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Initial temperature, T1 = 27°CLength of the brass wire at T1, l = 1.8 mFinal temperature, T2 = –39°CDiameter of the wire, d = 2.0 mm = 2 × 10–3 mTension developed in the wire = FCoefficient of linear expansion of brass, α = 2.0 × 10–5 K–1Young’s modulus of brass, Y = 0.91 × 1011 PaYoung’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 wireA = 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) = FL / [ π(d/2)2X Y ]F = α(T2 – T1)πY(d/2)2F = 2 × 10-5 × (-39-27) × 3.14 × 0.91 × 1011 × (2 × 10-3 / 2 )2= -3.8 × 102 N(The negative sign indicates that the tension is directed inward.)Hence, the tension developed in the wire is 3.8 ×102 N.
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