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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.
Solution 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.
Solution 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`
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