A potentiometer wire of length 300 cm is connected in series with a resistance 780 $Ω$ and a standard cell of emf 4V. A constant current flows through potentiometer wire. The length of the null point for cell of emf 20 mV is found to be 60 cm. The resistance of the potentiometer wire is ____ $Ω$ .

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answer is 20.

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Detailed Solution

Let resistance of potentiometers wire is R

$i = \frac{4}{R + 780}$

Potential difference across AB

$= \frac{4 R}{R + 780}$

Potential difference across AC

$= \frac{4 R \times 60}{(R + 780) \times 300} = \frac{4 R}{5 (R + 780)}$

This should be equal to 20 mV

$\begin{array}{l} \frac{4 R}{5 (R + 780)} = 20 \times 10^{- 3} = 2 \times 10^{- 2} \\ 4 R = 10^{- 1} (R + 780) \\ 4 R = \frac{R}{10} + 78 \\ 4 R - \frac{R}{10} = 78 \\ \frac{39 R}{10} = 78 \\ R = 20 Ω \end{array}$

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