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Question
An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod. It is heated in such a way that temperature of each rod increases by ∆T. Find change in the angle ABC. [Coeff. of linear expansion for Cu is α1, Coeff. of linear expansion for Al is α2]
Solution
We can solve this by using trigonometry
`cos theta = (1_1^2 + l_3^2 - l_2^2)/(2l_1l_3)`
`2l_1 1_3 = 1_1^2 + l_3^2 - l_2^2`
Now, differentiating both sides:
`2[d(1_1 1_3) xx cos theta + 1_1 1_3 d(cos theta)] = 2l_1dl_1 + 2l_3dl_3 - 2l_2dl_2`
`2[(1_1dl_3 + 1_3dl_1) cos theta - 1_1 1_3 sin theta d theta] = 2(1_1dl_1 + l_3dl_3 - 1_2dl_2)`
`(1_1dl_3 + 1_3dl_1) cos theta - 1_1 1_3 sin theta d theta = 1_1dl_1 + 1_3dl_3 - 1_2dl_2`
`L_t = L_0(1 + αΔt)`
`L_t = L_0 = L_0 αΔt`
`ΔL = Lα xx Δt`
`dl_1 = 1_1α_1 Δt, dl_3 = 1_2α_1Δt`
`dl_2 = 1_2α_2 Δt`
`1_1 = 1_2 = 1_3 = 1`
`dl_1 = 1l_1Δt, dl_3 = 1α_1 Δt`
`dl_2 = 1α_2 Δt` .....(i)
Now, substitute in (i)
`cos theta (1^2 xx α_1 Δt + 1^2α_1 Δt) - 1^2 sin theta d theta = l^2α_1 Δt + l^2 α_1 Δt - 1^2 α_1 Δt`
`2l^2α_1 Δt cos theta - l^2 [sin theta xx d theta]`
= `1^2 [α_1 + α_1 - α_2]Δt`
`1^2 [2α_1 Δ cos 60^circ - sin 60^circ d theta] = 1^2 [α_1 - α_2] Δt`
`2α_1 Δt xx 1/2 - 2α_1 Δt + α_2 Δt = sqrt(3)/2 d theta`
`sqrt(3)/2 d theta = [α_1 - 2α_1 + α_2] Δt`
`d theta = (2(α_2 - α_1)Δt)/sqrt(3)` .....[∵ Δt = ΔT]
`d theta = (2(α_2 - α_1)ΔT)/sqrt(3)`
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