The resistances in the two arms of the meter bridge are 5$\Omega$ and R$\Omega$, respectively. When the resistance R is shunted with an equal resistance, the new balance point is at 1.6l₁. The resistance ‘R’ is 

 

$\frac{5}{\mathrm{R}}=\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1}....\left(1\right)$
When R is shunted with equal resistance then its resistance will be R/2 then
$\begin{array}{l}\frac{5}{\mathrm{R}/2}=\frac{1.6{\mathrm{l}}_1}{100-1.6{\mathrm{l}}_1};2\times \frac{5}{\mathrm{R}}=\frac{1.6{\mathrm{l}}_1}{100-1.6{\mathrm{l}}_1}....\left(2\right)\\ \left(1\right)÷\left(2\right)\Rightarrow \frac{\frac{5}{\mathrm{R}}}{2\times \frac{5}{\mathrm{R}}}=\frac{\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1}}{\frac{1.6{\mathrm{l}}_1}{100-1.6{\mathrm{l}}_1}}\\ \frac{1}{2}=\frac{100-1.6{\mathrm{l}}_1}{1.6\left(100-{\mathrm{l}}_1\right)};200-3.2{\mathrm{l}}_1=160-1.6{\mathrm{l}}_1\\ 40=\mathrm{s}4.6{\mathrm{l}}_1;{\mathrm{l}}_1=\frac{40}{1.6}=25\\ \therefore \frac{5}{\mathrm{R}}=\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1}=\frac{25}{75}=\frac{1}{3};\mathrm{R}=15\mathrm{\Omega }\end{array}$