In the experiment of Ohm's law, a potential difference of 5.0 V is applied across the end of a conductor of length 10.0 cm and diameter of 5.00 mm. The measured current in the conductor is 2.00 A. The maximum permissible percentage error in the resistivity of the conductor is:-

The resistivity of a conductor is given by $$\rho = \frac{RA}{l} = \frac{V}{I} \cdot \frac{\pi d^2/4}{l} = \frac{V \pi d^2}{4Il}$$, where $$d$$ is the diameter, $$l$$ is the length, $$V$$ is the voltage, and $$I$$ is the current.

The maximum percentage error in $$\rho$$ is $$\frac{\Delta\rho}{\rho} \times 100 = \frac{\Delta V}{V} \times 100 + 2\frac{\Delta d}{d} \times 100 + \frac{\Delta I}{I} \times 100 + \frac{\Delta l}{l} \times 100$$.

The least count errors are: $$\Delta V = 0.1$$ V (since 5.0 V has least count 0.1 V), $$\Delta l = 0.1$$ cm (since 10.0 cm), $$\Delta d = 0.01$$ mm (since 5.00 mm), and $$\Delta I = 0.01$$ A (since 2.00 A).

Substituting: $$\frac{\Delta V}{V} \times 100 = \frac{0.1}{5.0} \times 100 = 2\%$$, $$\frac{\Delta l}{l} \times 100 = \frac{0.1}{10.0} \times 100 = 1\%$$, $$2 \times \frac{\Delta d}{d} \times 100 = 2 \times \frac{0.01}{5.00} \times 100 = 0.4\%$$, and $$\frac{\Delta I}{I} \times 100 = \frac{0.01}{2.00} \times 100 = 0.5\%$$. Total maximum percentage error $$= 2 + 1 + 0.4 + 0.5 = 3.9\%$$.