A given wire of resistance 1 Ω is stretched to double it's length. What will be it's new resistance?

When the wire is stretched to double it's length, it's area of cross section becomes half and it's length becomes double.

Let, a be the area of initial cross section and ρ be the specific resistance of the material of wire.

Then,

length = l,

R = 1 Ω,

new length = 2l,

new area = $\dfrac{a}{2}$

From relation

R = ρ $\dfrac{l}{a}$ = ρ $\dfrac{l}{πr^2}$

Initial resistance R₁ = 1 = ρ $\dfrac{l}{a}$    [Equation 1]

New resistance R_n = ρ $\dfrac{2l}{\dfrac{a}{2}}$ = ρ $\dfrac{4l}{a}$    [Equation 2]

On dividing eqn (ii) by (i), we get,

$\dfrac{R$n}{1} = \dfrac{ ρ \dfrac{4l}{a}}{ρ \dfrac{l}{a}} \\[0.5em] \Rightarrow Rn = 4 \varOmega

Hence, the new resistance = 4 Ω.