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Step-by-Step Solution
Step 1: Express the resistivity in terms of measurable quantities
Ohm’s law states that the voltage $V$ across a conductor is related to the current $I$ and resistance $R$ by
$V = IR$. The resistance $R$ for a wire of length $l$ and cross-sectional area $A$ is given by
$R = \frac{\rho \, l}{A}$,
where $\rho$ is the resistivity. Rearranging to solve for $\rho$:
$$
\rho \;=\; \frac{V A}{I \, l}.
$$
Step 2: Express the cross-sectional area in terms of diameter
The cross-sectional area $A$ of a wire of diameter $d$ (cylindrical cross-section) is
$$
A = \frac{\pi d^{2}}{4}.
$$
Therefore,
$$
\rho = \frac{V}{I\,l} \times \frac{\pi d^2}{4}
= \frac{\pi}{4} \;\frac{V\,d^{2}}{I \,l}.
$$
Step 3: Setup the formula for the percentage error in $\rho$
The total percentage error in a quantity that depends on multiple measured variables can be estimated (assuming small independent errors) by summing up the relative errors appropriately. For
$$
\rho = \frac{\pi}{4} \;\frac{V\,d^{2}}{I \,l},
$$
the differential (or relative) errors add as follows:
$$
\frac{\Delta \rho}{\rho}
= 2\,\frac{\Delta d}{d}
+ \frac{\Delta V}{V}
+ \frac{\Delta I}{I}
+ \frac{\Delta l}{l}.
$$
Note the factor of 2 multiplying $\frac{\Delta d}{d}$ because $d$ appears squared.
Step 4: Substitute the given errors to calculate the relative (percentage) error
From the question, the measurements and their uncertainties are:
Voltage, $V = 5.0\,\text{V}$ with $\Delta V = 0.1\,\text{V}$
Current, $I = 2.00\,\text{A}$ with $\Delta I = 0.01\,\text{A}$
Length, $l = 10.0\,\text{cm}$ with $\Delta l = 0.1\,\text{cm}$
Diameter, $d = 5.00\,\text{mm}$ with $\Delta d = 0.01\,\text{mm}$
Therefore:
$$
\frac{\Delta d}{d}
= \frac{0.01}{5.00}
= 0.002,
$$
$$
\frac{\Delta V}{V}
= \frac{0.1}{5.0}
= 0.02,
$$
$$
\frac{\Delta I}{I}
= \frac{0.01}{2.00}
= 0.005,
$$
$$
\frac{\Delta l}{l}
= \frac{0.1}{10.0}
= 0.01.
$$
Now sum them, remembering to multiply the diameter’s relative error by 2:
$$
\frac{\Delta \rho}{\rho}
= 2 \times 0.002
+ 0.02
+ 0.005
+ 0.01
= 0.004 + 0.02 + 0.005 + 0.01
= 0.039.
$$
Step 5: Convert to percentage
Converting the relative error to a percentage:
$$
0.039 \times 100\% = 3.9\%.
$$
Hence, the maximum permissible percentage error in the resistivity of the conductor is
3.9%.
