Question

The resistance of a wire of length 250m is 1 ohm. if the resistivity of the material of wire is 1.6*10^-8 ohm metre, find the area of cross section of wire. how much does the resistance change if the diameter is doubled?

Answer 1

The information provided in the question is as follows:
 Length of wire, l  = 250 m
 Resistance of wire, R = 1 Ω
Resistivity of wire, ρ = 1.6×10^-8 Ω-m

 The resistance is given as : R = ρ(l/A)
⇒ 1 = 1.6 × 10^-8 × 250/A
⇒ A = 1. 6 × 10−8 × 250 = 4 × 10^−6 m^2
    A=π(d)^2 
 When the diameter is doubled, then

 A' = π( d'/2)^2 = π(2d/2)^2 = 4π(d/2)^2 = 4A

 Keeping the same length, the resistance of the wire,
 R' = ρ(l/A) = ρ(l/4A) = R/4 = 1/4 = 0. 25 Ω


i hope it will help you
regards

Answer 2

hey bro/gal.

since, resistance = resistivity× length / area

=) 1 = 1.6 * 10^-8 * 250/ A

=) A = 400*10^-8 = 4*10^-6 m²

=) if diameter is doubled then area will become 4A.

=) new Area = A' = 4*4*10^-6 = 16*10^-6 m²

=) R' = 1.6*10^-8 * 250/ 16*10^-6

= 1/4 ohm.