A wire's resistance is directly proportional to its length and inversely proportional to its cross-sectional area. When a wire is folded, its length is halved, but its cross-sectional area doubles. This means that the wire's resistance will change.


To calculate the new resistance of the folded wire, we can use the formula:

R = (ρ * L) / A

Where:
- R = Resistance
- ρ = Resistivity of the wire material
- L = Length of the wire
- A = Cross-sectional area of the wire

When the wire is folded, its length is halved, so L/2 is used instead of L. The cross-sectional area doubles, so 2A is used instead of A.

Therefore, the new resistance of the folded wire is:

R' = (ρ * (L/2)) / (2A)

Simplifying this equation, we get:

R' = R / 4

This means that the resistance of the folded wire is one-fourth of the original wire's resistance.


When a wire is folded, its resistance changes due to changes in length and cross-sectional area. To calculate the new resistance, we use the formula R' = R/4, where R is the original resistance of the wire.

When a 6ohm resistance wire is fold double then its total resistance is we know that resistance is directly proportional to the length of the conductor then the total resistance of wire is 3ohm. I think answer is correct.