Current Electricity — Question 9

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,

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

Hence, the new resistance = 4 Ω.