To solve this problem, we need to calculate the resistance of the final combination of the wire segments.$\\$Given:$\\$• Total resistance of the wire $R = 100 \Omega. \\$• The wire is divided into 10 equal parts, so each part has a resistance $r. \\$Calculate the resistance of each part:$\\ r = \frac{R}{10} = \frac{100 \Omega}{10} = 10 \Omega. \\$First, connect the first 5 parts in series:$\\$• The resistance of parts in series is the sum of their resistances.$\\ R_{series} = 5 \times r = 5 \times 10 \Omega = 50 \Omega. \\$Next, connect the next 5 parts in parallel:$\\$• The resistance of parts in parallel is given by $\frac{1}{R_{parallel}} = \frac{1}{r_1} + \frac{1}{r_2} + \cdots + \frac{1}{r_n}. \\$• Since all resistances are equal, $\frac{1}{R_{parallel}} = 5 \times \frac{1}{r} = 5 \times \frac{1}{10 \Omega}. \\$Calculate $R_{parallel}: \\ \frac{1}{R_{parallel}} = \frac{5}{10 \Omega} = \frac{1}{2 \Omega}. \\ R_{parallel} = 2 \Omega. \\$Finally, connect the series and parallel combinations in series:$\\ R_{total} = R_{series} + R_{parallel} = 50 \Omega + 2 \Omega = 52 \Omega. \\$Therefore, the resistance of the final combination is $52 \Omega. \\$This corresponds to Option 4.