To solve this problem, we need to understand how the potentiometer works when cells are connected in series.$\\$Given:$\\$• Length of the potentiometer wire $L = 100$ cm.$\\$• Balance points when cells support each other: $l_1 = 50$ cm.$\\$• Balance points when cells oppose each other: $l_2 = 10$ cm.$\\$Let's denote:$\\$• $E_1$ and $E_2$ as the emfs of the two cells.$\\$When the cells support each other, the effective emf is $(E_1 + E_2). \\$The balance point is given by:$\\ E_1 + E_2 = k \cdot l_1 \\$where $k$ is the potential gradient.$\\$When the cells oppose each other, the effective emf is $(E_1 - E_2). \\$The balance point is given by:$\\ E_1 - E_2 = k \cdot l_2 \\$We have two equations:$\\ 1. \quad E_1 + E_2 = k \cdot 50 \\ 2. \quad E_1 - E_2 = k \cdot 10 \\$Add these two equations to eliminate $E_2: \\ (E_1 + E_2) + (E_1 - E_2) = k \cdot 50 + k \cdot 10 \\ 2E_1 = k \cdot 60 \\ E_1 = \frac{k \cdot 60}{2} \\ E_1 = 30k \\$Subtract the second equation from the first to eliminate $E_1: \\ (E_1 + E_2) - (E_1 - E_2) = k \cdot 50 - k \cdot 10 \\ 2E_2 = k \cdot 40 \\ E_2 = \frac{k \cdot 40}{2} \\ E_2 = 20k \\$The ratio of the emfs is:$\\ \frac{E_1}{E_2} = \frac{30k}{20k} = \frac{3}{2} \\$Therefore, the ratio of the emfs is $3:2$, which corresponds to Option 2.