Combination of Resistors question from JEE Main 2022 (25 Jul, Shift 1)

Answer: 4

1. One wire: $r_0 = \dfrac{4\rho l}{\pi d^2}$; eight in parallel: $R = \dfrac{r_0}{8} = \dfrac{\rho l}{2\pi d^2}$.
2. Single wire of length $2l$, diameter $D$: $R = \dfrac{8\rho l}{\pi D^2}$.
3. $\dfrac{8}{D^2} = \dfrac{1}{2d^2} \Rightarrow D^2 = 16d^2 \Rightarrow D = 4d$
4. Answer: 4