In an experiment with potentiometer to measure the internal resistance of a cell, when the cell is shunted by 5 $Ω$ , the null point is obtained at 2m. When cell is shunted by 20 $Ω$ the null point is obtained at 3m. The internal resistance of cell is

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2 $Ω$

4 $Ω$

6 $Ω$

8 $Ω$

answer is B.

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Detailed Solution

$\begin{array}{l} \frac{ℓ_{1}}{ℓ_{2}} = \frac{R_{1}}{R_{2}} (\frac{R_{2} + r}{R_{1} + r}); \frac{2}{3} = \frac{5}{20} (\frac{20 + r}{5 + r}) \\ 40 + 8 r = 60 + 3 r \Rightarrow 5 r = 20 \Rightarrow r = 4 Ω \end{array}$

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