Explanation:
The resistance of a wire having length L and cross-sectional area A is given by :
[tex]R=\dfrac{\rho L}{A}[/tex]
Where
[tex]\rho[/tex] is the resistivity of the material.
Let R' is the resistance of the wire when it is stretched to twice its original length, L' = 2L
Since, the volume remains constant. So,
[tex]AL=A'L'[/tex]
[tex]A'=\dfrac{A}{2}[/tex]
The new resistance is given by :
[tex]R'=\dfrac{\rho (2L)}{A/2}[/tex]..............(2)
[tex]\dfrac{R}{R'}=\dfrac{1}{4}[/tex]
R' = 4R
So, the resistance of the wire becomes four time of the original resistance. Hence, this is the required solution.
