The figure shows experimental set up of a meter bridge when the two unknown resistances X and Y are inserted, the null point D is obtained 40 cm from the end A. When a resistance of 10 Ω is connected in series with X, the null point D is obtained 10 cm. Find the position of the null point when the 10 Ω resistance is inserted in series with resistance Y. Determine the value of the resistances X and Y.

When two unknown resistances X and y are placed as shown in the fig. then, the null point D is obtained at a distance of 40 cm from A. 

i.e., length, l = 40 cm 

Therefore,
 
           $\frac{\mathrm{X}}{\mathrm{Y}}=\frac{40}{(100-40)}=\frac{40}{60}$ 

$\Rightarrow$                 $\mathrm{Y} = \frac{3}{2 }\mathrm{X}$                       ... (1)
Given that, the null point shifts by 10 cm when a resistance of 10 $\mathrm{\Omega }$ is connected in series with x. 
Length l₁ becomes (40+10) cm = 50 cm. 

       $\frac{\mathrm{X}+10}{\mathrm{Y}}= \frac{50}{(100-50)} = \frac{50 }{50}= 1$ 

$\Rightarrow$                 X+10 = Y                         ... (2) 

Now, on solving equation (1) and (2) we get, 

Resistance, X = 20 $\mathrm{\Omega } \mathrm{and} \mathrm{Y} = 30 \mathrm{\Omega }$

If, a resistance of 10 $\mathrm{\Omega }$ is connected in series with Y, new null point will be obtained at a distance of l₂ from A . 

i.e.,                 $\frac{\mathrm{X}}{\mathrm{Y}+10} = \frac{l_2}{100 - l_{2 }}$ 

$\Rightarrow$      $\frac{20}{30+10} = \frac{20}{40} = \frac{l_2}{100 - l_2}$

On solving the above equation, we get

               ${\mathrm{l}}_2 = \frac{100}{3} = 33.3 \mathrm{cm}$