Question

The formulae, for resistivity (p) and Young’s modules (Y) are, respectively,

\(p=\frac{\pi r^2R}{l}\) and Y = \(\frac{FL}{\pi^2l}\)

The relative errors, in p and Y, would then be, given, by the respective expressions.

(1) \((\frac{Δp}{p}=\frac{2Δr}{r}+\frac{ΔR}{R}+\frac{Δl}{l})\) and \((\frac{ΔY}{Y}=\frac{dF}{F}+\frac{ΔL}{L}+\frac{2Δr}{r}+\frac{Δl}{l})\)

(2) \((\frac{Δp}{p}=\frac{2Δr}{r}+\frac{ΔR}{R}-\frac{Δl}{l})\) and \((\frac{ΔY}{Y}=\frac{dF}{F}+\frac{ΔL}{L}+\frac{2Δr}{r}+\frac{Δl}{l})\)

(3) \((\frac{Δp}{p}=\frac{2Δr}{r^2}+\frac{ΔR}{R}-\frac{Δl}{l})\) and \((\frac{ΔY}{Y}=\frac{ΔF}{F}+\frac{ΔL}{L}+\frac{2Δr}{r^2}+\frac{Δl}{l})\)

(4) \((\frac{Δp}{p}=\frac{2Δr}{r^2}+\frac{ΔR}{R}+\frac{Δl}{l})\) and \((\frac{ΔY}{Y}=\frac{ΔF}{F}+\frac{ΔL}{L}-\frac{2Δr}{r^2}+\frac{Δl}{l})\)

Answer 1

(1) \((\frac{Δp}{p}=\frac{2Δr}{r}+\frac{ΔR}{R}+\frac{Δl}{l})\) and \((\frac{ΔY}{Y}=\frac{dF}{F}+\frac{ΔL}{L}+\frac{2Δr}{r}+\frac{Δl}{l})\)

The relative errors, in all the products as well as the quotients, have to be all added to get the relative error in the given quantity.

Note the relative error in r² is 2(Δr/r) and NOT 2(Δr/r²). Hence, the two expressions, given only in option (1), correspond to these rules.