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  • An equilateral triangle ABC is formed by two Cu rods AB and BC and one Al rod.

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 the temperature of each rod increases by \DeltaT. Find the change in the angle ABC.

Answers (1)

\cos \theta = \frac{l_1^{2}+l_3^{2}-l_2^{2}}{2l_1l_3}

 

2l_1l_3\cos \theta =l_1^{2}+l_3^{2}-l_2^{2}

Differentiating on both sides of the equation,

2[d (l_1 l_3) \cos \theta + l_1 l_3 d (\cos \theta)] = 2 l_1 dl_1 + 2 l_3 dl_3 - 2 l_2 dl_2

(l_1 dl_3 + l_3 dl_1) \cos \theta - l_1 l_3 \cos \theta d\theta = l_1 dl_1 + l_3 dl_3 - l_2 dl_2 --------------(1)

dl_1 = l_1 \alpha_ 1 \Delta t

dl_2 = l_2 \alpha_ 2 \Delta t

dl_3 = l_3 \alpha _3 \Delta t

L1 = L2 = L3

\\dl_1=dl_3=l\alpha_1\Delta t

dl_2 = l\alpha _2 \Delta t

Substituting in 1, we get

\cos \theta (l^{ 2} \alpha_ 1 \Delta t + l^{2} \alpha_ 1 \Delta t) - l^{2} \sin \theta d\theta = l^{2} \alpha_1 \Delta t + l^{2} \alpha_1 \Delta t - l^{2} \alpha_2 \Delta t

2 \alpha _1 \Delta t \times \frac{1}{2}- 2 \alpha_ 1 \Delta t + \alpha_ 2 \Delta t = \frac{\sqrt{3}}{2}d \theta

d\theta = 2 (\alpha 2-\alpha 1) \frac{\Delta t }{\sqrt{3}}

Posted by

infoexpert24

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