A 40 resistance wire is doubled on it. Calculate the new resistance of the wire.

Explanation

We are given, `R = 4 Omega`
When a wire is doubled on it, its length would become half and area of cross-section would be doubled. That is, a wire of length l becomes `(l)/(2)` and area of cross-section 2A.
We know that’s
`R= rho(l)/(A)`
`R_(1)= rho (l//2)/(2A)` where `R_(1)` is the new resistance.
Therefore, `(R_(1))/(R)= (rho (l//2)/(2A))/( rho(l)/(A))= (1)/(4)`
or `R_(1)= (R)/(4)= (4Omega)/(4)= 1Omega`
The new resistance of the wire is `1Omega`