A student measures the resistance, diameter, and length of a wire...

The formula for the electrical resistivity $\rho$ of a wire is given by:
$\rho = \dfrac{R A}{l} = \dfrac{R (\pi d^2 / 4)}{l}$

The maximum fractional error in resistivity is the sum of the fractional errors of the individual quantities (with the power of $d$ bringing a factor of 2):
$\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta R}{R} + 2\dfrac{\Delta d}{d} + \dfrac{\Delta l}{l}$

Calculating the individual percentage errors:
Percentage error in $R = \left(\dfrac{0.1}{10.0}\right) \times 100 = 1\%$
Percentage error in $d = \left(\dfrac{0.02}{2.00}\right) \times 100 = 1\%$
Percentage error in $l = \left(\dfrac{0.5}{50.0}\right) \times 100 = 1\%$

Substituting these into the error equation:
Maximum percentage error in $\rho = 1\% + 2(1\%) + 1\% = 4\%$

Answer: 4