Two wires have their lengths, diameters and resistivities, all in the ratio of 1: 2. If the resistance of the thinner wire is 10 ohms, the resistance of the thicker wire is

1. The resistance (R) of a wire is given by the formula R = ρL/A, where ρ is resistivity, L is length, and A is the cross-sectional area. 2. The cross-sectional area A is proportional to the square of the diameter (d), specifically A = π(d/2)^2 = (π/4)d^2. 3. Let the properties of the thinner wire be L1, d1, ρ1, and R1 = 10 ohms. Let the properties of the thicker wire be L2, d2, ρ2, and R2. 4. We are given that the ratios L1:L2, d1:d2, and ρ1:ρ2 are all 1:2. This means L2 = 2L1, d2 = 2d1, and ρ2 = 2ρ1. 5. The area of the thicker wire A2 is (π/4)d2^2 = (π/4)(2d1)^2 = (π/4)(4d1^2) = 4 * [(π/4)d1^2] = 4A1. 6. Now calculate the resistance of the thicker wire R2 = ρ2 * L2 / A2. 7. Substitute the values in terms of the thinner wire: R2 = (2ρ1) * (2L1) / (4A1). 8. Simplify: R2 = (4 * ρ1 * L1) / (4 * A1) = ρ1 * L1 / A1. 9. Since R1 = ρ1 * L1 / A1 = 10 ohms, then R2 = R1 = 10 ohms.