In the circuit (Fig) the galvanometer G shows 

zero deflection. If the batteries A and I have negligible internal resistance, the value of resistor R will be

Let i be the current flowing in first loop. Then
$(500+\mathrm{R})\mathrm{i}=12\text{ or }\mathrm{i}=\frac{12}{(500+\mathrm{R})}$
Potential difference across .E is given by
$\mathrm{V}=\mathrm{iR}=\frac{12\mathrm{R}}{(500+\mathrm{R})}$
The galvanometer will not show any deflection if
V=2 V Therefore,
$\frac{12\mathrm{R}}{500+\mathrm{R}}=2$
or 12 R = 1000+2R
or 10 R = 1000 or R= 100 $\mathrm{\Omega }$