The adjoining figure shows the connections of potentiometer experiment to determine internal resistance of of a leclanche cell. When the cell is on open circuit the balancing length of the potentiometer wire is $3.4\, m$ and on closing the key $K_2$ the balancing length becomes $1.7\, m$ . If the resistance $R$ through which current is drawn is $10\,\Omega $ then the internal resistance of the cell is .............. $\Omega$
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${\rm{r}} = \left( {\frac{{{l_1}}}{{{l_2}}} - 1} \right){\rm{R}}$
$=\left(\frac{3.4}{1.7}-1\right) \times 10=10 \,\Omega $
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