First slide

Thermal expansion

Question

A brass rod of length 50 cm and diameter 3.0 cm is joined to a steel rod of the same length and diameter. What is the change in length of the combined rod at 250°C, if the original lengths are at 40.0°C?
(Coefficient of linear expansion of brass = 2.0 x $10^{- 5}$ /°C, steel = 1.2 x $10^{- 5}$ /°C)

Moderate

A

0.27 cm
B

0.34 cm
C

0.21 cm
D

0.18 cm

Solution

Change in temperature of each rod,
$∆$ T = 250 - 40 = 210°C
Clearly, change-in length of the brass bar
$∆ L_{b} = α_{0} L ∆ T$ = (2.0 x $10^{- 5}$ ) x 50 x 210 = 0.21 cm
and change in length of steel rod
$∆ L_{s} = α_{s} L ∆ T = (1.2 \times 10^{- 5}) \times 50 \times 210 = 0.126 cm$
Since the ends of the rod are free to expand, change in the length of the combined rod
$∆ L = ∆ L_{b} + ∆ L_{s} = 0.21 + 0.13 = 0.34 cm$

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