To solve this problem, we need to use the formula for resistance:
R=ρ⋅LAR = frac{
ho cdot L}{A}R=Aρ⋅L
where: R = resistance ρ (rho) = resistivity of the material L = length of the wire A = cross-sectional area of the wire
Let's assume that the resistivity of the material remains constant. We'll also assume that the resistivity is not provided and is not necessary for solving the problem.
For the first wire: Length, L₁ = 1 cm = 0.01 m Resistance, R₁ = 0.852 ohms
For the second wire: Length, L₂ = 50 cm = 0.5 m (coiled wire) Diameter, d₂ = 2 × d₁ (twice the diameter of the first wire) Radius, r₂ = d₂ / 2 = 2 × (d₁ / 2) = 2r₁
Now, we need to find the relationship between the cross-sectional areas of the two wires:
A2=πr22=π(2r1)2=4πr12=4A1A₂ = pi r₂^2 = pi (2r₁)^2 = 4pi r₁^2 = 4A₁A2=πr22=π(2r1)2=4πr12=4A1
So, the cross-sectional area of the second wire is four times that of the first wire.
Now, let's substitute the values into the resistance formula for the second wire:
R2=ρ⋅L2A2=ρ⋅0.54A1R₂ = frac{
ho cdot L₂}{A₂} = frac{
ho cdot 0.5}{4A₁}R2=A2ρ⋅L<span class="sizing reset-size3 size1