$A$ battery of $\mathrm{emf}$ $E_0 = 12\, V$ is connected across a $4\,m$ long uniform wire having resistance $4\,\Omega /m$. The cells of small $\mathrm{emfs}$ $\varepsilon_1 = 2\,V$ and $\varepsilon_2 = 4\,V$ having internal resistance $2\Omega$ and $6\Omega$ respectively, are connected as shown in the figure. If galvanometer shows no deflection at the point $N$, the distance of point $N$ from the point $A$ is equal to 
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$E_{e q}=\frac{E_{2} r_{2}-E_{2} r_{1}}{r_{1}+r_{2}}=\frac{2 \times 6-4 \times 2}{6+2}=\frac{1}{2}$ volt

At balancing length $l$

$E_{e q}=\frac{E_{0}}{R+16} \times 4 \lim p l i e s \frac{1}{2}=\frac{12}{8+16} \times 4 \times l$

$l=25 c m$

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