In the potentiometer circuit as shown in the figure, the balance length $Al = 60\ cm$ when switch $S$ is open. When switch $S$ is closed and the value of $R$ is $5W$ the balance length $Al' = 50 cm.$ The internal resistance of the cell $C'$ is : .............. $\Omega$
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Here $I_{1}=60 c m I_{2}=50 c m, R=5 \Omega$
$e=k l_{1}$
where $\mathrm{k}$ is the potential gradient ac ross the wire
$V=k l_{2}$
$\therefore \varepsilon v=\frac{I_{1}}{I_{2}}=\frac{60}{50}=\frac{6}{5}$
Internal resistance of the cell is
$r=\left(\frac{\varepsilon}{V}-1\right) R=\left(\frac{6}{5}-1\right)=1.0 \Omega$
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