A potentiometer circuit has been set up for finding the internal resistance of a given cell. The main battery, used across the potentiometer wire, has an emf of $2.0\,V$ and a negligible internal resistance. The potentiometer wire itself is $4\,m$ long. When the resistance $R,$ connected across the given cell, has values of $(i)$ infinity $(ii)$ $9.5\,\Omega$ the balancing lengths on the potentiometer wire are found to be $3\,m$ and $2.85\,m,$ respectively. The value of internal resistance of the cell is ............... $\Omega$
AIPMT 2014, Medium
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The internal resistance of the cell is
$r=\left(\frac{l_{1}}{l_{2}}-1\right) R$
Here, $l_{1}=3\, \mathrm{m}, l_{2}=2.85\, \mathrm{m}, R=9.5\, \Omega$
$\therefore \quad r=\left(\frac{3}{2.85}-1\right)(9.5 \,\Omega)=\frac{0.15}{2.85} \times 9.5\, \Omega=0.5\, \Omega$
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