A copper wire has diameter 0.5 mm and resistivity of 1.6 x 10-8 Ω m. What will be the length of this wire to make its resistance 10 Ω ? How much does the resistance change if the diameter is doubled

To solve this problem, we first need to understand the relationship between resistance (R), resistivity (ρ), length (L), and cross-sectional area (A) of a wire. This relationship is given by the formula:

R = ρ * (L/A)

Where:

• R is the resistance
• ρ is the resistivity
• L is the length of the wire
• A is the cross-sectional area of the wire

The cross-sectional area (A) of a wire can be calculated using the formula for the area of a circle (since the wire is cylindrical), which is:

A = π * (d/2)^2

Where:

• A is the area
• d is the diameter of the wire

Now, let's solve the problem step by step:

1. Calculate the cross-sectional area of the wire:

   A = π * (0.5 mm / 2)^2 = 0.196 mm^2 = 0.196 x 10^-6 m^2 (since 1 mm = 10^-3 m)

2. Substitute the values into the resistance formula and solve for L:

   10 Ω = 1.6 x 10^-8 Ω m * (L / 0.196 x 10^-6 m^2)

   Solving for L gives: L = 10 Ω * 0.196 x 10^-6 m^2 / 1.6 x 10^-8 Ω m = 1225 m

So, the length of the wire should be 1225 m to make its resistance 10 Ω.

For the second part of the question, if the diameter is doubled, the new diameter is 1 mm.

1. Calculate the new cross-sectional area:

   A = π * (1 mm / 2)^2 = 0.785 mm^2 = 0.785 x 10^-6 m^2

2. Substitute the values into the resistance formula and solve for the new resistance:

   R = 1.6 x 10^-8 Ω m * (1225 m / 0.785 x 10^-6 m^2) = 25 Ω

So, if the diameter is doubled, the resistance will increase to 25 Ω.