Question

A certain copper wire has a resistance of 12.0 Ω . At some point along its length the wire was cut so that the resistance of one piece is 5.0 times the resistance of the other.

Part A. Determine the length of the short piece.

Part B. Determine the resistance of the short piece.

Part C. Determine the resistance of the long piece.

Answer 1

Solution-

Resistance is given by

R= rho* (L/A)

here rho is the resistivity , ,

L is the length ,

A is area of cross section

now,

consider the area and resistivity of the wire is constant

so , now resistance is propertional to length.

so we can write R=K*L where K is a constant

now ,

let the length of original pience be L

now let it be cut into a short piece of x and remaining long piece of (L-x)

now it is given that the resistance of long piece is 5 times of short piece

so ,

K* ( L-x) = 5*K*x

=> L-x = 5x

=> L= 6x

=> x= L/6

b) now , we have K*L = 12 ohms

so short piece will have resistance , K*x = K*(L/6) = 2 ohms

c) resistance of long piece = K* ( 5L/6) = 10 ohms