A conducting wire of length 1 has resistance R. If its length is increased to nl by stretching it uniformly, what would be the new resistance of wire ?
(Assume that there is no change in the volume of the wire when it is stretched.)

Explanation

Original length of the wire = l
and area of cross-section=A.
New length of the wire l.= nl
New area of the cross-section of wire = A.
Here, Al-A.l. = A. (nl)
( `:.` It is given that the volume of the wire remains constant when it is stretched.)
`:.` New area of cross-section of the wire,
`A. = A/n`
Oiginal resistance of the wire, `R = rho . l/A`
Newresistance of the wire.
`R. = rho. (l.)/(A.)`
`= rho . (nl)/(A/n) = n^2 (rho . l/A)`
`:. R. = n^2 R`
(Note: If the length of the wire is doubled (n=2), the new resistance of the wire will be `R. = (2)^2 R = 4R`.]

* You need not memorise these values.You can use these values for solving numerical problems.