Physics · Current Electricity

JEE Main 2026 — 23 January, Morning Shift — Question 34

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 $(\mathrm{AOB})$ as shown in figure. The equivalent resistance between points A and B is $\_\_\_\_$ $\Omega$ .

Option A:
$\frac{3 \pi \lambda \mathrm{r}}{8}$
Option B:
$(\pi+1) 2 \mathrm{r} \lambda$
Option C:
$\frac{6 \pi \lambda \mathrm{r}}{3 \pi+16}$
Correct
Option D:
$2 \pi \lambda \mathrm{r}$

Answer: C

Step-by-step solution

$\frac{1}{R_{A B}}=\frac{2}{\lambda \pi r}+\frac{1}{\lambda .2 r}+\frac{2}{\lambda .3 \pi r}$ $=\frac{1}{\lambda \mathrm{r}}\left[\frac{2}{\pi}+\frac{1}{2}+\frac{2}{3 \pi}\right]$ $=\frac{1}{\lambda \mathrm{r}}\left(\frac{12+3 \pi+4}{6 \pi}\right)=\frac{1}{\lambda \mathrm{r}} \cdot\left(\frac{16+3 \pi}{6 \pi}\right)$ $\mathrm{R}_{\mathrm{AB}}=\lambda \mathrm{r}\left(\frac{6 \pi}{16+3 \pi}\right)$

Answer key and solution verified before publishing.

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Exam: JEE Main 2026
Paper: 23 January, Morning Shift
Subject: Physics
Chapter: Current Electricity
Topic: Combination of Resistors and cells, Wheatstone Bridge