Current Electricity Part 4

Concept:The equivalent resistance between two points in a network is found by identifying parallel paths and calculating the combined resistance using the reciprocal formula.Explanation:The wire has resistance per unit length $\lambda\ \Omega/\mathrm{m}$.Resistance $R = \lambda \times \text{length}$.Three paths exist between A and B, all in parallel.Path 1 (minor arc AB) – a quarter of the circle: length $= \frac{1}{4} \times 2\pi r = \frac{\pi r}{2}$.$R_1 = \lambda \left(\frac{\pi r}{2}\right)$.Path 2 (major arc AOB) – three-quarters of the circle: length $= \frac{3}{4} \times 2\pi r = \frac{3\pi r}{2}$.$R_2 = \lambda \left(\frac{3\pi r}{2}\right)$.Path 3 (straight wire AOB): given total length $= 2r$ (AO + OB = $r + r$).$R_3 = \lambda (2r)$.These three resistances are in parallel, so$\frac{1}{R_{\text{eq}}} = \frac{1}{R_1} + \frac{1}{R_2} + \frac{1}{R_3}$.$\frac{1}{R_{\text{eq}}} = \frac{2}{\lambda \pi r} + \frac{2}{3\lambda \pi r} + \frac{1}{2\lambda r}$.$\frac{1}{R_{\text{eq}}} = \frac{1}{\lambda r} \left( \frac{2}{\pi} + \frac{2}{3\pi} + \frac{1}{2} \right)$.$\frac{1}{R_{\text{eq}}} = \frac{1}{\lambda r} \left( \frac{12 + 4 + 3\pi}{6\pi} \right) = \frac{16 + 3\pi}{6\pi \lambda r}$.$R_{\text{eq}} = \frac{6\pi \lambda r}{16 + 3\pi}$.Answer:The equivalent resistance is $\frac{6 \pi \lambda r}{3\pi + 16}\ \Omega$, which corresponds to option (B).