A silver wire and a platinum wire of the same length have the same resistance. What is the ratio of the diameter of the silver wire to that of the platinum wire.
I did 10.6E-8/1.59E-8 but am not getting correct answer?
To find the ratio of the diameter of the silver wire to that of the platinum wire, we can make use of the fact that the resistance of a wire is inversely proportional to its cross-sectional area.
Resistance is given by the formula: R = (ρ * L) / A,
where R is the resistance, ρ is the resistivity of the material, L is the length of the wire, and A is the cross-sectional area of the wire.
Since the resistance is the same for both wires, we can write:
(ρ1 * L) / A1 = (ρ2 * L) / A2,
where ρ1 and ρ2 are the resistivities of silver and platinum respectively, A1 and A2 are the cross-sectional areas of the silver and platinum wires respectively.
Now, let's assume that the diameter ratio of the silver wire to the platinum wire is D1 / D2, where D1 is the diameter of the silver wire, and D2 is the diameter of the platinum wire. We know that the cross-sectional area of a wire is given by the formula: A = π * (D/2)^2.
Substituting this into our equation, we get:
(ρ1 * L) / (π * (D1/2)^2) = (ρ2 * L) / (π * (D2/2)^2).
Now, we can simplify the equation further by canceling out the length (L) and the π:
(ρ1 / (D1/2)^2) = (ρ2 / (D2/2)^2).
To find the ratio of the diameters, we can cross-multiply and solve for it:
(D2/2)^2 / (D1/2)^2 = ρ2 / ρ1.
Taking the square root of both sides, we get:
D2 / D1 = √(ρ2 / ρ1).
Now, we need the resistivity values for silver and platinum. The resistivity of silver (ρ1) is approximately 1.59 x 10^-8 Ω·m, and the resistivity of platinum (ρ2) is approximately 10.6 x 10^-8 Ω·m.
Calculating the ratio:
D2 / D1 = √(10.6 x 10^-8) / √(1.59 x 10^-8) = √(10.6 / 1.59) ≈ 1.42.
Therefore, the ratio of the diameter of the silver wire to that of the platinum wire is approximately 1.42.
It seems like you inverted the resistivity values when calculating the ratio.