Question

Coefficient of linear expansion of brass and steel rods are ${\alpha }_1$and ${\alpha }_2$ . Length of brass and steel rods are $l_1$ and $l_2$ respectively. If $l_2–l_1$ is maintained same at all temperatures, then find the ratio of ${\alpha }_1$ to ${\alpha }_2$.

• A. $\frac{l_2}{l_1}$
• B. $\frac{l_1}{l_2}$
• C. $\frac{{\alpha }_2}{l_1}$
• D. $\frac{l_2}{{\alpha }_1}$

Answer 1

The correct option is A $\frac{l_2}{l_1}$

Let $l_1$ and $l_2$ be length of the rods at temperature $T$.
For any temperature $t$, length of rods can be given as
$l_1{}^{\prime }=l_1(1+{\alpha }_1\Delta T)$
$l_2{}^{\prime }=l_2(1+{\alpha }_2\Delta T)$
Given:
$l_2{}^{\prime }-l_1{}^{\prime }=$ constant
$l_2(1+{\alpha }_2\Delta T)-l_1(1+{\alpha }_1\Delta T)=\text{constant}$
$l_2(1+{\alpha }_2\times (t-T\left)\right)-l_1(1+{\alpha }_1\times (t-T\left)\right)=\text{constant}$
Differetiating both side w.r.t. $t$
$l_2{\alpha }_2=l_1{\alpha }_1$
i.e $\frac{{\alpha }_1}{{\alpha }_2}=\frac{l_2}{l_1}$