Circuit diagram of Metre bridge is shown in figure. The null point is found at a distance of 40cm from A. If now a resistance of $12\Omega$  is connected in parallel with S, the null point occurs at 64cm. Determine the value of R and S.
 

When12 ohm resistance is not connected in parallel to S than we can write,

$\begin{array}{l}\mathrm{R}=\mathrm{S}\frac{{\mathrm{l}}_1}{100-{\mathrm{l}}_1}\\ {\mathrm{l}}_1=40\mathrm{cm}\\ \therefore \mathrm{R}=\mathrm{S}\left(\frac{40}{60}\right)=\frac{2}{3}\mathrm{S}\end{array}$

Now when 12 Ω is connected in parallel to S, then the effective resistance can be written as

$\frac{1}{S_1}=\frac{1}{S}+\frac{1}{12}=\frac{12+5}{12S}\Rightarrow S_1=\frac{12S}{(12+5)}$

$\mathrm{R}=\mathrm{S}\frac{{\mathrm{l}}^{\mathrm{\prime }}}{100-{\mathrm{l}}^{\mathrm{\prime }}}\Rightarrow {\mathrm{l}}^{\mathrm{\prime }}=64\mathrm{cm}$

$\text{ i.e. }\mathrm{R}=\frac{12\mathrm{S}}{(12+\mathrm{S})}\times \frac{64}{36}=\frac{16}{9}\frac{12\mathrm{S}}{12+\mathrm{S}}\text{ \_(II) }$

From the above equations

$\begin{array}{l}\frac{2}{3}\mathrm{S}=\frac{12\mathrm{S}}{\left(12+\mathrm{S}\right)}\frac{16}{9}\\ \Rightarrow \mathrm{S}=20 \mathrm{\Omega }\\ \text{ and }\mathrm{R}=\frac{2}{3}\times 20=\frac{40}{3}\mathrm{\Omega }\end{array}$