A wire of uniform cross section having resistance $8 Ω$ 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 $1 Ω$ . 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

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$(2 - \sqrt{2}) Ω$

$(\frac{\sqrt{2} + 1}{4}) Ω$

$(\sqrt{2} - 1) Ω$

$(\sqrt{5} - 1) Ω$

answer is B.

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Detailed Solution

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

As B and P are at same potential,

$\frac{(2 + R) \times 1}{1 + (2 + R)} = (2 - R)$
$\Rightarrow \frac{2 + R}{3 + R} = 2 - R$
$\Rightarrow 2 + R = 6 - 3 R - R^{2} + 2 R$
$\Rightarrow R^{2} + 2 R - 4 = 0$
$\Rightarrow R = \sqrt{5} - 1$

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