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:
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → 18 : 1
The resistance of a wire is \( R = \rho\,\frac{L}{A} \) with \( A = \frac{\pi d^{2}}{4} \). For the same material \( \rho \) is common, so \( R \propto \frac{L}{d^{2}} \).
The thinner wire has \( L_1 = 2,\ d_1 = 1 \) and the thicker wire has \( L_2 = 1,\ d_2 = 3 \) (in the given ratios).
\( \displaystyle \frac{R_1}{R_2} = \frac{L_1}{L_2}\times\left(\frac{d_2}{d_1}\right)^{2} = \frac{2}{1}\times\left(\frac{3}{1}\right)^{2} = 18 \)
So \( R_1 : R_2 = 18 : 1 \).