Two resistance \(\mathrm{X}\) and \(\mathrm{Y}\) are connected in the two gaps of a meterbridge and the null points is obtained at \(20 \mathrm{~cm}\) from zero end. When the resistance of \(20 \Omega\) is connected in series with the smaller of the two resistance \(\mathrm{X}\) and \(\mathrm{Y}\), the null point shifts to \(40 \mathrm{~cm}\) from left end. The value of smaller resistance in ohm is

For a meterbridge, \(\frac{\mathrm{X}}{\mathrm{Y}}=\frac{l}{100-l}\)
In the first case, \(l=20 \mathrm{~cm}\)
\(\frac{\mathrm{X}}{\mathrm{Y}} =\frac{20}{100-20}=\frac{20}{80}=\frac{1}{4}\)
\(\therefore 4 \mathrm{X} =\mathrm{Y}\)
In the second case, \(l^{\prime}=40 \mathrm{~cm}\)
\(\therefore \frac{\mathrm{X}^{\prime}}{\mathrm{Y}^{\prime}}=\frac{40}{100-40}= \)
\( \text {But, } \mathrm{X}^{\prime}=\mathrm{X}+2 \)
\( \therefore \frac{\mathrm{X}+20}{\mathrm{Y}}=\frac{2}{3} \)
\( \therefore \frac{\mathrm{X}+20}{4 \mathrm{X}}=\frac{2}{3} \)
\( \therefore 8 \mathrm{X}=3 \mathrm{X}+60 \)
\( \therefore \mathrm{X}=12 \Omega\)