In the given figure, battery $E $ is balanced on $55\, cm$ length of potentiometer wire but when a resistance of $10 \,\Omega$ is connected in parallel with the battery then it balances on $50\, cm$ length of the potentiometer wire then internal resistance $r$ of the battery is ............. $\Omega $
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(a) $r = \left( {\frac{{{l_1} - {l_2}}}{{{l_2}}}} \right) \times R'$$ \Rightarrow r = \left( {\frac{{55 - 50}}{{50}}} \right) \times 10 = 1\,\Omega $.
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