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

nitial 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 = "Stress"/"Strain" = (F/A)/((triangleL)/L)`

`triangleL = (FxxL)/(AxxY)`   ...(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:

`alphaL(T_2-T_2 )= (FL)/(pi(d/2)^2 xx y)`

`F = alpha(T_2 - T_1)piY (d/2)^2`

`F = 2xx 10^(-5)xx(-39-27) xx 3.14 xx 0.91xx 10^(11)xx((2xx10^(-3))/2)^2`

`= -3.8 xx 10^2 N`

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

Hence, the tension developed in the wire is 3.8 ×10² N.

Answer 2

Here l = 1.8 m

`triangle t  = (-39 - 27) ""^@C = -66 ""^@C`

`alpha = 2.0 xx 10^(-5) K^(-1)`

`Y  = 0.91 xx 10^11 Pa`

`A = (piD^2)/4 = 22/7 xx1/4 (2xx10^(-3))^2 m^2`

Now, `Y = "Fl"/(AtriangleL) => trianglel =  (Fl)/(AY) or lalphatrianglet =  (Fl)/(AY)`

or `F = -YA alpha triangle t`

or `F = -0.91 xx10^11 xx 22/7xx1/4(2xx10^(-3))^2 xx 2.0xx10^(-5) xx 66 N`

`= -3.77 xx 10^2 N`