The resistance of a wire is `20Omega`. What will be new resistance, if it is stretched uniformly 8 times its original length?

Explanation

`R_(1)=20Omega, R_(2)=?`
Let the original length `(l_(1))` be `l`.
The new length, `l_(2)=8l_(1)(i.e) l_(2)=8l`
The original resistance,
`R_(1)=rho (l_(1))/(A_(1))`
The new resistance `R_(2)=rho(l_(2))/(A_(2))= (rho (8l))/(A_(2))`
Through the wire is stretched, its volume is unchanged.
Initial volume = Final volume
`A_(1)l_(1)=A_(2)l_(2), A_(1)l=A_(2)8l`
`(A_(1))/(A_(2))=(8l)/(l)=8`
By dividing equation `R_(2)` by equation `R_(1)`, we get
`(R_(2))/(R_(1))=(rho (8l))/(A_(2))xx(A_(1))/(rhol) rArr (R_(2))/(R_(1))=(A_(1))/(A_(2))xx8`
Substituting the value of `(A_(1))/(A_(2))`, we get
`(R_(2))/(R_(1))=8xx8=64 rArr R_(2)=64xx20=1280Omega`
Hence, stretching the length of the wire has increased it resistance.