Question

An equilateral triangle ABC is formed by two Cu rods AB and BC and one A1 rod. It is heated in such a way that temperature of each rod increases by $\Delta T$The change in the angle ABC is

(Here, coefficient of linear expansion for Cu is$\alpha_{1}$ coefficient of linear expansion for A1 is$\alpha_{2}$ )

• A. $\frac{\sqrt{3}}{2}(\alpha_{2} - \alpha_{1})\Delta T$
• B. $\frac{2}{\sqrt{3}}(\alpha_{2} - \alpha_{1})\Delta T$
• C. $\frac{1}{\sqrt{3}}(\alpha_{2} - \alpha_{1})\Delta T$
• D. $\frac{1}{2}(\alpha_{2} - \alpha_{1})\Delta T$

Answer 1

Answer: (B)

$\frac{2}{\sqrt{3}}(\alpha_{2} - \alpha_{1})\Delta T$

Answer 2

: The situation is as shown in the figure.

Let $AB = L_{1},AC = L_{2}andBC = L_{3}$

According to cosine formula

$L_{2}^{2} = L_{1}^{2} + L_{3}^{2} - 2L_{1}L_{3}\cos\theta$

$\cos\theta = \frac{L_{1}^{2} + L_{3}^{2} - L_{2}^{2}}{2L_{1}L_{3}}$

$2L_{1}L_{3}\cos\theta = L_{1}^{2} + L_{3}^{2} - L_{2}^{2}$

Differentiating both sides, we get

$2\lbrack L_{1}dL_{3} + L_{3}dL_{1}\rbrack\cos\theta - 2L_{1}L_{3}\sin\theta d\theta$

=$2L_{1}dL_{1} + 2L_{3}dL_{3} - 2L_{2}dL_{2}$

$(L_{1}dL_{3} + L_{3}d_{1})Cos\theta - L_{1}L_{3}\sin\theta d\theta$

$= L_{1}dl_{1} + L_{3}dL_{3} - L_{2}dL_{2}$ ……(i)

Here, $L_{1} = L_{2} = L_{3} = L$

$\therefore dL_{1} = L\alpha_{1}\Delta T$

$dL_{2} = L\alpha_{2}\Delta T$

$dL_{3} = L\alpha_{1}\Delta T$

Substituting these values in Eq. (i) we get

$\lbrack L^{2}\alpha_{1}\Delta T + L^{2}\alpha_{1}\Delta T\rbrack\cos\theta - L^{2}\sin\theta d\theta$

$= L^{2}\alpha_{1}\Delta T + L^{2}\alpha_{1}\Delta T - L^{2}\alpha_{2}\Delta T\rbrack$

}{= 2\alpha_{1}\Delta T(\cos\theta - 1) + \alpha_{2}\Delta T}$$ Using $\theta = 60^{0},$ we get $$\frac{\sqrt{3}}{2}d\theta = 2\alpha_{1}\Delta T\left( \frac{1}{2} - 1 \right) + \alpha_{2}\Delta T = - \alpha_{1}\Delta T + \alpha_{2}\Delta T$$ $$d\theta = \frac{2}{\sqrt{3}}(\alpha_{2} - \alpha_{1})\Delta T$$