Question

At $50^{\circ }C$, a brass rod has a length $50cm$ and a diameter $2mm$. It is joined to a steel rod of the same length and diameter at the same temperature. The change in the length of the composite rod when it is heated to $250^{\circ }C$ is (Coefficient of linear expansion of brass $=2.0\times 10^{-5}{}^{\circ }{C}^{-1}$, coefficient of linear expansion of steel $=1.2\times 10^{-5}{C}^{-1}$).

• A. $0.28cm$
• B. $0.30cm$
• C. $0.32cm$
• D. $0.34cm$

Answer 1

Answer: (C)

Change in length of the brass rod is
$\Delta L_b={\alpha }_b{L}_b\Delta T$
$=2.0\times 10^{-5}{}^{\circ }{C}^{-1}\times 50cm\times \left(250{}^{\circ }C-50{}^{\circ }C\right)$
$=0.2cm$
Change in length of the steel rod is
$\Delta L_s={\alpha }_s{L}_s\Delta T$
$=1.2\times 10^{-5}{}^{\circ }{C}^{-1}\times 50cm\times \left(250{}^{\circ }C-50{}^{\circ }C\right)$
$=0.12cm$
$\therefore$ Change in length of the combined rod
$=\Delta L_b+\Delta L_s=0.2cm+0.12cm=0.32cm$