JEE Main · 2019 · 10 Jan 2019 · Shift 1 · Current Electricity

The driver cell $D$ of emf $\epsilon$ and internal resistance $r$ sends a steady current down the whole wire of resistance $12r$, so the main-loop current is $I=\frac{\epsilon }{12r+r}=\frac{\epsilon }{13r}$. The test cell $C$ taps in at $J$ through the galvanometer, and at the balance point that branch carries no current, which is exactly why the $3r$ internal resistance of $C$ never enters. The drop across the length $AJ$, the fraction $x/L$ of the wire, must match the emf of $C$: $I\cdot (\frac{x}{L}\cdot 12r)=\frac{\epsilon }{2}$. Putting $I=\epsilon /13r$ gives $\frac{12x}{13L}=\frac{1}{2}$, so $x=\frac{13L}{24}$. The tempting slip is to fold $3r$ into the main loop and use $15r$; at null that branch is dead, so the balance length is $\frac{13}{24}L$.