To solve this problem, we will use the principle of a metre bridge (or Wheatstone bridge).$\\$Given:$\\$• Resistance in the right gap $R_2 = 10 \, \Omega. \\$• The bridge wire is divided in the ratio $3:2. \\$Let the resistance in the left gap be $R_1. \\$The balance condition for a metre bridge is given by:$\\ \frac{R_1}{R_2} = \frac{l_1}{l_2} \\$Where $l_1$ and $l_2$ are the lengths of the bridge wire on the left and right sides, respectively.$\\$Given that the ratio $\frac{l_1}{l_2} = \frac{3}{2},$ we can write:$\\ \frac{R_1}{10} = \frac{3}{2} \\$Solving for $R_1: \\ R_1 = 10 \times \frac{3}{2} \\ R_1 = 15 \, \Omega \\$The resistance wire has a total length of $1.5 \,$m.$\\$We need to find the length of the wire that corresponds to $1 \, \Omega. \\$The resistance of a wire is directly proportional to its length.$\\$Let $L$ be the length of wire that corresponds to $1 \, \Omega. \\$Then, we can write:$\\ \frac{L}{1.5} = \frac{1}{15} \\$Solving for $L: \\ L = \frac{1.5}{15} \\ L = 0.1 \,$m$\\$Therefore, the length of $1 \, \Omega$ of the resistance wire is $0.1 \,$m.$\\$This corresponds to Option 1.