Answer:
The length of wire remaining on the spool is 46.15 meters.
Explanation:
It is given that,
Length of the wire, L = 60 m
Current, I = 2 A
Some weeks later, after cutting off various lengths of wire for use in repairs,
New current, I' = 2.6 A
Let l' is the length of wire remaining on the spool. We know that he resistance of wire in terms of length and area is given by :
[tex]R=\dfrac{\rho L}{A}[/tex]
If V is the voltage, then, V = IR
Current, [tex]I=\dfrac{AV}{\rho L}[/tex]
So, current is inversely proportional to the length of the wire. So,
[tex]\dfrac{I}{I'}=\dfrac{L'}{L}[/tex]
L' is the length of wire remaining on the spool
[tex]\dfrac{IL}{I'}=L'[/tex]
[tex]\dfrac{2\times 60}{2.6}=L'[/tex]
L' = 46.15 meters
So, the length of wire remaining on the spool is 46.15 meters.