Assignment 1.2: A physicist calculated the wall width of half brick thickness (the brick is laid in a flat position, lengthwise called stretcher position), as (13.6 ± 0.1) cm. And one brick thickness (the brick is placed in flat position, lengthwise orthogonal to wall, called header position), as (23.6 ± 0.1) cm. Calculate difference in width of walls with uncertainty in it.
Solution
Width of the first brick (half brick wall), w1 = (13.6 ± 0.1) cm
Width of the second brick (full brick wall), w2 = (23.6 ± 0.1) cm
For finding the uncertainty in the difference, we add the uncertainties in the two measurements in the result (sum).
Difference in the widths of walls = (23.6 ± 0.1) cm – (13.6 ± 0.1) cm = (23.6 – 13.6)cm ± (0.1 + 0.1)cm = 10 cm ± 0.2 cm. (Answer)