

*A* = Area of cross-section of the wire.

Δ*L* = Change in the length, given by the relation:

Δ*L* = *α**L*(*T*_{2} – *T*_{1}) … **(ii)**

Equating equations **(i)** and **(ii)**, we get:

α*L*(*T*_{2} – *T*_{1}) = *FL* / [ π(*d*/2)^{2} *X Y* ]

*F* = α(T_{2} – T_{1})πY(d/2)^{2}

*F* = 2 × 10^{-5} × (-39-27) × 3.14 × 0.91 × 10^{11} × (2 × 10^{-3} / 2 )^{2}

= -3.8 × 10^{2} N

(The negative sign indicates that the tension is directed inward.)

Hence, the tension developed in the wire is 3.8 ×10^{2} N.