Combination of Resistors question from JEE Main 2013 (25 Apr, Shift 1)

Answer: (b)

1. Side PQ $= r$, side PR $= \dfrac{r}{3}$ (three in parallel), side QR $= \dfrac{r}{2}$ (two in parallel). Total $= \dfrac{11r}{6}$.
2. For a triangle, $R_{XY} = \dfrac{(\text{side } XY)(\text{rest})}{\text{total}}$.
3. $R_{PQ} = \dfrac{r \cdot \frac{5r}{6}}{\frac{11r}{6}} = \dfrac{5r}{11}$, $R_{QR} = \dfrac{\frac r2 \cdot \frac{4r}{3}}{\frac{11r}{6}} = \dfrac{4r}{11}$, $R_{PR} = \dfrac{\frac r3 \cdot \frac{3r}{2}}{\frac{11r}{6}} = \dfrac{3r}{11}$
4. Maximum is between P and Q. So the answer is (b).