Two unknown resistances X and  Y are connected in the left and right gaps of a meter bridge and the balancing point is obtained at $60\mathrm{cm}$ from the left. When a $20\mathrm{\Omega }$ resistance is connected in parallel to X  the balance point is $50\mathrm{cm}.$ Calculate X and Y (in ohm)

Solution:

Resistance R = X Resistance  S = Y Balancing length $l=60\mathrm{cm}$ Balancing length $l=50\mathrm{cm}$ Balancing condition for meter bridge $\frac{\mathrm{R}}{\mathrm{S}}=\frac{\mathrm{l}}{100-\mathrm{l}}\frac{\mathrm{X}}{\mathrm{Y}}=\frac{60}{40}\Rightarrow \frac{3}{2}$ ...(1) 

When a 20 $\Omega$ resistance is connected in parallel to X 
$\frac{X}{Y}=\frac{l^{\mathrm{\prime }}}{100-l^{\mathrm{\prime }}}\frac{\frac{20X}{20+X}}{Y}=\frac{50}{50}\Rightarrow 1\frac{20}{20+X}\left(\frac{3}{2}\right)=1\Rightarrow 30=20+X\therefore X=10\mathrm{\Omega }\therefore Y=\frac{20}{3}\mathrm{\Omega }$