In the circuit shown  $R_1=R_2=10\Omega$ and resistance per unit length of wire  $PQ=1\Omega /cm$ and length  $PQ=10cm.$ If  $R_2$ is made $20\Omega$ the to get zero deflection in galvanometer.  $S$ is midpoint of wire $PQ.$

equivalent diagram is as shown is  $P$ is moved  $2cm$ right them  $R_1=12,R_3=3$
$\frac{R_1}{R_3}=\frac{R_2}{R_4}$  (Hence wheat stone will be balanced)
If s is moved left  $\frac{5}{3}cm$ then $R_3=\frac{10}{3}$  and  $R_4=\frac{20}{3}$ hence  $\frac{R_1}{R_3}=\frac{R_2}{R_4}$  (hence wheat stone will be balanced)