A metal wire of resistance `6Omega` is stretched so that its length is increased to twice its original length. Calculate its new resistance.

Explanation

Given`: R _(1) = 6 Omega, I _(1) = l`
`l _(2) = 2 l_(1) = 2l, R_(2) = ?`
`R _(1) = rho ( l _(1))/( A _(1)) and R _(2) = rho ( l _(2))/( A _(2))`
The volume of metal wire remains same.
`therefore A _(1) l _(1) = A _(2) l _(2)`
`(l _(2))/( l _(1)) = (A_(1))/( A _(2))`
Now, `(R _(2))/(R_(1)) = ( l _(2))/(A_(2)) xx (A_(1))/( l _(2)) = ( l _(2))/( l _(1)) xx (l_(2))/( l _(1))`
`implies R _(2) = R _(1) (( l _(2))/( l _(1))) ^(2) = 6 ((2l )/(l)) ^(2) = 24 Omega`