To find the internal resistance of the cell, we will use the potentiometer principle.$\\$Given:$\\$• Emf of the main battery $E = 2.0 \, \mathrm{V} \\$• Length of the potentiometer wire $L = 4 \, \mathrm{m} \\$• Balancing length when $R = \infty$ (open circuit) is $l_1 = 3 \, \mathrm{m} \\$• Balancing length when $R = 9.5 \, \Omega$ is $l_2 = 2.85 \, \mathrm{m} \\$The potential gradient along the potentiometer wire is given by:$\\ k = \frac{E}{L} = \frac{2.0}{4} = 0.5 \, \mathrm{V/m} \\$The emf of the cell $\varepsilon$ is balanced by the potential difference across $l_1: \\ \varepsilon = k \cdot l_1 = 0.5 \cdot 3 = 1.5 \, \mathrm{V} \\$When the resistance $R = 9.5 \, \Omega$ is connected, the potential difference across $l_2$ is:$\\ V = k \cdot l_2 = 0.5 \cdot 2.85 = 1.425 \, \mathrm{V} \\$The internal resistance $r$ of the cell can be found using the formula:$\\ \frac{\varepsilon}{V} = 1 + \frac{r}{R} \\$Substitute the known values:$\\ \frac{1.5}{1.425} = 1 + \frac{r}{9.5} \\$Simplify and solve for $r: \\ 1.0526 = 1 + \frac{r}{9.5} \\ 0.0526 = \frac{r}{9.5} \\ r = 0.0526 \times 9.5 \\ r = 0.5 \, \Omega \\$Therefore, the internal resistance of the cell is $0.5 \, \Omega$, which corresponds to Option 3.