Question

The variation of lengths of two metal rods $A$ and $B$ with change in temperature are as shown in the figure. The ratio of $\frac{\alpha_{B}}{\alpha_{A}}$ is
(Here $\alpha_{A}=$ Linear coefficient of thermal expansion of rod $A, \alpha_{B}=$ Linear coefficient of thermal expansion of rod $B$ )

Answer 1

$l=l_{0}(1+\alpha \Delta t)$
$\Rightarrow l-l_{0}=l_{0} \alpha \Delta T$
$\Rightarrow \frac{l_{0} \alpha_{A} \Delta T}{l_{0} \alpha_{B} \Delta T}=\frac{104-100}{106-100}=\frac{4}{6}$
$\Rightarrow \frac{\alpha_{A}}{\alpha_{B}}=\frac{2}{3}$