A steel metre scale measures the correct length at 16^ C . Evaluate the percentage error if this scale is employed on a winter day when the temperature is 6^ C . The coefficient of linear expansion of steel is given as 11 10⁻⁶ ^ C⁻¹ .

Step-by-step solution

The true length of an object is L . The scale is calibrated at T₁ = 16^ C . At T₂ = 6^ C , the change in temperature is T = 6^ C - 16^ C = -10^ C . Due to thermal contraction, the length of one division on the scale becomes d' = d(1 + T) . The measured length L_m is given by L_m = L 1 + T L(1 - T) . The error in measurement is L = L_m - L = -L T . Percentage error = L L 100 = - T 100 . Substituting the given values: Percentage error = -(11 10⁻⁶) (-10) 100 = 11 10⁻³ = 0.011 % .