Q55 / 75 · JEE Main 2026 (23 January Shift 1) · Physics
A wire of uniform resistance $\lambda \Omega / \mathrm{m}$ is bent into a circle of radius $r$ and another piece of wire with length $2 r$ is connected between points $A$ and $B(A O B)$ as shown in figure. The equivalent resistance between points $A$ and $B$ is $\_\_\_\_$ $\Omega$.
- A$\frac{3 \pi \lambda r}{8}$
- Correct answerB$\frac{6 \pi \lambda r}{3 \pi+16}$
- C$2 \pi \lambda r$
- D$(\pi+1) 2 r \lambda$
Solution
The wire bent into a circle of radius $r$ has resistance $\lambda \cdot \pi r$ for the semicircular arcs, and the straight wire of length $2r$ has resistance $\lambda \cdot 2r$. The three paths between A and B (two semicircular arcs of the circle and one straight wire through center) are in parallel.
$\frac{1}{R_{AB}} = \frac{2}{\lambda \pi r} + \frac{1}{\lambda \cdot 2r} + \frac{2}{\lambda \cdot 3\pi r}$
$= \frac{1}{\lambda r} \left[\frac{2}{\pi} + \frac{1}{2} + \frac{2}{3\pi}\right]$
$= \frac{1}{\lambda r} \left(\frac{12 + 3\pi + 4}{6\pi}\right) = \frac{1}{\lambda r} \cdot \frac{16 + 3\pi}{6\pi}$
$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right) = \frac{6\pi \lambda r}{3\pi + 16}$
$\frac{1}{R_{AB}} = \frac{2}{\lambda \pi r} + \frac{1}{\lambda \cdot 2r} + \frac{2}{\lambda \cdot 3\pi r}$
$= \frac{1}{\lambda r} \left[\frac{2}{\pi} + \frac{1}{2} + \frac{2}{3\pi}\right]$
$= \frac{1}{\lambda r} \left(\frac{12 + 3\pi + 4}{6\pi}\right) = \frac{1}{\lambda r} \cdot \frac{16 + 3\pi}{6\pi}$
$R_{AB} = \lambda r \left(\frac{6\pi}{16 + 3\pi}\right) = \frac{6\pi \lambda r}{3\pi + 16}$
Hence, the correct answer is option: 'B'.