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 $0^{\circ }C$ and their difference in length ($l_2-l_1$) will remain the same at all temperatures if

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

Answer 1

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

at any temperature, length of brass = $L_1(1+{\alpha }_{1}T$
and length of steel = $L_2(1+{\alpha }_{2}T$
now difference of lengths = $L_1(1+{\alpha }_{1}T-L_2(1+{\alpha }_{2}T$
= $(L_1-L_2)+T(L_1{\alpha }_1-L_2{\alpha }_2)$
this is independent of temperature if coefficient of $T=0$
that means $L_1{\alpha }_1=L_2{\alpha }_2$