To solve this problem, we will use the principle of the meter bridge (Wheatstone bridge).$\\$Initially, the balance condition is given by:$\\ \frac{5}{R} = \frac{l_1}{100 - l_1} \\$When the resistance $R$ is shunted with an equal resistance, the effective resistance becomes:$\\ R_{shunted} = \frac{R \cdot R}{R + R} = \frac{R}{2} \\$The new balance condition is:$\\ \frac{5}{\frac{R}{2}} = \frac{1.6 l_1}{100 - 1.6 l_1} \\$Simplify the equation:$\\ \frac{10}{R} = \frac{1.6 l_1}{100 - 1.6 l_1} \\$From the initial condition:$\\ \frac{5}{R} = \frac{l_1}{100 - l_1} \\$We can express $R$ in terms of $l_1: \\ R = \frac{5(100 - l_1)}{l_1} \\$Substitute $R$ in the new balance condition:$\\ \frac{10}{\frac{5(100 - l_1)}{l_1}} = \frac{1.6 l_1}{100 - 1.6 l_1} \\$Simplify:$\\ \frac{2 l_1}{100 - l_1} = \frac{1.6 l_1}{100 - 1.6 l_1} \\$Cross-multiply:$\\ 2 l_1 (100 - 1.6 l_1) = 1.6 l_1 (100 - l_1) \\$Expand both sides:$\\ 200 l_1 - 3.2 l_1^2 = 160 l_1 - 1.6 l_1^2 \\$Rearrange:$\\ 40 l_1 = 1.6 l_1^2 \\$Divide by $l_1: \\ 40 = 1.6 l_1 \\ l_1 = \frac{40}{1.6} = 25 \\$Substitute $l_1$ back to find $R: \\ R = \frac{5(100 - 25)}{25} \\ R = \frac{5 \times 75}{25} \\ R = 15 \, \Omega \\$Therefore, the resistance $R$ is $15 \, \Omega. \\$This corresponds to Option 2.