Answer:
the reading on the scale is 158.5 mm.
Explanation:
Given;
upper fixed point of the temperature scale, x₁ = 260 mm
lower fixed point of the temperature scale, x₂ = 50 mm
upper temperature scale, T₁ = 212 °F
lower temperature scale, T₂ = 32 °F
thermometer reading, t = 125 °F, let the reading on the scale = x
Interpolate as follows to determine the value of "x"
![\frac{X_1 - X}{X_1-X_2} = \frac{T_1-t}{T_1-T_2} \\\\\frac{260-X}{260-50} = \frac{212-125}{212-32} \\\\\frac{260 -X}{210} =\frac{87}{180} \\\\180(260-X) = 87(210)\\\\46800 -180 X = 18270 \\\\180X = 46800-18270\\\\180X = 28530\\\\X = \frac{28530}{180} \\\\X = 158.5 \ mm](https://tex.z-dn.net/?f=%5Cfrac%7BX_1%20-%20X%7D%7BX_1-X_2%7D%20%3D%20%5Cfrac%7BT_1-t%7D%7BT_1-T_2%7D%20%5C%5C%5C%5C%5Cfrac%7B260-X%7D%7B260-50%7D%20%3D%20%5Cfrac%7B212-125%7D%7B212-32%7D%20%5C%5C%5C%5C%5Cfrac%7B260%20-X%7D%7B210%7D%20%3D%5Cfrac%7B87%7D%7B180%7D%20%5C%5C%5C%5C180%28260-X%29%20%3D%2087%28210%29%5C%5C%5C%5C46800%20-180%20X%20%3D%2018270%20%5C%5C%5C%5C180X%20%3D%2046800-18270%5C%5C%5C%5C180X%20%3D%2028530%5C%5C%5C%5CX%20%3D%20%5Cfrac%7B28530%7D%7B180%7D%20%5C%5C%5C%5CX%20%3D%20158.5%20%5C%20mm)
Therefore, the reading on the scale is 158.5 mm.