A metallic wire has a definite resistance . IF the wire is stretched in such a way that its length becomes doubled then what will be the resistance of the wire? Consider the volume and the resistivity of the wire remains unchanged.

Explanation

Keeping the volume unchanged , if the length of the wire is doubled the area of cross section is halved (since volume= length `times` area of cross section) If initial length and area of cross section of the wire and l and A respectively. Then its final length and area of cross section will be 2l and `A/2` respectively.
Therefore, initial resistance `R_1=rho1/A` and final resistance,
`R_2=rho(2l)/(A/2)=rho(4l)/A=4R_1`