a steel tape is calibrated at 20 C. on a cold day when the temperature is -15C , percentage error in the tape will be (

α_steel = 1.2 * 10^-5 C^-1

let the length of the steel tape at 20C be = L

now, the change in length of the tape will be given as

dL = α_L.L.dT

here

α_L = coefficient of linear expansion = 1.2 x 10^-5 C^-1

dT = change in temperature = 15C + 20C = 35C

so,

dL = L x [ 1.2 x 10^-5 x 35]

or

change in length

dL = 4.2 x 10^-4 x L

so,

the percentage error will be

 = (dL / L) x 100

 = [( 4.2 x 10^-4 x L) / L] x 100

 = 4.2 x 10^-2

thus,

% error in measurement = 0.042 %