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]

Answer 1

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)`