Question

An isosceles triangle is formed with a thin rod of length $l_{1}$ and coefficient of linear expansion $\alpha_{1}$, as the base and two thin rods each of length $l_{2}$ and coefficient of linear expansion $\alpha_{2}$ as the two sides. The distance between the apex and the midpoint of the base remain unchanged as the temperature is varied. The ratio of lengths $\frac{l_{1}}{l_{2}}$ is

• A. $2 \sqrt{\frac{\alpha_{1}}{\alpha_{2}}}$
• B. $\sqrt{\frac{\alpha_{2}}{\alpha_{1}}}$
• C. $2 \sqrt{\frac{\alpha_{2}}{\alpha_{1}}}$
• D. $\frac{1}{2} \sqrt{\frac{\alpha_{2}}{\alpha_{1}}}$

Answer 1

Answer: (C)

The distance between the apex and the midpoint of the base, using Pythagoras theorem
$l=\sqrt{\left(l_{2}\right)^{2}-\left(\frac{l_{1}}{2}\right)^{2}}$
or $l^{2}=\left(l_{2}\right)^{2}-\left(\frac{l_{1}}{2}\right)^{2}$ ...(i)
Differentiating (i) w.r.t. temperature

$0=2 l_{2} \times \frac{d l_{2}}{d T}-2\left(\frac{l_{1}}{2}\right) \cdot \frac{1}{2} \times \frac{d l_{1}}{d T}$
$\frac{l_{1}}{2} \times l_{1} \alpha_{1}=2 l_{2} \times l_{2} \alpha_{2}$
$\left(\frac{l_{1}}{l_{2}}\right)^{2}=4 \frac{\alpha_{2}}{\alpha_{1}}$
$\Rightarrow \frac{l_{1}}{l_{2}} =2 \sqrt{\frac{\alpha_{2}}{\alpha_{1}}}$