Question

At $50^{\circ}C$, a brass rod has a length $50\, cm$ and a diameter $2\, mm$. 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. $0.28 \,cm$
• B. $0.30 \,cm$
• C. $0.32 \,cm$
• D. $0.34 \,cm$

Answer 1

Answer: (C)

Change in length of the brass rod is
$\Delta L_{b}=\alpha_{b}L_{b}\Delta T$
$=2.0 \times 10^{-5}\,{}^{\circ}C^{-1} \times50\,cm \times \left(250\,{}^{\circ}C-50\,{}^{\circ}C\right)$
$=0.2\,cm$
Change in length of the steel rod is
$\Delta L_{s}=\alpha_{s}L_{s}\Delta T$
$=1.2 \times 10^{-5}\,{}^{\circ}C^{-1} \times 50\,cm \times \left(250\,{}^{\circ}C-50\,{}^{\circ}C\right)$
$=0.12\,cm$
$\therefore $ Change in length of the combined rod
$= \Delta L_{b} + \Delta L_{s} = 0.2 \,cm + 0.12\, cm = 0.32\,cm$