JEE Main · 2026 · 21 Jan 2026 · Shift 2 · Current Electricity

The cell drives current through the uniform wire $AB$, so the full terminal voltage $V$ appears across its 50 cm length, while the series pair $6\Omega$ and $4\Omega$ is joined across the same ends $A$ and $B$. At the null point the galvanometer carries no current, which requires the potential at the junction of the two resistors to equal the potential at $P$ on the wire. The drop from $A$ to the junction is the divider fraction $\frac{6}{6+4}V=0.6V$, while the drop from $A$ to $P$ along the uniform wire is $\frac{{\ell }_{AP}}{50}V$. Equating, $\frac{{\ell }_{AP}}{50}=0.6$, so ${\ell }_{AP}=30$ cm.