JEE Main · 2026 · 23 Jan 2026 · Shift 1 · Current Electricity

The wire has resistance $\lambda$ per unit length. From the figure $A$ and $B$ are separated by a quarter of the circle, so the minor arc ($90^{\circ }$) has resistance $R_1=\lambda \cdot \frac{\pi r}{2}$ and the major arc ($270^{\circ }$) has $R_2=\lambda \cdot \frac{3\pi r}{2}$. The connecting path $A\text{-}O\text{-}B$ is made of two radii, total length $2r$, with $R_3=2\lambda r$. The three paths are in parallel between $A$ and $B$: $\frac{1}{R_{AB}}=\frac{2}{\pi \lambda r}+\frac{2}{3\pi \lambda r}+\frac{1}{2\lambda r}=\frac{1}{\lambda r}\cdot \frac{16+3\pi }{6\pi }$. Therefore $R_{AB}=\frac{6\pi \lambda r}{3\pi +16}$.