Question

A metre scale made of steel is calibrated at 20°C to give correct reading. Find the distance between the 50 cm mark and the 51 cm mark if the scale is used at 10°C. Coefficient of linear expansion of steel is 1.1 × 10^–5 °C^–1.

Answer 1

Given:
Temperature at which the steel metre scale is calibrated, t₁ = 20^oC
Temperature at which the scale is used, t₂ = 10^oC
So, the change in temperature, $\Delta$t = (20^o$-$10^o) C
The distance to be measured by the metre scale, L_o = (51$-$50) = 1 cm = 0.01 m
Coefficient of linear expansion of steel, ${\alpha }_{steel}$ = 1.1 × 10^–5 °C^–1
Let the new length measured by the scale due to expansion of steel be L_​2, Change in length is given by,
$$\begin{gathered} ∆L=L_1{\alpha }_{steel}\Delta t \\ \Rightarrow ∆L=1\times 1.1\times {10}^{–5}\times 10 \\ \Rightarrow ∆L=0.00011\mathrm{cm} \end{gathered}$$
As the temperature is decreasing, therefore length will decrease by $∆L$.
Therefore, ​the new length measured by the scale due to expansion of steel (L₂) will be,
L₂ = 1 cm $-$ 0.00011 cm = 0.99989 cm