A piece of wire of resistance is 20 Ω is drawn out so that its length is increased to twice its original length the new resistance of the wire will be

Given that,
Resistance

= 20 Ω

And length of the cross section is l and its area of cross section is A and its resistivity is

ρ

R = \frac{ρ \times l}{A}

…………….(1)
When length will become twice its area will

A / 2

on doubling the wire by folding the new resistance will be

R = \frac{ρ \times 2 l}{A / 2}

……………..(2)

\frac{R}{20} = \frac{ρ \times l \times A}{\frac{A}{2} \times ρ \times 2 l}

After solving,

R = 80 Ω

Thus, the value of resistance will be

80 Ω

A piece of wire of resistance is $20 Ω$ is drawn out so that its length is increased to twice its original length the new resistance of the wire will be