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

% of error =`((ΔL)/L × 100) %`

= `((L∝Δθ)/L × 100)%`

=[1.1 × 10^-5 × 30 ×100]%

=3.3 × 10^-2 %

(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​

`text/"% of  error" = (ΔL/L × 100)%`

\[ = \left( \frac{L\alpha \Delta\theta}{L} \times 100 \right) % \]

\[ = 1 . 1 \times {10}^{- 5} \times 10 \times 100 % \]

\[ = 1 . 1 \times {10}^{- 2} \]