We have a wire of uniform cross section and its electrical resistance is R . This wire is cut in n equal parts and each part is stretched so that its length becomes equal to the length of the original wire. Now all these parts are connected in parallel. Calculate the equivalent resistance now.

Explanation

Resistance of wire =R, length of wire =l On cutting the wire into n equal parts, resistance of each part of wire becomes `R_(i) = (R )/(n)`, and length becomes `(l = (l)/(n))`. When each part remains unchanged. V=Final volume=Initial volume
Therefore, `V= A_(F)l_(F) = A_(i) l_(i)`
Now, the initial length will be `l_(i) = (l)/(n)` and final length `=l_(P) = l`
`A_(P) l= (A_(i)l)/(n)`
`(A_(F))/(A_(i)) = (1)/(n)`
`:. (R_(F))/(R_(i)) = (rho l_(F)A_(i))/(A_(F)rho l_(i)) = (l_(F))/(l_(i)) xx (A_(i))/(A_(f))`
`R_(F) = n^(2)R_(i) ( :. R_(i) = (R )/(n))`
`:. R_(F) = n^(2)(R )/(n) = nR`
Hence, resistance of each stretched part= nR
On connecting these parts in parallel, the resultant resistance will be
`(1)/(R_(P)) = (1)/(nR) + (1)/(nR) + (1)/(nR) + ...n` times
`(1)/(R_(P)) = (n)/(nR)`
`rArr R_(P) = R`