JEE Main · 2005 · 08 May 2005 · Offline · Current Electricity

Zero deflection means no current flows through the branch holding cell $B$, so cell $B$'s emf exactly matches the potential dropped across $R$ by the current that cell $A$ drives round the outer loop. That current is $\frac{E_A}{R_0+R}$ through the series resistor $R_0$ and $R$, and balance requires this times $R$ to equal $E_B$. With the figure's values this reads $\frac{12R}{500+R}=2$, so $12R=1000+2R$, giving $10R=1000$ and $R=100\Omega$.