The given figure represents an arrangement of potentiometer for the calculation of internal resistance $(r)$ of the unknown battery $(E)$. The balance length is $70.0\, cm$ with the key opened and $60.0\,cm$ with the key closed. $R$ is $132.40 \Omega $. The internal resistance $(r)$ of the unknown cell will be.......$\Omega$ (Given $E_o > E$) :-
Medium
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Internal resistance $r=\frac{R\left[l-l^{\prime}\right]}{l^{\prime}}$
where $l=$ balanced length with key open.
$l^{\prime}=$ balanced length with key $K$ closed
$\therefore r=\frac{132.40[70-60]}{60}=\frac{132.40 \times 10}{60}$
$=21.06\, \Omega \approx 22.1\, \Omega$
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