Two resistance X and Y are connected in the two gaps of a...

For a meterbridge, $\frac{\mathrm{X}}{\mathrm{Y}}=\frac{l}{100-l}$ In the first case, $l=20 \mathrm{~cm}$$\begin{aligned}\frac{\mathrm{X}}{\mathrm{Y}}& =\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4}\\ \therefore \quad 4 \mathrm{X}& =\mathrm{Y} \end{aligned}$In the second case, $l^{\prime}=40 \mathrm{~cm}\begin{array}{ll} \therefore& \frac{\mathrm{X}^{\prime}}{\mathrm{Y}^{\prime}}=\frac{40}{100-40}=\\ & \text { But, } \mathrm{X}^{\prime}=\mathrm{X}+2 \\ \therefore \quad & \frac{\mathrm{X}+20}{\mathrm{Y}}=\frac{2}{3}\\ \therefore \quad & \frac{\mathrm{X}+20}{4 \mathrm{X}}=\frac{2}{3}\\ \therefore \quad & 8 \mathrm{X}=3 \mathrm{X}+60 \\ \therefore \quad & \mathrm{X}=12 \Omega \end{array}$$