First slide

Thermal expansion

Question

A steel scale measures the length of a copper wire as 80.0 cm, when both are at 20°C, the calibration temperature for the scale. What would the scale read for the length of the wire when both are at 40°C ? (Given : $α$ for steel =11 x10^-6 per °C and $α$ for Cu = 17 x 10^-6 per °C)

Moderate

A

80.0096 cm
B

80.0272 cm
C

1 cm
D

25.2 cm

Solution

We know that
$\begin{array}{l} 4 = l_{0} (1 + αt) \\ = 1 [1 + (11 \times 10^{- 6}) \times (40^{\circ} - 20^{\circ})] \\ = 1 \cdot 00022 cm \end{array}$
Now the length of copper rod at 40°C
$\begin{array}{l} l_{1}^{'} = l_{0}^{'} (1 + α^{'} t) \\ = 80 [1 + (17 \times 10^{- 6}) \times (40 - 20)] \\ = 80 \cdot 0272 cm \end{array}$
Number of cms observed on the scale
$= \frac{80 \cdot 0272}{1 \cdot 00022} = 80 \cdot 0096$

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