Question

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

Answer 1

let resistivity of the wire be p

let area of crossection of the wire be A

let the lenght of the wire be L

let the resistance of the wire be R

area of crossection of the wire, A = $\pi r^{2}$

                                                        = 3.14 x (5 x 10^-4)/2 x (5 x 10^-4)/2  

                                                      (here we are taking radius in meters)

                                                        = 3.14 x 2.5 x 10^-4 x 2.5 x 10^-4

                                                        = 19.625 x 10^-8 m^2

also R = (p x L)/A

Therefore L = (R x p)/A

L = (10 x 1.6 x 10^-8 )/19.625 x 10^-8

  = 0.815 m

  = 81.5 cm

if diameter is doubled

area of crossection a wire A' = $\pi r^{2}$

                                                        = 3.14 x (10 x 10^-4)/2 x (10 x 10^-4)/2  

                                                      (here we are taking radius in meters)

                                                        = 3.14 x 5 x 10^-4 x 5 x 10^-4

                                                        = 78.5 x 10^-8 m^2

also R = (p x L)/A

Therefore L' = (R x p)/A

L' = (10 x 1.6 x 10^-8 )/78.5 x 10^-8

L ' = 0.203 m

L' = 20.3 cm