JEE Main 2025 — Current Electricity Question with Solution

Length of the wire, $L=25\mathrm{m}$

Cross‐sectional area, $A=5\mathrm{m}{\mathrm{m}}^2=5\times 10^{-6}{\mathrm{m}}^2$

Resistivity, $\rho =2\times 10^{-6}\Omega \cdot m$

Resistance of the entire wire:

$$R_{\mathrm{total}}=\frac{\rho L}{A}=\frac{2\times 10^{-6}\times 25}{5\times 10^{-6}}=\frac{50\times 10^{-6}}{5\times 10^{-6}}=10\Omega$$

When bent into a circle, points diametrically opposite divide the ring into two equal halves.

Each half-circle has length $L/2=12.5\mathrm{m}$, so its resistance is half of $R_{\mathrm{total}}$:

$$R_{\mathrm{half}}=\frac{R_{\mathrm{total}}}{2}=\frac{10}{2}=5\Omega$$

These two $5\Omega$ halves are in parallel between the opposite points.

Equivalent resistance $R$ is given by:

$$\frac{1}{R}=\frac{1}{R_{\mathrm{half}}}+\frac{1}{R_{\mathrm{half}}}=\frac{1}{5}+\frac{1}{5}=\frac{2}{5}$$

$$R=\frac{5}{2}=2.5\Omega$$

Answer: 2.5 Ω (Option B)