For a cell of $e.m.f.$ $2\,V$, a balance is obtained for $50\, cm$ of the potentiometer wire. If the cell is shunted by a $2\,\Omega $ resistor and the balance is obtained across $40\, cm$ of the wire, then the internal resistance of the cell is ............. $\Omega $
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(b) $r = \left( {\frac{{{l_1} - {l_2}}}{{{l_1}}}} \right)R = 0.5\,\Omega $.
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