Masses of three wires of copper are in the ratio of 1 : 3 : 5 and their lengths are in the ratio of 5 : 3 : 1. The ratio of their electrical resistance are:
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 125 : 15 : 1
The resistance is \( R = \rho\,\frac{L}{A} \). The wires are of the same material, so \( \rho \) and the density \( d \) are common. The mass is fixed, so the volume \( V = AL = m/d \) fixes the area: \( A = \frac{m}{dL} \).
\( \displaystyle R = \frac{\rho L}{A} = \frac{\rho d\,L^{2}}{m} \Rightarrow R \propto \frac{L^{2}}{m} \)
\( \displaystyle R_1 : R_2 : R_3 = \frac{5^{2}}{1} : \frac{3^{2}}{3} : \frac{1^{2}}{5} = 25 : 3 : 0.2 \)
Multiplying by 5, the ratio is \( 125 : 15 : 1 \).