A wire of uniform cross-section having resistance $8\Omega$ is bent to form a square ABCD. End A is connected to a point P on the side DC by a wire AP of resistance $1\Omega$ . When V potential difference is applied between A and C, the resistance of part DP if the points B and P are to be at same potential

Resistances CB and BA are in series. Since resistances of CB and BA are equal, therefore potential drops across them is same.

As B and P are at same potential, with some analysis, we can see that potential drops across branches CP and between PA must be same.

Let R be the resistance of portion DP. Then resistance of portion PC = 2 – R.

$R_{PA}=R_{CP}$
$\frac{\left(2+R\right)\times 1}{1+\left(2+R\right)}=\left(2-R\right)$$\Rightarrow \frac{2+R}{3+R}=2-R$
$\Rightarrow 2+R=6-3R-R^2+2R$
$\Rightarrow R^2+2R-4=0$
$\Rightarrow R=\left(\sqrt{5}-1\right)\Omega$