In the figure, the potentiometer wire $AB$ of length $L$ and resistance $9r$ is joined to the cell $D$ of $\mathrm{emf}$ $\varepsilon$ and internal resistance $r$. The cell $C’s$ $\mathrm{emf}$ is $\varepsilon /2$ and its internal resistance is $2r$. The galvanometer $G$ will show no deflection when the length $AJ$ is
Diffcult
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Current through potentiometer $=\frac{\varepsilon}{r+9 r}=\frac{\varepsilon}{10 r}$
Potential difference across potentiometer $=i \times 9 r=V$
$=\frac{\varepsilon}{10 r} \times 9 r=V$
$\frac{9 \varepsilon}{10}=V$
Potential gradient $\frac{V}{L}=\frac{9 \varepsilon}{10 L}=\mathrm{K}$
For no deflection
$\frac{\varepsilon}{2}=k(A J)$
$\frac{\varepsilon}{2}=\frac{9 \varepsilon}{10 L} \times A J$
$\mathrm{AJ}=\frac{5}{9} L$
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