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 or Cu is α₁, coeff. of linear expansion for Al is α₁.

Answer 1

Due to temperature increase, length of each side will changes, so the angle corresponding to any vertex also changes.

Let us consider the following diagram.

Let l₁ = AB, l₂ = AC, l₃ = BC

cos θ = \(\frac{l^2_3+l^2_1-l^2_2}{2l_3l_1}\)

or 2l₃l₁ cos θ = \(l^2_3 + l^2_1 +l^2_2\)

differentiating the following equation

2(l₃dl₁ + l₁dl₃) cos θ − 2l₁l₃ sin θ dθ = 2l₃dl₃ + 2l₁dl₁ − 2l₂dl₂ [∵ ∆t = change in temperature]

Now,

dl₁ = l₁α₁∆t

dl₂ = l₂α₁∆t

dl₃ = l₃α₂∆t

and l₁ = l₂ = l₃ = l

(l²α₁∆t + l²α₁∆t) cos θ + l² sin θ dθ = l²α₁∆t + l²α₁∆t − l²α₂∆t

sin θ dθ = 2α₁∆t(1 − cos θ) − α₂∆t

Putting θ = 60°

dθ × sin θ° = 2α₁∆t(1 − cos 60°) − α₂∆t)

dθ × \(\frac{\sqrt {3}}{2}\) = 2α₁∆t × \(\frac{1}{2}\) − α₂∆t

dθ × \(\frac{\sqrt {3}}{2}\) = (α₁ − α₂)∆t

Changing in angle, dθ = \(\frac{(α_1 − α_2)Δt}{\frac{\sqrt{3}}{2}}\)

= \(\frac{2(α_1 − α_2)Δt}{\sqrt{3}}\)