Question

A piece of wire of resistance is 2o ohm is drawn out so that its length is increased to twice its original length. Calculate the resistance of the wire in the new resistance?

Answer 1

R = ρ L/A

R = ρ L/A * (L/L)

R = ρ L^2 / V

Volume will be constant

R ∝ L^2

R1/R2 = (L1/L2)^2

20/R2 = (L1/(2L1))^2

20/R2 = 1/4

R2 = 80 Ohm

Resistance of the wire in new resistance is 80 Ohm

Answer 2

Let the initial resistance be R.
therefore,
R = P× l/A
(here P is the constant of proportionality. )
A=l^2
=>R= P ×l/l^2

now the length Is increased in such a way that it becomes double of the original length.
let the new resistance be R2.

therefore,
R2 = P×2l/A ( A=l^2)
=>R2 = P×2l/(2l)^2
=>R2 = P2l/4l^2

therfore,
R/R2=(Pl/l^2)÷(P2l/4l^2)
=>20/R2=2
=>R2=10

the new resistance of the wire will be 10 ohms.