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

Because the cross-section is non-uniform, resistance is not proportional to length; instead $\frac{dR}{d\ell }\propto \frac{1}{\sqrt{\ell }}$, so $\frac{dR}{d\ell }=\frac{c}{\sqrt{\ell }}$ for some constant $c$. Integrating from one end, the resistance up to length $\ell$ is $R(\ell )=2c\sqrt{\ell }$, and the whole metre of wire has resistance $R(1)=2c$. The two external resistances are equal, so balance of the bridge requires the segment $AP$ to carry half the total wire resistance, $R_{AP}=c$. Setting $2c\sqrt{\ell }=c$ gives $\sqrt{\ell }=\frac{1}{2}$, so the balancing length is $\ell =\frac{1}{4}=0.25\text{m}$.