Two identical potentiometer $P_{1}$ and $P_{2}$ of equal length $l$ , resistance $R_{1}, R_{2}$ are connected with a battery of emf $ε_{0}$ and internal resistance $1 Ω$ , through two switches $S_{1}$ and $S_{2}$ . A battery of emf $ε$ is balanced on these potentiometer wires one by one. If resistance of wire $R_{1}$ is $3.68 Ω$ and balancing length is $\frac{l}{2}$ on it, when $S_{1}$ is closed and $S_{2}$ is open. On closing $S_{2}$ and opening $S_{1}$ the balancing length on $P_{2}$ is found to be $\frac{2 l}{3}$ , then the resistance of wire $R_{2}$ is $k \times 10^{- 2} Ω$ . Find the value of k.

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answer is 143.75.

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Detailed Solution

When $S_{1}$ is closed and $S_{2}$ is open
Current through $R_{1}$
$I_{1} = \frac{ε_{0}}{1 + 3.68} = \frac{ε_{0}}{4.68}$
Potential difference across $\frac{l}{2}$ length $= (\frac{ε_{0}}{4.68}) \times 1.84 = ε$
Similarly in the second case
$I_{2} = \frac{ε_{0}}{R_{2} + 1} = \frac{468 ε}{104 (R_{2} + 1)}$
Potential difference across $\frac{2 l}{3}$ length of $R_{2} = \frac{2}{3} R_{2} . I_{2} = \frac{2}{3} \times \frac{468 ε R_{2}}{184 (R_{2} + 1)}$
$\frac{2}{3} \times \frac{468 ε R_{2}}{184 (R_{2} + 1)} = ε$
$64 R_{2} = 92$
$\Rightarrow R_{2} = 1.4375 Ω$