Question

The coefficient of linear expansion of brass and steel are
${\alpha }_1$ and ${\alpha }_2$. If we take a brass rod of length $l_1$ and steel rod of length $l_{2}at{0}^{\circ }C,$ their difference in length $(l_2-l_1)$will remain the same at a temperature if

• A. ${\alpha }_1{l}_2={\alpha }_2{l}_1$
• B. ${\alpha }_1{l}_2^2={\alpha }_2{l}_1^2$
• C. ${\alpha }_1^2{l}_1={\alpha }_2^2{l}_2$
• D. ${\alpha }_1{l}_1={\alpha }_2{l}_2$

Answer 1

The correct option is D ${\alpha }_1{l}_1={\alpha }_2{l}_2$

$d)L_2=l_2(l+{\alpha }_2△\theta )$ and $L_1=l_1(l+{\alpha }_1△\theta )$
$\Rightarrow (L_2-L_1)=(l_2-l_1)+△\theta (l_2{\alpha }_2-l_1{\alpha }_1)$
Now $(L_2-L1)=(l_2-l_1)$ so, $l_2{\alpha }_2-l_1{\alpha }_1=0$