The resistivity of a wire is found from ρ=(π d^2V)/(4IL). The voltmeter reads V=2.00±0.02 V and the ammeter I=0.50±0.01 A. The length is L=50.0±0.5…

The true diameter is the observed reading minus the zero error: $d=0.52-0.02=0.50\ \mathrm{mm}$. The zero correction is exact, so the uncertainty in $d$ is the least count, $0.01\ \mathrm{mm}$. For $\rho=\pi d^2V/(4IL)$ the maximum fractional error is $\dfrac{\Delta\rho}{\rho}=2\dfrac{\Delta d}{d}+\dfrac{\Delta V}{V}+\dfrac{\Delta I}{I}+\dfrac{\Delta L}{L}$. The individual terms are $2\times\dfrac{0.01}{0.50}=4\%$, $\dfrac{0.02}{2.00}=1\%$, $\dfrac{0.01}{0.50}=2\%$ and $\dfrac{0.5}{50.0}=1\%$. Adding the errors of products and quotients, the maximum error is $4+1+2+1=8\%$. Note that using the raw reading $0.52\ \mathrm{mm}$ would give $3.85\%$ for the diameter term and the wrong total; the zero error must be removed before computing the relative error, and the diameter appears squared, which doubles its contribution.