Question

The mass of the three wire of copper are in the ratio 1 : 3 : 5. And their lengths are in ratio 5 : 3 : 1 . The ratio of their electrical resistance is

• A. 1 : 3 : 5
• B. 5 : 3 : 1
• C. 1 : 15 : 125
• D. 125 : 15 : 1

Answer 1

Answer: (D)

125 : 15 : 1

Answer 2

R = $\frac{\rho\mathcal{l}}{A} = \frac{\rho\mathcal{l}^{2}}{A\mathcal{l}} = \frac{\rho\mathcal{l}^{2}}{V} = \frac{\rho\mathcal{l}^{2}}{m/d}$R = $\frac{\rho d\mathcal{l}^{2}}{m}$ or R ∝ $\frac{\mathcal{l}^{2}}{m}$

R₁ : R₂ : R₃ = $\frac{\mathcal{l}_{1}^{2}}{m_{1}}:\frac{\mathcal{l}_{2}^{2}}{m_{2}}:\frac{\mathcal{l}_{3}^{2}}{m_{3}}$ = $\frac{25}{1}:\frac{9}{3}:\frac{1}{5}$

= 125 : 15 : 1, Hence correct answer is