Question

At ${50}^{o}C$, 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 ${250}^{o}C$ is (coefficient of linear expansion of brass $=2.0\times {10}^{-5}/C$ , coefficient of linear expansion of steel $=1.2\times {10}^{-5}/C$ )

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

Answer 1

Answer: (C)

0.32 cm

change in length of the brass rod is
$\Delta L_b={\alpha }_b{L}_b\Delta T$
$=2\times {10}^{-5}\times 50\times (250-50)$
$=0.2$
Change in length of the steel rod is
$\Delta L_s={\alpha }_s{L}_s\Delta T$
$=1.2\times {10}^{-5}\times 50(250-50)$
$=0.12$
$herefore$ Change in length of the combined rod
$\Delta L_b+\Delta L_s=0.2+0.12=0.32$