Question

A metal wire of length 1cm and of uniform diameter has a resistance of 0.852. What is the resistance of a 50cm wire coil made of the same material but having twice its diameter? A. 25052 B. 1252 C. 402 D. 102?

Answer 1

To solve this problem, we need to use the formula for resistance:

$R=\frac{\rho \cdot L}{A}$

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:

$A_2=\pi r_2^2=\pi (2r_1{)}^2=4\pi r_1^2=4A_1$

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​=A2​ρ⋅L<span class="sizing reset-size3 size1