A steel metre scale reads the correct length at 16^ C . Determine the percentage error if this scale is utilised on a summer day when the temperature is 46^ C . The coefficient of linear expansion of steel is 11 10⁻⁶ ^ C⁻¹ .

Step-by-step solution

The change in temperature is T = 46^ C - 16^ C = 30^ C . When the temperature increases, the scale expands. The measured length L' of an object of true length L is given by: L' = L(1 - T) The error in measurement is: L = L' - L = -L T The percentage error is given by: Percentage error = L L 100 = - T 100 Substituting the given values = 11 10⁻⁶ ^ C⁻¹ and T = 30^ C : Percentage error = -11 10⁻⁶ 30 100 = -0.033 %