An iron rod of length 50m and at a temperature of 60oC is heated to 70oC. Calculate its new length. (Linear expansivity of iron = 1.2 x 10-5K-1)

An iron rod of length 50m and at a temperature of 60^oC is heated to 70^oC. Calculate its new length. (Linear expansivity of iron = 1.2 x 10^-5K^-1)

Answer Details

The formula for linear thermal expansion is: ΔL = αLΔT Where: ΔL = change in length α = coefficient of linear expansion L = original length ΔT = change in temperature Using the values given: L = 50 m α = 1.2 x 10^-5 K^-1 ΔT = 70°C - 60°C = 10°C Substituting the values in the formula: ΔL = (1.2 x 10^-5 K^-1)(50 m)(10°C) ΔL = 0.006 m The new length of the iron rod is the sum of the original length and the change in length: New length = 50 m + 0.006 m = 50.006 m Therefore, the correct answer is: 50.006m.