JEE Main · 2019 · 12 Jan 2019 · Shift 1 · Current Electricity

The primary loop has the $4\text{V}$ battery driving current through the series resistance $R$ and the potentiometer wire of resistance $5\Omega$, so $i=\frac{4}{5+R}$. The potential dropped across the full one-metre wire is $i\times 5=\frac{20}{5+R}$, and across $10\text{cm}$, one-tenth of the wire, it is one-tenth of that, $\frac{2}{5+R}$. We need this to equal $5\text{mV}$: $\frac{2}{5+R}=5\times 10^{-3}$, so $5+R=\frac{2}{5\times 10^{-3}}=400$, giving $R=395\Omega$.