A steel metre scale is calibrated to provide correct readings at 20^ C . Determine the distance between the 50 cm mark and the 51 cm mark when the scale is used at 10^ C . The coefficient of linear expansion of steel is 1.1 10⁻⁵ ^ C⁻¹ .

Step-by-step solution

The scale is calibrated at 20^ C , which means the physical distance between the 50 cm mark and the 51 cm mark is exactly 1 cm at this temperature. Let L₀ = 1 cm be the distance at T₀ = 20^ C . When the temperature is changed to T = 10^ C , the scale undergoes thermal contraction. The new physical distance L between the marks is given by: L = L₀ [1 + (T - T₀)] Substituting the given values: L = 1 [1 + 1.1 10⁻⁵ (10 - 20)] L = 1 [1 - 1.1 10⁻⁵ 10] L = 1 - 0.00011 L = 0.99989 cm