Question

A metre scale is made up of steel and measures correct length at 16°C. What will be the percentage error if this scale is used (a) on a summer day when the temperature is 46°C and (b) on a winter day when the temperature is 6°C? Coefficient of linear expansion of steel = 11 × 10^–6 °C^–1.

Answer 1

(a) Let the correct length measured by a metre scale made up of steel 16 °C be L.
Initial temperature, t₁ = 16 °C
Temperature on a hot summer day, t₂ = 46 °C
​So, change in temperature, Δθ = t₂ $-$ t₁ = 30 °C
Coefficient of linear expansion of steel, $\alpha$ = 1.1 × 10^–5 °C^​-1
Therefore, change in length,
ΔL = L αΔθ = L × 1.1 × 10^–5 × 30
$$\begin{gathered} \%\mathrm{of}\mathrm{error}=\left(\frac{∆L}{L}\times 100\right)\% \\ =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\% \\ =\left[1.1\times {10}^{-5}\times 30\times 100\right]\% \\ =3.3\times {10}^{-2}\% \end{gathered}$$

(b) Temperature on a winter day, t₂ = 6 °C​
​So, change in temperature, Δθ = t₁ $-$ t₂ = 10 °C​
ΔL = L_​2 $-$ L₁ = L αΔθ = L × 1.1 × 10^–5 × 10​
$$\begin{gathered} \%\mathrm{of}\mathrm{error}=\left(\frac{∆L}{L}\times 100\right)\% \\ =\left(\frac{L\alpha \Delta \theta }{L}\times 100\right)\% \\ =1.1\times {10}^{-5}\times 10\times 100\% \\ =1.1\times {10}^{-2} \end{gathered}$$