Explanation:
A We know that,
$\text { Resistance }(\mathrm{R})=\frac{\rho \mathrm{L}}{\mathrm{A}}$
$\mathrm{R}=\frac{\rho \mathrm{L}}{\frac{\pi}{4} \mathrm{~d}^{2}}=\frac{4 \rho \mathrm{L}}{\pi \mathrm{d}^{2}}$
$\mathrm{R} \propto \frac{\mathrm{L}}{\mathrm{d}^{2}}$
For option (a)
Length $=2 \mathrm{~L}$, diameter $=\mathrm{d} / 2$
$\mathrm{R}_{1}=\frac{2 \mathrm{~L}}{(\mathrm{~d} / 2)^{2}}=\frac{8 \mathrm{~L}}{\mathrm{~d}^{2}}$
For option (b)
Length $=\mathrm{L} / 2$, diameter $=2 \mathrm{~d}$
$\mathrm{R}_{2}=\frac{\mathrm{L} / 2}{(2 \mathrm{~d})^{2}}=\frac{\mathrm{L}}{8 \mathrm{~d}^{2}}$
For option (C)
Length $=\mathrm{L}$, diameter $=\mathrm{d}$
$\mathrm{R}_{3}=\frac{\mathrm{L}}{\mathrm{d}^{2}}$
For option (d)
Length $=2 \mathrm{~L}$, diameter $=\mathrm{d}$
$\mathrm{R}_{4}=\frac{2 \mathrm{~L}}{\mathrm{~d}^{2}}$
Hence, at length $=2 \mathrm{~L}$ and diameter $=\mathrm{d} / 2$, the resistance of wire is maximum.