The external and internal diameter of a tube are measured as (32 ± 2)mm and (21 ± 1)mm respectively. Determine the percentage error in the thickness of the tube.
The thickness of the tube is given by:
thickness = (external diameter − internal diameter) / 2
= (32 − 21) / 2
= 11 / 2
= 5.5 mm
Let t = thickness = \(\frac{(D − d)}{2}\), where D = external diameter and d = internal diameter.
The absolute error in thickness is found using the rules for propagation of errors (for subtraction and division by a constant):
Δt = \(\frac{(ΔD + Δd)}{2}\)
= \(\frac{(2 + 1)}{2}\) = \(\frac{3}{2}\)
=1.5 mm.
The percentage error in thickness is:
Percentage error = \(\frac{Δt }{\text{t}}\) × 100%
= (\(\frac{1.5}{ 5.5}\)) × 100%
= (\(\frac{15}{ 55}\)) × 100%
= (\(\frac{3}{11}\)) × 100% ≈ 27.27%
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