Two cylindrical wires of the same material have their lengths in the ratio 2 : 1 and their diameters in the ratio 1 : 3. The ratio of the resistance of thinner wire to that of thicker wire is:

18 : 1

The correct answer is Option (3) → 18 : 1

The resistance of a given wire is -

$R=ρ\frac{L}{A}$

where,

$ρ$ = Resistivity (same of both wires)

$L$ = length of wire

$A$ = Cross-sectional Area of the wire

$\frac{R_1}{R_2}=\frac{L_1}{A_1}×\frac{A_2}{L_2}=\frac{2×9}{1.1}=\frac{18}{1}$