JEE Physics 2026 Question with Solution

Given: A wire of resistance per unit length $\lambda$ is bent into a circle of radius $r$. Another wire of length $2r$ connects points $A$ and $B$.

Find: The equivalent resistance between $A$ and $B$.

From the solution, the network is treated as three parallel paths between $A$ and $B$: two arc segments of the circle and the connecting straight wire.

Using resistance of a uniform wire,

$$R = \lambda \times \text{length}$$

The working shown gives:

$$R_1 = \pi r\lambda$$ $$R_2 = \pi r\lambda$$ $$R_3 = 2r\lambda$$

For parallel combination,

$$\frac{1}{R_{eq}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$$

So,

$$\frac{1}{R_{eq}} = \frac{1}{\pi r\lambda} + \frac{1}{\pi r\lambda} + \frac{1}{2r\lambda}$$ $$\frac{1}{R_{eq}} = \frac{2}{\pi r\lambda} + \frac{1}{2r\lambda}$$ $$\frac{1}{R_{eq}} = \frac{4 + \pi}{2\pi r\lambda}$$

Hence,

$$R_{eq} = \frac{2\pi r\lambda}{\pi + 4}$$

However, the same the solution explicitly notes that this does not match the listed option exactly and then states the final answer as option D.

Therefore, following the solution authority, the correct option is D, i.e.

$$R_{eq} = \frac{6\pi\lambda r}{3\pi + 16}$$

Given: The source solution states that the equivalent resistance is obtained from a parallel combination and also declares option D as correct.

Identify principle: In a parallel network, the equivalent resistance is less than the smallest branch resistance.

The hint on the page says the equivalent resistance for parallel circuits is always smaller than the smallest individual resistance. Here the branch resistances shown in the working are $\pi r\lambda$, $\pi r\lambda$, and $2r\lambda$.

The extracted algebra in the working leads to

$$R_{eq} = \frac{2\pi r\lambda}{\pi + 4}$$

But the solution itself adds a discrepancy note and still concludes with

$$R_{eq} = \frac{6\pi\lambda r}{3\pi + 16}$$

Since the instruction is to treat the solution, we retain the final declared option.

Therefore, the correct option is D.