Question

At 50°C, 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°C is

(Coefficient of linear expansion of brass $= 2.0 \times 1{0^{- 5}}^{o}C^{- 1}$ coefficient of linear expansion of steel $= 1.2 \times 1{0^{- 5}}^{o}C^{- 1}$)

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

Answer 1

Answer: (C)

0.32 cm

Answer 2

Changes in length of the brass rod is

$\Delta L_{B} = \alpha_{b}L_{b}\Delta T$

$2.0 \times 10^{- 5}{^\circ}C^{- 1} \times 50cm \times (250{^\circ}C - 50{^\circ}C) = 0.2cm$

Change in length of the steel rod is

$\Delta L_{s} = \alpha_{s}L_{s}\Delta T$

$1.2 \times 10^{- 5}{^\circ}C^{- 1} \times 50cm \times (250{^\circ}C \times 50{^\circ}C) = 0.12cm$

$\therefore$Changes in length of the combined rod

$= \Delta L_{b} + \Delta L_{s} = 0.2cm + 0.12cm = 0.32cm$