To solve this problem, we need to determine the resistance that must be connected in series with the potentiometer wire to achieve a specific potential gradient.$\\$Given:$\\$• Length of the potentiometer wire $L = 4$ m$= 400$ cm$\\$• Resistance of the potentiometer wire $R_{wire} = 8 \, \Omega \\$• Emf of the accumulator $E = 2 \,$V$\\$• Desired potential gradient $k = 1 \,$mV/cm$= 0.001 \,$V/cm$\\$The potential gradient $k$ is given by:$\\ k = \frac{V_{total}}{L} \\$where $V_{total}$ is the total potential difference across the potentiometer wire.$\\$We want $k = 0.001 \,$V/cm$,$ so:$\\ 0.001 = \frac{V_{total}}{400} \\ V_{total} = 0.001 \times 400 = 0.4 \,$V$\\$The total potential difference across the wire is $V_{total} = 0.4 \,$V$. \\$The remaining potential drop across the series resistance $R_s$ is:$\\ V_s = E - V_{total} = 2 - 0.4 = 1.6 \,$V$\\$The current $I$ through the circuit is given by Ohm's law:$\\ I = \frac{V_{total}}{R_{wire}} = \frac{0.4}{8} = 0.05 \,$A$\\$Using Ohm's law for the series resistance $R_s: \\ V_s = I \cdot R_s \\ 1.6 = 0.05 \cdot R_s \\ R_s = \frac{1.6}{0.05} = 32 \, \Omega \\$Therefore, the resistance that must be connected in series is $32 \, \Omega. \\$This corresponds to Option 4.